size_categories:n<1K format:parquet modality:text library:datasets library:pandas library:mlcroissant library:polars library:distilabel region:us synthetic distilabel rlaif
---
size_categories: n<1K
dataset_info:
  features:
  - name: solution
    dtype: string
  - name: problem
    dtype: string
  - name: cot_type
    dtype: string
  - name: source_type
    dtype: string
  - name: metadata
    dtype: string
  - name: cot
    dtype: 'null'
  - name: thinking_trajectories
    dtype: string
  - name: attempt
    dtype: string
  - name: generation
    dtype: string
  - name: distilabel_metadata
    struct:
    - name: raw_input_text_generation_0
      list:
      - name: content
        dtype: string
      - name: role
        dtype: string
    - name: raw_output_text_generation_0
      dtype: string
    - name: statistics_text_generation_0
      struct:
      - name: input_tokens
        dtype: int64
      - name: output_tokens
        dtype: int64
  - name: model_name
    dtype: string
  splits:
  - name: train
    num_bytes: 599715
    num_examples: 10
  download_size: 280674
  dataset_size: 599715
configs:
- config_name: default
  data_files:
  - split: train
    path: data/train-*
tags:
- synthetic
- distilabel
- rlaif
---

<p align="left">
  <a href="https://github.com/argilla-io/distilabel">
    <img src="https://raw.githubusercontent.com/argilla-io/distilabel/main/docs/assets/distilabel-badge-light.png" alt="Built with Distilabel" width="200" height="32"/>
  </a>
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# Dataset Card for s1K-deepseek-r1-qwen-14b-test

This dataset has been created with [distilabel](https://distilabel.argilla.io/).



## Dataset Summary

This dataset contains a `pipeline.yaml` which can be used to reproduce the pipeline that generated it in distilabel using the `distilabel` CLI:

```console
distilabel pipeline run --config "https://huggingface.co/datasets/Rickxz06/s1K-deepseek-r1-qwen-14b-test/raw/main/pipeline.yaml"
```

or explore the configuration:

```console
distilabel pipeline info --config "https://huggingface.co/datasets/Rickxz06/s1K-deepseek-r1-qwen-14b-test/raw/main/pipeline.yaml"
```

## Dataset structure

The examples have the following structure per configuration:


<details><summary> Configuration: default </summary><hr>

```json
{
    "attempt": "Let $P(x) = 1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x$.\nSince $P(x) \\geq 0$ for all real $x$, by Fejer-Riesz theorem, there exists a polynomial $Q(z) = c_0 + c_1 z + c_2 z^2$ such that $P(x) = |Q(e^{ix})|^2$.\nWe have $|Q(e^{ix})|^2 = (c_0 + c_1 e^{ix} + c_2 e^{2ix})(\\bar{c_0} + \\bar{c_1} e^{-ix} + \\bar{c_2} e^{-2ix})$.\nExpanding this, we get\n\\begin{align*} |Q(e^{ix})|^2 \u0026= |c_0|^2 + |c_1|^2 + |c_2|^2 \\\\ \u0026+ (c_0 \\bar{c_1} + c_1 \\bar{c_2}) e^{ix} + (\\bar{c_0} c_1 + \\bar{c_1} c_2) e^{-ix} \\\\ \u0026+ c_0 \\bar{c_2} e^{2ix} + \\bar{c_0} c_2 e^{-2ix} \\\\ \u0026= |c_0|^2 + |c_1|^2 + |c_2|^2 \\\\ \u0026+ (c_0 \\bar{c_1} + \\bar{\\bar{c_1}} c_2 + \\overline{c_0 \\bar{c_1} + \\bar{\\bar{c_1}} c_2}) \\cos x + i (c_0 \\bar{c_1} + \\bar{\\bar{c_1}} c_2 - \\overline{c_0 \\bar{c_1} + \\bar{\\bar{c_1}} c_2}) \\sin x \\\\ \u0026+ (c_0 \\bar{c_2} + \\bar{c_0} c_2) \\cos 2x + i (c_0 \\bar{c_2} - \\bar{c_0} c_2) \\sin 2x \\\\ \u0026= |c_0|^2 + |c_1|^2 + |c_2|^2 + 2 \\text{Re}(c_0 \\bar{c_1} + c_1 \\bar{c_2}) \\cos x - 2 \\text{Im}(c_0 \\bar{c_1} + c_1 \\bar{c_2}) \\sin x + 2 \\text{Re}(c_0 \\bar{c_2}) \\cos 2x - 2 \\text{Im}(c_0 \\bar{c_2}) \\sin 2x\\end{align*}\nComparing the coefficients of $P(x)$ with $|Q(e^{ix})|^2$, we have:\n$1 = |c_0|^2 + |c_1|^2 + |c_2|^2$\n$-a = 2 \\text{Re}(c_0 \\bar{c_1} + c_1 \\bar{c_2})$\n$-b = -2 \\text{Im}(c_0 \\bar{c_1} + c_1 \\bar{c_2})$\n$-A = 2 \\text{Re}(c_0 \\bar{c_2})$\n$-B = -2 \\text{Im}(c_0 \\bar{c_2})$\nFrom the last two equations, $A^2 + B^2 = 4 (\\text{Re}(c_0 \\bar{c_2})^2 + \\text{Im}(c_0 \\bar{c_2})^2) = 4 |c_0 \\bar{c_2}|^2 = 4 |c_0|^2 |c_2|^2$.\nSince $1 = |c_0|^2 + |c_1|^2 + |c_2|^2$, we have $|c_0|^2 \\leq 1$ and $|c_2|^2 \\leq 1$.\nThus $A^2 + B^2 = 4 |c_0|^2 |c_2|^2 \\leq 4 \\cdot 1 \\cdot 1 = 4$. This is not the desired result.\n\nConsider $Q(z) = 1 - \\alpha z - \\beta z^2$. Then $|Q(e^{ix})|^2 = |1 - \\alpha e^{ix} - \\beta e^{2ix}|^2$.\n$|1 - \\alpha e^{ix} - \\beta e^{2ix}|^2 = (1 - \\alpha e^{ix} - \\beta e^{2ix})(1 - \\bar{\\alpha} e^{-ix} - \\bar{\\beta} e^{-2ix})$\n$= 1 - \\bar{\\alpha} e^{-ix} - \\bar{\\beta} e^{-2ix} - \\alpha e^{ix} + |\\alpha|^2 - \\alpha \\bar{\\beta} e^{-ix} - \\beta e^{2ix} - \\beta \\bar{\\alpha} e^{ix} + |\\beta|^2$\n$= 1 + |\\alpha|^2 + |\\beta|^2 - (\\alpha + \\alpha \\bar{\\beta} + \\bar{\\alpha}) \\cos x + i (\\alpha + \\alpha \\bar{\\beta} - \\bar{\\alpha}) \\sin x - (\\beta + \\beta \\bar{\\alpha} + \\bar{\\beta}) \\cos 2x + i (\\beta + \\beta \\bar{\\alpha} - \\bar{\\beta}) \\sin 2x$.\nCompare with $1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x$.\n$a = \\alpha + \\bar{\\alpha} + \\alpha \\bar{\\beta} + \\bar{\\alpha} \\beta = 2 \\text{Re}(\\alpha) + 2 \\text{Re}(\\alpha \\bar{\\beta})$.\n$b = -i (\\alpha - \\bar{\\alpha} + \\alpha \\bar{\\beta} - \\bar{\\alpha} \\beta) = 2 \\text{Im}(\\alpha) + 2 \\text{Im}(\\alpha \\bar{\\beta})$.\n$A = \\beta + \\bar{\\beta} = 2 \\text{Re}(\\beta)$.\n$B = -i (\\beta - \\bar{\\beta}) = 2 \\text{Im}(\\beta)$.\n$A^2+B^2 = 4 (\\text{Re}(\\beta)^2 + \\text{Im}(\\beta)^2) = 4 |\\beta|^2$.\n$a^2+b^2 = |2\\alpha + 2\\alpha \\bar{\\beta}|^2 = 4 |\\alpha|^2 |1+\\bar{\\beta}|^2$.\n$1 = 1 + |\\alpha|^2 + |\\beta|^2$. So $|\\alpha|^2 + |\\beta|^2 = 0$, so $\\alpha = \\beta = 0$.\n\nFinal Answer: The final answer is $\\boxed{a^2 + b^2 \\leq 2, A^2 + B^2 \\leq 1}$",
    "cot": null,
    "cot_type": "math",
    "distilabel_metadata": {
        "raw_input_text_generation_0": [
            {
                "content": "You will be given a problem. Please reason step by step, and put your final answer within \boxed{}:\nLet  $a,b,A,B$  be given reals. We consider the function defined by \\[ f(x) = 1 - a \\cdot \\cos(x) - b \\cdot \\sin(x) - A \\cdot \\cos(2x) - B \\cdot \\sin(2x). \\] Prove that if for any real number  $x$  we have  $f(x) \\geq 0$  then  $a^2 + b^2 \\leq 2$  and  $A^2 + B^2 \\leq 1.$ ",
                "role": "user"
            }
        ],
        "raw_output_text_generation_0": "\u003cthink\u003e\nOkay, so I have this problem where I\u0027m given a function f(x) that\u0027s defined as:\n\nf(x) = 1 - a\u00b7cos(x) - b\u00b7sin(x) - A\u00b7cos(2x) - B\u00b7sin(2x)\n\nAnd I need to prove that if f(x) is always greater than or equal to zero for any real number x, then two inequalities must hold: a\u00b2 + b\u00b2 \u2264 2 and A\u00b2 + B\u00b2 \u2264 1.\n\nHmm, okay. Let\u0027s break this down. The function f(x) is a combination of cosine and sine terms with different coefficients. Since it\u0027s given that f(x) is non-negative for all real x, that must mean that the function doesn\u0027t dip below zero anywhere on the real line.\n\nI remember that for functions involving cosines and sines, especially with multiple angles, sometimes you can express them as a single trigonometric function or use some identities to simplify. Maybe that can help here.\n\nFirst, let me write down the function again:\n\nf(x) = 1 - a\u00b7cos(x) - b\u00b7sin(x) - A\u00b7cos(2x) - B\u00b7sin(2x)\n\nI need to ensure that this is always non-negative. So, perhaps I can rewrite f(x) in a way that makes it easier to analyze. Maybe group the terms with cos(x) and sin(x) together and similarly for cos(2x) and sin(2x).\n\nSo, let\u0027s see:\n\nf(x) = 1 - [a\u00b7cos(x) + b\u00b7sin(x)] - [A\u00b7cos(2x) + B\u00b7sin(2x)]\n\nHmm, okay. So, this is 1 minus two terms, each of which is a linear combination of sine and cosine. I know that expressions like c\u00b7cos(x) + d\u00b7sin(x) can be rewritten as R\u00b7cos(x - \u03c6) or R\u00b7sin(x + \u03c6), where R is the amplitude. Maybe that\u0027s useful here.\n\nLet me recall that any linear combination of sine and cosine can be written as a single sine or cosine with a phase shift. Specifically, for constants c and d, we have:\n\nc\u00b7cos(x) + d\u00b7sin(x) = \u221a(c\u00b2 + d\u00b2)\u00b7cos(x - \u03c6), where \u03c6 = arctan(d/c) or something like that.\n\nSimilarly, for the 2x terms:\n\nA\u00b7cos(2x) + B\u00b7sin(2x) = \u221a(A\u00b2 + B\u00b2)\u00b7cos(2x - \u03b8), where \u03b8 is some angle.\n\nSo, if I rewrite both of these, f(x) becomes:\n\nf(x) = 1 - \u221a(a\u00b2 + b\u00b2)\u00b7cos(x - \u03c6) - \u221a(A\u00b2 + B\u00b2)\u00b7cos(2x - \u03b8)\n\nHmm, okay. So now, f(x) is 1 minus two cosine terms with different frequencies and amplitudes.\n\nNow, since f(x) is non-negative for all x, the minimum value of f(x) must be at least zero.\n\nTherefore, the minimum value of 1 - \u221a(a\u00b2 + b\u00b2)\u00b7cos(x - \u03c6) - \u221a(A\u00b2 + B\u00b2)\u00b7cos(2x - \u03b8) must be \u2265 0.\n\nTo find the minimum of this expression, I might need to analyze the maximum possible value that the sum of the two cosine terms can take, because subtracting that from 1 needs to stay non-negative.\n\nWait, actually, since cosine functions can have maximum values of 1 and minimum values of -1, but since they\u0027re subtracted, the overall expression can be thought of as 1 minus something. So, to ensure f(x) is non-negative, the maximum of the subtracted term must be \u2264 1.\n\nSo, the maximum of [\u221a(a\u00b2 + b\u00b2)\u00b7cos(x - \u03c6) + \u221a(A\u00b2 + B\u00b2)\u00b7cos(2x - \u03b8)] must be \u2264 1.\n\nHmm, that seems a bit tricky because the two cosine terms have different frequencies (one is cos(x), the other is cos(2x)). Their sum can have a maximum that\u0027s not straightforward.\n\nAlternatively, maybe I can consider f(x) as a sum of cosines and sines and think about its minimum value.\n\nWait, another approach could be to use the concept of positive semi-definite functions or something related to Fourier series, but I\u0027m not sure if that\u0027s necessary here.\n\nAlternatively, maybe I can use calculus to find the minima of f(x). That is, take the derivative, set it to zero, and find critical points.\n\nBut that might get complicated because f(x) is a combination of cos(x) and cos(2x). Let me try that approach.\n\nFirst, compute the derivative f\u0027(x):\n\nf\u0027(x) = a\u00b7sin(x) - b\u00b7cos(x) + 2A\u00b7sin(2x) + 2B\u00b7cos(2x)\n\nWait, let\u0027s double-check that:\n\nf(x) = 1 - a\u00b7cos(x) - b\u00b7sin(x) - A\u00b7cos(2x) - B\u00b7sin(2x)\n\nSo, f\u0027(x) = a\u00b7sin(x) - b\u00b7cos(x) + 2A\u00b7sin(2x) + 2B\u00b7cos(2x)\n\nYes, that\u0027s correct. Now, to find critical points, set f\u0027(x) = 0:\n\na\u00b7sin(x) - b\u00b7cos(x) + 2A\u00b7sin(2x) + 2B\u00b7cos(2x) = 0\n\nHmm, this seems complicated because it\u0027s a transcendental equation with multiple sine and cosine terms. Not sure if this is the right path.\n\nMaybe another approach is needed. Let me think about the function f(x) as a combination of multiple harmonics.\n\nAlternatively, perhaps I can write f(x) in terms of multiple angles or use some trigonometric identities to combine the terms.\n\nWait, let\u0027s think about f(x). It has cos(x), sin(x), cos(2x), sin(2x). Maybe I can express it as a sum of squares or something that can be bounded below.\n\nAlternatively, perhaps I can square f(x) and use some inequalities, but that might complicate things.\n\nWait, another idea: since f(x) is non-negative for all x, then f(x) can be considered as a sum of squares, which is always non-negative. Maybe I can represent f(x) as a sum of squares of some functions.\n\nBut that might be a stretch because f(x) is a combination of different trigonometric functions.\n\nAlternatively, perhaps I can use the method of expressing f(x) as a single sinusoidal function with some amplitude and phase shift, but with multiple frequencies, that\u0027s tricky.\n\nWait, maybe I can use the concept of the envelope of the function. Since f(x) is 1 minus two oscillating terms, the maximum possible dip would occur when both cosine terms are at their maximum positive values. So, if both \u221a(a\u00b2 + b\u00b2) and \u221a(A\u00b2 + B\u00b2) are positive, then their sum could subtract from 1.\n\nTherefore, to ensure that 1 - [\u221a(a\u00b2 + b\u00b2) + \u221a(A\u00b2 + B\u00b2)] \u2265 0, we must have \u221a(a\u00b2 + b\u00b2) + \u221a(A\u00b2 + B\u00b2) \u2264 1.\n\nWait, but that seems too restrictive because the two cosine terms might not reach their maximums at the same x. That is, when \u221a(a\u00b2 + b\u00b2)\u00b7cos(x - \u03c6) is at its maximum, \u221a(A\u00b2 + B\u00b2)\u00b7cos(2x - \u03b8) might not be at its maximum. So, maybe their sum doesn\u0027t actually reach the sum of their individual maxima.\n\nSo, perhaps the maximum of their sum is less than or equal to the sum of their maxima, but I\u0027m not sure.\n\nAlternatively, maybe I can consider specific values of x where f(x) could attain its minimum.\n\nFor example, suppose x is such that cos(x) = 1 and cos(2x) = 1. Is that possible? Let\u0027s see.\n\nIf cos(x) = 1, then x = 2\u03c0k for integer k. Then, cos(2x) = cos(4\u03c0k) = 1. So, yes, at x = 0, 2\u03c0, etc., both cos(x) and cos(2x) are 1.\n\nSimilarly, sin(x) and sin(2x) would be zero at these points.\n\nSo, plugging x = 0 into f(x):\n\nf(0) = 1 - a\u00b7cos(0) - b\u00b7sin(0) - A\u00b7cos(0) - B\u00b7sin(0) = 1 - a - A\n\nSince f(x) is non-negative everywhere, f(0) must be \u2265 0:\n\n1 - a - A \u2265 0 =\u003e a + A \u2264 1\n\nSimilarly, if we take x = \u03c0, cos(x) = -1, sin(x) = 0; cos(2x) = cos(2\u03c0) = 1, sin(2x) = 0.\n\nThen, f(\u03c0) = 1 - a\u00b7(-1) - b\u00b70 - A\u00b71 - B\u00b70 = 1 + a - A\n\nSo, f(\u03c0) = 1 + a - A \u2265 0 =\u003e -a + A \u2264 1\n\nSo, from f(0) and f(\u03c0), we get:\n\na + A \u2264 1\n\nand\n\n-a + A \u2264 1\n\nHmm, adding these two inequalities:\n\n(a + A) + (-a + A) \u2264 1 + 1 =\u003e 2A \u2264 2 =\u003e A \u2264 1\n\nSimilarly, subtracting the second inequality from the first:\n\n(a + A) - (-a + A) \u2264 1 - 1 =\u003e 2a \u2264 0 =\u003e a \u2264 0\n\nWait, is that right? Let me check:\n\nWait, f(0) gives a + A \u2264 1\n\nf(\u03c0) gives -a + A \u2264 1\n\nIf we add them, we get 2A \u2264 2, so A \u2264 1\n\nIf we subtract the second inequality from the first:\n\n(a + A) - (-a + A) = 2a \u2264 0, so a \u2264 0\n\nSimilarly, if we take x = \u03c0/2, then cos(x) = 0, sin(x) = 1; cos(2x) = cos(\u03c0) = -1, sin(2x) = sin(\u03c0) = 0\n\nSo, f(\u03c0/2) = 1 - a\u00b70 - b\u00b71 - A\u00b7(-1) - B\u00b70 = 1 - b + A\n\nSince f(\u03c0/2) \u2265 0, we have -b + A \u2264 1\n\nSimilarly, x = 3\u03c0/2: cos(x) = 0, sin(x) = -1; cos(2x) = cos(3\u03c0) = -1, sin(2x) = sin(3\u03c0) = 0\n\nf(3\u03c0/2) = 1 - a\u00b70 - b\u00b7(-1) - A\u00b7(-1) - B\u00b70 = 1 + b + A\n\nSo, f(3\u03c0/2) = 1 + b + A \u2265 0 =\u003e b + A \u2264 1\n\nWait, so from x = \u03c0/2, we have A - b \u2264 1\n\nFrom x = 3\u03c0/2, we have A + b \u2264 1\n\nSo, combining with previous inequalities, we have:\n\na \u2264 0\n\nA \u2264 1\n\na + A \u2264 1\n\nA - b \u2264 1\n\nA + b \u2264 1\n\nHmm, interesting. Let me collect all these inequalities:\n\n1. a + A \u2264 1\n\n2. -a + A \u2264 1\n\n3. A - b \u2264 1\n\n4. A + b \u2264 1\n\n5. A \u2264 1\n\n6. a \u2264 0\n\nWait, but how does this help me get to a\u00b2 + b\u00b2 \u2264 2 and A\u00b2 + B\u00b2 \u2264 1? Maybe I need a different approach.\n\nAlternatively, perhaps I can use the fact that f(x) is non-negative for all x, so its minimum is zero. So, I can consider f(x) as a sum of squares or in a way that relates to the amplitude of the trigonometric polynomials.\n\nWait, another thought: since f(x) is a sum of cosines and sines, it\u0027s a trigonometric polynomial, and if it\u0027s non-negative for all x, then perhaps it can be expressed as a sum of squares of trigonometric polynomials.\n\nBut I\u0027m not sure about the exact theorem here. Maybe I can consider using the Fejer\u0027s theorem or something similar, but I might be overcomplicating.\n\nWait, perhaps I can use the concept of the function being a square, but it\u0027s more complicated because it\u0027s not a perfect square.\n\nAlternatively, maybe I can consider f(x) as 1 minus a linear combination of sine and cosine terms and then use the concept of the minimum value.\n\nWait, another idea: maybe I can use the fact that the function f(x) is non-negative, so its Fourier coefficients must satisfy certain conditions.\n\nBut I\u0027m not too familiar with the exact conditions. Maybe I can use the concept of positive definiteness or something like that.\n\nWait, maybe it\u0027s better to consider the function f(x) as a sum of squares. Let me think about that.\n\nWait, here\u0027s another approach: let\u0027s consider that f(x) is non-negative for all x, so the integral of f(x) over any interval should be non-negative. But I\u0027m not sure if that helps.\n\nAlternatively, perhaps I can evaluate f(x) at specific points and get some inequalities on a, b, A, B.\n\nEarlier, I evaluated f(0), f(\u03c0), f(\u03c0/2), f(3\u03c0/2) and got some inequalities. Maybe if I evaluate f(x) at more points, I can get more inequalities.\n\nBut I already have 4 inequalities from 4 points. Maybe I can combine them with Cauchy-Schwarz or something.\n\nWait, let me think: from f(0) = 1 - a - A \u2265 0 =\u003e a + A \u2264 1\n\nFrom f(\u03c0) = 1 + a - A \u2265 0 =\u003e -a + A \u2264 1\n\nFrom f(\u03c0/2) = 1 - b + A \u2265 0 =\u003e A - b \u2264 1\n\nFrom f(3\u03c0/2) = 1 + b + A \u2265 0 =\u003e A + b \u2264 1\n\nSo, I have:\n\n1. a + A \u2264 1\n\n2. -a + A \u2264 1\n\n3. A - b \u2264 1\n\n4. A + b \u2264 1\n\nAlso, from x = \u03c0/4, let\u0027s compute f(\u03c0/4):\n\ncos(\u03c0/4) = \u221a2/2, sin(\u03c0/4) = \u221a2/2\n\ncos(2*(\u03c0/4)) = cos(\u03c0/2) = 0, sin(2*(\u03c0/4)) = sin(\u03c0/2) = 1\n\nSo, f(\u03c0/4) = 1 - a*(\u221a2/2) - b*(\u221a2/2) - A*0 - B*1\n\nSimplify:\n\nf(\u03c0/4) = 1 - (a + b)*(\u221a2/2) - B\n\nSince f(\u03c0/4) \u2265 0, we have:\n\n1 - (a + b)*(\u221a2/2) - B \u2265 0 =\u003e (a + b)*(\u221a2/2) + B \u2264 1\n\nSimilarly, f(5\u03c0/4) = 1 - a*(\u221a2/2) - b*(-\u221a2/2) - A*0 - B*(-1)\n\nWait, cos(5\u03c0/4) = -\u221a2/2, sin(5\u03c0/4) = -\u221a2/2, cos(2*(5\u03c0/4)) = cos(5\u03c0/2) = 0, sin(2*(5\u03c0/4)) = sin(5\u03c0/2) = 1\n\nSo, f(5\u03c0/4) = 1 - a*(-\u221a2/2) - b*(-\u221a2/2) - A*0 - B*1\n\nSimplify:\n\nf(5\u03c0/4) = 1 + (a + b)*(\u221a2/2) - B\n\nWhich gives:\n\n1 + (a + b)*(\u221a2/2) - B \u2265 0 =\u003e (a + b)*(\u221a2/2) - B \u2265 -1\n\nHmm, not sure if that helps.\n\nAlternatively, perhaps I can use multiple angles or represent the function in terms of exponentials. Maybe Euler\u0027s formula.\n\nBut that might complicate things further.\n\nWait, another thought: f(x) can be written as 1 minus [a cos x + b sin x] minus [A cos 2x + B sin 2x]. Let me denote the first term as T1 = a cos x + b sin x, and the second term as T2 = A cos 2x + B sin 2x.\n\nSo, f(x) = 1 - T1 - T2\n\nSince f(x) \u2265 0 for all x, we have T1 + T2 \u2264 1 for all x.\n\nSo, T1 + T2 \u2264 1 for all x.\n\nNow, T1 and T2 are both sinusoidal functions. So, their sum is another sinusoidal function, but with potentially different amplitude.\n\nWait, but T1 has frequency x, and T2 has frequency 2x, so their sum is not a single sinusoid but a combination.\n\nHmm, perhaps I can bound T1 + T2.\n\nWait, but T1 and T2 are functions with different frequencies. So, their maximums may not align.\n\nHmm, but maybe I can use the Cauchy-Schwarz inequality or something.\n\nWait, think about T1 and T2 separately. For T1, the maximum value is \u221a(a\u00b2 + b\u00b2), and for T2, the maximum value is \u221a(A\u00b2 + B\u00b2). So, if I can bound T1 + T2, maybe I can say that the maximum of T1 + T2 is \u2264 \u221a(a\u00b2 + b\u00b2) + \u221a(A\u00b2 + B\u00b2). But if their maxima don\u0027t align, perhaps the maximum of T1 + T2 is less than that.\n\nBut to ensure that T1 + T2 \u2264 1 for all x, it\u0027s necessary that the maximum of T1 + T2 is \u2264 1.\n\nSo, the maximum of T1 + T2 is \u2264 1.\n\nBut as I thought earlier, since T1 and T2 have different frequencies, their maximums might not occur at the same x. Therefore, the maximum of their sum might be less than the sum of their individual maxima.\n\nBut how can I relate that?\n\nWait, actually, maybe using the concept of the maximum of T1 + T2.\n\nBut perhaps another idea: use the fact that for all x, T1 + T2 \u2264 1. Then, using some optimization techniques, maybe set up an optimization problem where we maximize T1 + T2 and set that maximum to be \u2264 1.\n\nBut how do I find the maximum of T1 + T2? It\u0027s a function involving cos(x) and cos(2x). Maybe I can write it in terms of multiple angles.\n\nAlternatively, perhaps I can use some trigonometric identities.\n\nWait, cos(2x) can be written as 2 cos\u00b2x - 1, and sin(2x) is 2 sinx cosx.\n\nSo, let me try that.\n\nExpress T2 in terms of cos\u00b2x and sinx cosx:\n\nT2 = A cos(2x) + B sin(2x) = A(2 cos\u00b2x - 1) + B(2 sinx cosx)\n\nSo, T2 = 2A cos\u00b2x - A + 2B sinx cosx\n\nTherefore, T1 + T2 = [a cosx + b sinx] + [2A cos\u00b2x - A + 2B sinx cosx]\n\nSimplify:\n\nT1 + T2 = 2A cos\u00b2x + [a + 2B sinx] cosx + b sinx - A\n\nHmm, that\u0027s a quadratic in cosx and sinx. Maybe I can write it as a quadratic in cosx.\n\nBut it\u0027s still a bit messy because of the sinx term.\n\nAlternatively, perhaps I can write everything in terms of cosx and sinx:\n\nLet me denote t = x, so cosx and sinx are variables, but they are dependent since cos\u00b2x + sin\u00b2x = 1.\n\nBut that might complicate things. Alternatively, maybe parametrize using t = tan(x/2), but that might not be helpful here.\n\nWait, maybe another approach: consider T1 + T2 as a function in terms of x, then find its maximum.\n\nBut as I thought earlier, it\u0027s difficult because of the different frequencies.\n\nWait, maybe I can use the Cauchy-Schwarz inequality on T1 and T2.\n\nBut T1 and T2 are different functions, so I\u0027m not sure.\n\nAlternatively, maybe I can use the fact that for any x, T1 + T2 \u2264 1, so we can write:\n\nT1 + T2 = a cosx + b sinx + A cos2x + B sin2x \u2264 1\n\nThen, square both sides:\n\n(T1 + T2)^2 \u2264 1\n\nWhich gives:\n\n(a cosx + b sinx + A cos2x + B sin2x)^2 \u2264 1\n\nExpanding the left side:\n\n= [a cosx + b sinx]^2 + [A cos2x + B sin2x]^2 + 2(a cosx + b sinx)(A cos2x + B sin2x)\n\nSo,\n\n= (a\u00b2 cos\u00b2x + 2ab cosx sinx + b\u00b2 sin\u00b2x) + (A\u00b2 cos\u00b22x + 2AB cos2x sin2x + B\u00b2 sin\u00b22x) + 2(aA cosx cos2x + aB cosx sin2x + bA sinx cos2x + bB sinx sin2x)\n\nThat\u0027s a bit messy, but maybe we can use some trigonometric identities to simplify.\n\nFirst, note that cos\u00b2x = (1 + cos2x)/2, sin\u00b2x = (1 - cos2x)/2, cos\u00b22x = (1 + cos4x)/2, sin\u00b22x = (1 - cos4x)/2.\n\nAlso, 2ab cosx sinx = ab sin2x, 2AB cos2x sin2x = AB sin4x.\n\nSimilarly, the cross terms: 2aA cosx cos2x, 2aB cosx sin2x, 2bA sinx cos2x, 2bB sinx sin2x.\n\nThese can be simplified using product-to-sum identities.\n\nFor example, cosx cos2x = [cos(x - 2x) + cos(x + 2x)] / 2 = [cos(-x) + cos3x]/2 = [cosx + cos3x]/2\n\nSimilarly, cosx sin2x = [sin(2x + x) - sin(2x - x)] / 2 = [sin3x - sinx]/2\n\nSimilarly, sinx cos2x = [sin(x + 2x) - sin(x - 2x)] / 2 = [sin3x - sin(-x)] / 2 = [sin3x + sinx]/2\n\nAnd sinx sin2x = [cos(x - 2x) - cos(x + 2x)] / 2 = [cos(-x) - cos3x]/2 = [cosx - cos3x]/2\n\nSo, substituting all these into the expanded square:\n\nFirst term:\n\na\u00b2 cos\u00b2x + 2ab cosx sinx + b\u00b2 sin\u00b2x = a\u00b2*(1 + cos2x)/2 + ab sin2x + b\u00b2*(1 - cos2x)/2\n\n= (a\u00b2 + b\u00b2)/2 + (a\u00b2 - b\u00b2)/2 cos2x + ab sin2x\n\nSecond term:\n\nA\u00b2 cos\u00b22x + 2AB cos2x sin2x + B\u00b2 sin\u00b22x = A\u00b2*(1 + cos4x)/2 + AB sin4x + B\u00b2*(1 - cos4x)/2\n\n= (A\u00b2 + B\u00b2)/2 + (A\u00b2 - B\u00b2)/2 cos4x + AB sin4x\n\nThird term:\n\n2(aA cosx cos2x + aB cosx sin2x + bA sinx cos2x + bB sinx sin2x)\n\nFirst, compute each part:\n\n2aA cosx cos2x = aA [cosx + cos3x]\n\n2aB cosx sin2x = aB [sin3x - sinx]\n\n2bA sinx cos2x = bA [sin3x + sinx]\n\n2bB sinx sin2x = bB [cosx - cos3x]\n\nSo, combining all together:\n\n= aA cosx + aA cos3x + aB sin3x - aB sinx + bA sin3x + bA sinx + bB cosx - bB cos3x\n\nNow, group like terms:\n\n= (aA + bB) cosx + (aB - bB) sinx + (aA - bB) cos3x + (aB + bA) sin3x\n\n= (aA + bB) cosx + B(a - b) sinx + (aA - bB) cos3x + (aB + bA) sin3x\n\nSo, putting everything back together, the square of (T1 + T2) is:\n\n[(a\u00b2 + b\u00b2)/2 + (a\u00b2 - b\u00b2)/2 cos2x + ab sin2x] +\n\n[(A\u00b2 + B\u00b2)/2 + (A\u00b2 - B\u00b2)/2 cos4x + AB sin4x] +\n\n[(aA + bB) cosx + B(a - b) sinx + (aA - bB) cos3x + (aB + bA) sin3x] \u2264 1\n\nSo, the entire expression is:\n\n= (a\u00b2 + b\u00b2 + A\u00b2 + B\u00b2)/2 +\n\n(a\u00b2 - b\u00b2)/2 cos2x + ab sin2x +\n\n(A\u00b2 - B\u00b2)/2 cos4x + AB sin4x +\n\n(aA + bB) cosx + B(a - b) sinx +\n\n(aA - bB) cos3x + (aB + bA) sin3x \u2264 1\n\nNow, since this inequality must hold for all x, the maximum value of the left-hand side must be \u2264 1.\n\nBut the left-hand side is a trigonometric polynomial, and for it to be \u2264 1 for all x, all the coefficients of the cosine and sine terms must be zero, except for the constant term, and the constant term must be \u2264 1.\n\nWait, is that true? Wait, actually, no. Because if you have a trigonometric polynomial that is bounded above by a constant, it doesn\u0027t necessarily mean all the coefficients must be zero except the constant term. For example, a function like cos(x) + cos(2x) can have maximum value \u221a2, but it\u0027s not just the constant term.\n\nWait, but in our case, the entire expression must be \u2264 1 for all x. So, the maximum of this trigonometric polynomial must be \u2264 1.\n\nBut trigonometric polynomials can achieve their maxima at various points, so unless all the oscillating terms cancel out, the maximum can be larger than the constant term.\n\nSo, perhaps the only way for the entire expression to be \u2264 1 for all x is that all the coefficients of cos(kx) and sin(kx) for k \u2260 0 are zero, and the constant term is \u2264 1.\n\nWait, that\u0027s a standard result in Fourier series: if a trigonometric polynomial is bounded and has zero coefficients for all non-constant terms, then it\u0027s just a constant. Otherwise, if there are non-zero coefficients, the maximum can be higher.\n\nTherefore, to ensure that the entire expression is \u2264 1 for all x, the coefficients of all cosine and sine terms must be zero, and the constant term must be \u2264 1.\n\nTherefore, setting the coefficients of cosx, sinx, cos2x, sin2x, cos3x, sin3x, cos4x, sin4x to zero.\n\nSo, let\u0027s write down the equations:\n\n1. Coefficient of cosx: (aA + bB) = 0\n\n2. Coefficient of sinx: B(a - b) = 0\n\n3. Coefficient of cos2x: (a\u00b2 - b\u00b2)/2 = 0\n\n4. Coefficient of sin2x: ab = 0\n\n5. Coefficient of cos3x: (aA - bB) = 0\n\n6. Coefficient of sin3x: (aB + bA) = 0\n\n7. Coefficient of cos4x: (A\u00b2 - B\u00b2)/2 = 0\n\n8. Coefficient of sin4x: AB = 0\n\nAnd the constant term:\n\n(a\u00b2 + b\u00b2 + A\u00b2 + B\u00b2)/2 \u2264 1\n\nSo, from equations 3 and 7:\n\n(a\u00b2 - b\u00b2)/2 = 0 =\u003e a\u00b2 = b\u00b2\n\n(A\u00b2 - B\u00b2)/2 = 0 =\u003e A\u00b2 = B\u00b2\n\nFrom equations 4 and 8:\n\nab = 0\n\nAB = 0\n\nFrom equations 1, 2, 5, 6:\n\nEquation 1: aA + bB = 0\n\nEquation 2: B(a - b) = 0\n\nEquation 5: aA - bB = 0\n\nEquation 6: aB + bA = 0\n\nLet\u0027s analyze these equations step by step.\n\nFirst, from equation 3: a\u00b2 = b\u00b2, so b = \u00b1a\n\nSimilarly, from equation 7: A\u00b2 = B\u00b2, so B = \u00b1A\n\nFrom equation 4: ab = 0\n\nIf ab = 0, either a = 0 or b = 0.\n\nSimilarly, from equation 8: AB = 0, so either A = 0 or B = 0.\n\nCase 1: a = 0 and A = 0\n\nThen, from equation 3: a\u00b2 = b\u00b2 =\u003e 0 = b\u00b2 =\u003e b = 0\n\nFrom equation 7: A\u00b2 = B\u00b2 =\u003e 0 = B\u00b2 =\u003e B = 0\n\nSo, all coefficients a, b, A, B are zero.\n\nThen, f(x) = 1, which is non-negative, so that\u0027s acceptable. But in this case, a\u00b2 + b\u00b2 = 0 \u2264 2 and A\u00b2 + B\u00b2 = 0 \u2264 1.\n\nCase 2: a = 0 and B = 0\n\nFrom equation 4: ab = 0 =\u003e since a = 0, b can be anything? Wait, no, equation 3 says a\u00b2 = b\u00b2, so a = 0 =\u003e b\u00b2 = 0 =\u003e b = 0.\n\nSimilarly, from equation 8: AB = 0. If A \u2260 0, then B = 0, but if A = 0, then B can be anything? Wait, no, from equation 7, A\u00b2 = B\u00b2. So, if A \u2260 0, then B = \u00b1A. But equation 8: AB = 0. If A \u2260 0, then B must be 0, which would require A = 0, a contradiction.\n\nTherefore, if a = 0, then b = 0, and if A = 0, then B = 0.\n\nSo, this reduces to case 1.\n\nCase 3: b = 0 and A = 0\n\nSimilarly, from equation 3: a\u00b2 = b\u00b2 =\u003e a\u00b2 = 0 =\u003e a = 0\n\nFrom equation 7: A\u00b2 = B\u00b2 =\u003e 0 = B\u00b2 =\u003e B = 0\n\nAgain, all coefficients zero.\n\nCase 4: b = 0 and B = 0\n\nFrom equation 4: ab = 0 =\u003e either a = 0 or b = 0. Here, b = 0, so a can be anything? Wait, no: equation 3: a\u00b2 = b\u00b2 = 0 =\u003e a = 0\n\nFrom equation 8: AB = 0. Since A could be non-zero, but equation 7: A\u00b2 = B\u00b2. If B = 0, then A\u00b2 = 0 =\u003e A = 0\n\nSo again, all coefficients zero.\n\nSo, the only solution when all the coefficients of the trigonometric polynomial are zero is when a = b = A = B = 0, which gives f(x) = 1.\n\nBut in the problem statement, the function is f(x) = 1 - a cosx - b sinx - A cos2x - B sin2x, so it\u0027s possible that not all coefficients are zero, but the function is still non-negative.\n\nWait, but according to our earlier reasoning, if all the coefficients of cosx, sinx, etc., are zero, then f(x) must be constant, which is 1, hence non-negative.\n\nBut in the problem statement, we have a function that is non-negative for all x, but may not be constant.\n\nWait, perhaps my earlier conclusion is incorrect because if the function f(x) is non-negative for all x, the square of f(x) is also non-negative, but the expansion leads to a trigonometric polynomial which must be \u2264 1 for all x. But that would require all the oscillating terms to cancel out, leading to f(x) being a constant function. But that\u0027s not necessarily the case because f(x) is non-negative, but not necessarily constant.\n\nWait, maybe my mistake was in squaring f(x). Because f(x) is non-negative, but squaring it doesn\u0027t necessarily give us a direct relationship. Because f(x) could be something like 1 - something that oscillates but is always \u2264 1, so f(x) is always \u2265 0.\n\nBut when I squared f(x), I got an expression that must be \u2264 1 for all x, which is a stronger condition, implying that f(x) must be constant.\n\nBut that contradicts the problem statement, since the problem allows for a, b, A, B to be non-zero, as long as f(x) is non-negative.\n\nTherefore, my approach must be flawed.\n\nWait, perhaps instead of squaring f(x), I should consider f(x) as a function that is always non-negative, and use another method, maybe optimization.\n\nLet me think: since f(x) is non-negative for all x, the minimum value of f(x) is zero.\n\nTherefore, the minimum of f(x) over all x is zero.\n\nTo find the minimum, perhaps I can find the critical points by setting the derivative to zero, as I tried earlier.\n\nBut solving f\u0027(x) = 0 seems difficult because of the different frequencies.\n\nAlternatively, maybe I can use the method of Lagrange multipliers or something similar, but I\u0027m not sure.\n\nWait, maybe another idea: express f(x) in terms of multiple angles and combine them.\n\nWait, another thought: Use the fact that for any x, f(x) \u2265 0, so for all x, 1 \u2265 a cosx + b sinx + A cos2x + B sin2x.\n\nSo, the maximum of a cosx + b sinx + A cos2x + B sin2x is \u2264 1.\n\nSo, if I can find the maximum value of this expression, it must be \u2264 1.\n\nSo, what\u0027s the maximum of a cosx + b sinx + A cos2x + B sin2x?\n\nHmm, this is a combination of sinusoidal functions with different frequencies. It might be challenging to find the maximum.\n\nBut perhaps I can use the Cauchy-Schwarz inequality in some way.\n\nWait, let me think about the expression as a sum of two functions: the first is a cosx + b sinx, which has amplitude \u221a(a\u00b2 + b\u00b2), and the second is A cos2x + B sin2x, which has amplitude \u221a(A\u00b2 + B\u00b2).\n\nSo, the maximum of their sum is \u2264 \u221a(a\u00b2 + b\u00b2) + \u221a(A\u00b2 + B\u00b2). But as I thought earlier, their maxima may not align, so the actual maximum might be less.\n\nBut to have the maximum of their sum \u2264 1, I must have \u221a(a\u00b2 + b\u00b2) + \u221a(A\u00b2 + B\u00b2) \u2264 1.\n\nWait, but that\u0027s a sufficient condition, not a necessary one.\n\nBut in the problem",
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    "generation": "\u003cthink\u003e\nOkay, so I have this problem where I\u0027m given a function f(x) that\u0027s defined as:\n\nf(x) = 1 - a\u00b7cos(x) - b\u00b7sin(x) - A\u00b7cos(2x) - B\u00b7sin(2x)\n\nAnd I need to prove that if f(x) is always greater than or equal to zero for any real number x, then two inequalities must hold: a\u00b2 + b\u00b2 \u2264 2 and A\u00b2 + B\u00b2 \u2264 1.\n\nHmm, okay. Let\u0027s break this down. The function f(x) is a combination of cosine and sine terms with different coefficients. Since it\u0027s given that f(x) is non-negative for all real x, that must mean that the function doesn\u0027t dip below zero anywhere on the real line.\n\nI remember that for functions involving cosines and sines, especially with multiple angles, sometimes you can express them as a single trigonometric function or use some identities to simplify. Maybe that can help here.\n\nFirst, let me write down the function again:\n\nf(x) = 1 - a\u00b7cos(x) - b\u00b7sin(x) - A\u00b7cos(2x) - B\u00b7sin(2x)\n\nI need to ensure that this is always non-negative. So, perhaps I can rewrite f(x) in a way that makes it easier to analyze. Maybe group the terms with cos(x) and sin(x) together and similarly for cos(2x) and sin(2x).\n\nSo, let\u0027s see:\n\nf(x) = 1 - [a\u00b7cos(x) + b\u00b7sin(x)] - [A\u00b7cos(2x) + B\u00b7sin(2x)]\n\nHmm, okay. So, this is 1 minus two terms, each of which is a linear combination of sine and cosine. I know that expressions like c\u00b7cos(x) + d\u00b7sin(x) can be rewritten as R\u00b7cos(x - \u03c6) or R\u00b7sin(x + \u03c6), where R is the amplitude. Maybe that\u0027s useful here.\n\nLet me recall that any linear combination of sine and cosine can be written as a single sine or cosine with a phase shift. Specifically, for constants c and d, we have:\n\nc\u00b7cos(x) + d\u00b7sin(x) = \u221a(c\u00b2 + d\u00b2)\u00b7cos(x - \u03c6), where \u03c6 = arctan(d/c) or something like that.\n\nSimilarly, for the 2x terms:\n\nA\u00b7cos(2x) + B\u00b7sin(2x) = \u221a(A\u00b2 + B\u00b2)\u00b7cos(2x - \u03b8), where \u03b8 is some angle.\n\nSo, if I rewrite both of these, f(x) becomes:\n\nf(x) = 1 - \u221a(a\u00b2 + b\u00b2)\u00b7cos(x - \u03c6) - \u221a(A\u00b2 + B\u00b2)\u00b7cos(2x - \u03b8)\n\nHmm, okay. So now, f(x) is 1 minus two cosine terms with different frequencies and amplitudes.\n\nNow, since f(x) is non-negative for all x, the minimum value of f(x) must be at least zero.\n\nTherefore, the minimum value of 1 - \u221a(a\u00b2 + b\u00b2)\u00b7cos(x - \u03c6) - \u221a(A\u00b2 + B\u00b2)\u00b7cos(2x - \u03b8) must be \u2265 0.\n\nTo find the minimum of this expression, I might need to analyze the maximum possible value that the sum of the two cosine terms can take, because subtracting that from 1 needs to stay non-negative.\n\nWait, actually, since cosine functions can have maximum values of 1 and minimum values of -1, but since they\u0027re subtracted, the overall expression can be thought of as 1 minus something. So, to ensure f(x) is non-negative, the maximum of the subtracted term must be \u2264 1.\n\nSo, the maximum of [\u221a(a\u00b2 + b\u00b2)\u00b7cos(x - \u03c6) + \u221a(A\u00b2 + B\u00b2)\u00b7cos(2x - \u03b8)] must be \u2264 1.\n\nHmm, that seems a bit tricky because the two cosine terms have different frequencies (one is cos(x), the other is cos(2x)). Their sum can have a maximum that\u0027s not straightforward.\n\nAlternatively, maybe I can consider f(x) as a sum of cosines and sines and think about its minimum value.\n\nWait, another approach could be to use the concept of positive semi-definite functions or something related to Fourier series, but I\u0027m not sure if that\u0027s necessary here.\n\nAlternatively, maybe I can use calculus to find the minima of f(x). That is, take the derivative, set it to zero, and find critical points.\n\nBut that might get complicated because f(x) is a combination of cos(x) and cos(2x). Let me try that approach.\n\nFirst, compute the derivative f\u0027(x):\n\nf\u0027(x) = a\u00b7sin(x) - b\u00b7cos(x) + 2A\u00b7sin(2x) + 2B\u00b7cos(2x)\n\nWait, let\u0027s double-check that:\n\nf(x) = 1 - a\u00b7cos(x) - b\u00b7sin(x) - A\u00b7cos(2x) - B\u00b7sin(2x)\n\nSo, f\u0027(x) = a\u00b7sin(x) - b\u00b7cos(x) + 2A\u00b7sin(2x) + 2B\u00b7cos(2x)\n\nYes, that\u0027s correct. Now, to find critical points, set f\u0027(x) = 0:\n\na\u00b7sin(x) - b\u00b7cos(x) + 2A\u00b7sin(2x) + 2B\u00b7cos(2x) = 0\n\nHmm, this seems complicated because it\u0027s a transcendental equation with multiple sine and cosine terms. Not sure if this is the right path.\n\nMaybe another approach is needed. Let me think about the function f(x) as a combination of multiple harmonics.\n\nAlternatively, perhaps I can write f(x) in terms of multiple angles or use some trigonometric identities to combine the terms.\n\nWait, let\u0027s think about f(x). It has cos(x), sin(x), cos(2x), sin(2x). Maybe I can express it as a sum of squares or something that can be bounded below.\n\nAlternatively, perhaps I can square f(x) and use some inequalities, but that might complicate things.\n\nWait, another idea: since f(x) is non-negative for all x, then f(x) can be considered as a sum of squares, which is always non-negative. Maybe I can represent f(x) as a sum of squares of some functions.\n\nBut that might be a stretch because f(x) is a combination of different trigonometric functions.\n\nAlternatively, perhaps I can use the method of expressing f(x) as a single sinusoidal function with some amplitude and phase shift, but with multiple frequencies, that\u0027s tricky.\n\nWait, maybe I can use the concept of the envelope of the function. Since f(x) is 1 minus two oscillating terms, the maximum possible dip would occur when both cosine terms are at their maximum positive values. So, if both \u221a(a\u00b2 + b\u00b2) and \u221a(A\u00b2 + B\u00b2) are positive, then their sum could subtract from 1.\n\nTherefore, to ensure that 1 - [\u221a(a\u00b2 + b\u00b2) + \u221a(A\u00b2 + B\u00b2)] \u2265 0, we must have \u221a(a\u00b2 + b\u00b2) + \u221a(A\u00b2 + B\u00b2) \u2264 1.\n\nWait, but that seems too restrictive because the two cosine terms might not reach their maximums at the same x. That is, when \u221a(a\u00b2 + b\u00b2)\u00b7cos(x - \u03c6) is at its maximum, \u221a(A\u00b2 + B\u00b2)\u00b7cos(2x - \u03b8) might not be at its maximum. So, maybe their sum doesn\u0027t actually reach the sum of their individual maxima.\n\nSo, perhaps the maximum of their sum is less than or equal to the sum of their maxima, but I\u0027m not sure.\n\nAlternatively, maybe I can consider specific values of x where f(x) could attain its minimum.\n\nFor example, suppose x is such that cos(x) = 1 and cos(2x) = 1. Is that possible? Let\u0027s see.\n\nIf cos(x) = 1, then x = 2\u03c0k for integer k. Then, cos(2x) = cos(4\u03c0k) = 1. So, yes, at x = 0, 2\u03c0, etc., both cos(x) and cos(2x) are 1.\n\nSimilarly, sin(x) and sin(2x) would be zero at these points.\n\nSo, plugging x = 0 into f(x):\n\nf(0) = 1 - a\u00b7cos(0) - b\u00b7sin(0) - A\u00b7cos(0) - B\u00b7sin(0) = 1 - a - A\n\nSince f(x) is non-negative everywhere, f(0) must be \u2265 0:\n\n1 - a - A \u2265 0 =\u003e a + A \u2264 1\n\nSimilarly, if we take x = \u03c0, cos(x) = -1, sin(x) = 0; cos(2x) = cos(2\u03c0) = 1, sin(2x) = 0.\n\nThen, f(\u03c0) = 1 - a\u00b7(-1) - b\u00b70 - A\u00b71 - B\u00b70 = 1 + a - A\n\nSo, f(\u03c0) = 1 + a - A \u2265 0 =\u003e -a + A \u2264 1\n\nSo, from f(0) and f(\u03c0), we get:\n\na + A \u2264 1\n\nand\n\n-a + A \u2264 1\n\nHmm, adding these two inequalities:\n\n(a + A) + (-a + A) \u2264 1 + 1 =\u003e 2A \u2264 2 =\u003e A \u2264 1\n\nSimilarly, subtracting the second inequality from the first:\n\n(a + A) - (-a + A) \u2264 1 - 1 =\u003e 2a \u2264 0 =\u003e a \u2264 0\n\nWait, is that right? Let me check:\n\nWait, f(0) gives a + A \u2264 1\n\nf(\u03c0) gives -a + A \u2264 1\n\nIf we add them, we get 2A \u2264 2, so A \u2264 1\n\nIf we subtract the second inequality from the first:\n\n(a + A) - (-a + A) = 2a \u2264 0, so a \u2264 0\n\nSimilarly, if we take x = \u03c0/2, then cos(x) = 0, sin(x) = 1; cos(2x) = cos(\u03c0) = -1, sin(2x) = sin(\u03c0) = 0\n\nSo, f(\u03c0/2) = 1 - a\u00b70 - b\u00b71 - A\u00b7(-1) - B\u00b70 = 1 - b + A\n\nSince f(\u03c0/2) \u2265 0, we have -b + A \u2264 1\n\nSimilarly, x = 3\u03c0/2: cos(x) = 0, sin(x) = -1; cos(2x) = cos(3\u03c0) = -1, sin(2x) = sin(3\u03c0) = 0\n\nf(3\u03c0/2) = 1 - a\u00b70 - b\u00b7(-1) - A\u00b7(-1) - B\u00b70 = 1 + b + A\n\nSo, f(3\u03c0/2) = 1 + b + A \u2265 0 =\u003e b + A \u2264 1\n\nWait, so from x = \u03c0/2, we have A - b \u2264 1\n\nFrom x = 3\u03c0/2, we have A + b \u2264 1\n\nSo, combining with previous inequalities, we have:\n\na \u2264 0\n\nA \u2264 1\n\na + A \u2264 1\n\nA - b \u2264 1\n\nA + b \u2264 1\n\nHmm, interesting. Let me collect all these inequalities:\n\n1. a + A \u2264 1\n\n2. -a + A \u2264 1\n\n3. A - b \u2264 1\n\n4. A + b \u2264 1\n\n5. A \u2264 1\n\n6. a \u2264 0\n\nWait, but how does this help me get to a\u00b2 + b\u00b2 \u2264 2 and A\u00b2 + B\u00b2 \u2264 1? Maybe I need a different approach.\n\nAlternatively, perhaps I can use the fact that f(x) is non-negative for all x, so its minimum is zero. So, I can consider f(x) as a sum of squares or in a way that relates to the amplitude of the trigonometric polynomials.\n\nWait, another thought: since f(x) is a sum of cosines and sines, it\u0027s a trigonometric polynomial, and if it\u0027s non-negative for all x, then perhaps it can be expressed as a sum of squares of trigonometric polynomials.\n\nBut I\u0027m not sure about the exact theorem here. Maybe I can consider using the Fejer\u0027s theorem or something similar, but I might be overcomplicating.\n\nWait, perhaps I can use the concept of the function being a square, but it\u0027s more complicated because it\u0027s not a perfect square.\n\nAlternatively, maybe I can consider f(x) as 1 minus a linear combination of sine and cosine terms and then use the concept of the minimum value.\n\nWait, another idea: maybe I can use the fact that the function f(x) is non-negative, so its Fourier coefficients must satisfy certain conditions.\n\nBut I\u0027m not too familiar with the exact conditions. Maybe I can use the concept of positive definiteness or something like that.\n\nWait, maybe it\u0027s better to consider the function f(x) as a sum of squares. Let me think about that.\n\nWait, here\u0027s another approach: let\u0027s consider that f(x) is non-negative for all x, so the integral of f(x) over any interval should be non-negative. But I\u0027m not sure if that helps.\n\nAlternatively, perhaps I can evaluate f(x) at specific points and get some inequalities on a, b, A, B.\n\nEarlier, I evaluated f(0), f(\u03c0), f(\u03c0/2), f(3\u03c0/2) and got some inequalities. Maybe if I evaluate f(x) at more points, I can get more inequalities.\n\nBut I already have 4 inequalities from 4 points. Maybe I can combine them with Cauchy-Schwarz or something.\n\nWait, let me think: from f(0) = 1 - a - A \u2265 0 =\u003e a + A \u2264 1\n\nFrom f(\u03c0) = 1 + a - A \u2265 0 =\u003e -a + A \u2264 1\n\nFrom f(\u03c0/2) = 1 - b + A \u2265 0 =\u003e A - b \u2264 1\n\nFrom f(3\u03c0/2) = 1 + b + A \u2265 0 =\u003e A + b \u2264 1\n\nSo, I have:\n\n1. a + A \u2264 1\n\n2. -a + A \u2264 1\n\n3. A - b \u2264 1\n\n4. A + b \u2264 1\n\nAlso, from x = \u03c0/4, let\u0027s compute f(\u03c0/4):\n\ncos(\u03c0/4) = \u221a2/2, sin(\u03c0/4) = \u221a2/2\n\ncos(2*(\u03c0/4)) = cos(\u03c0/2) = 0, sin(2*(\u03c0/4)) = sin(\u03c0/2) = 1\n\nSo, f(\u03c0/4) = 1 - a*(\u221a2/2) - b*(\u221a2/2) - A*0 - B*1\n\nSimplify:\n\nf(\u03c0/4) = 1 - (a + b)*(\u221a2/2) - B\n\nSince f(\u03c0/4) \u2265 0, we have:\n\n1 - (a + b)*(\u221a2/2) - B \u2265 0 =\u003e (a + b)*(\u221a2/2) + B \u2264 1\n\nSimilarly, f(5\u03c0/4) = 1 - a*(\u221a2/2) - b*(-\u221a2/2) - A*0 - B*(-1)\n\nWait, cos(5\u03c0/4) = -\u221a2/2, sin(5\u03c0/4) = -\u221a2/2, cos(2*(5\u03c0/4)) = cos(5\u03c0/2) = 0, sin(2*(5\u03c0/4)) = sin(5\u03c0/2) = 1\n\nSo, f(5\u03c0/4) = 1 - a*(-\u221a2/2) - b*(-\u221a2/2) - A*0 - B*1\n\nSimplify:\n\nf(5\u03c0/4) = 1 + (a + b)*(\u221a2/2) - B\n\nWhich gives:\n\n1 + (a + b)*(\u221a2/2) - B \u2265 0 =\u003e (a + b)*(\u221a2/2) - B \u2265 -1\n\nHmm, not sure if that helps.\n\nAlternatively, perhaps I can use multiple angles or represent the function in terms of exponentials. Maybe Euler\u0027s formula.\n\nBut that might complicate things further.\n\nWait, another thought: f(x) can be written as 1 minus [a cos x + b sin x] minus [A cos 2x + B sin 2x]. Let me denote the first term as T1 = a cos x + b sin x, and the second term as T2 = A cos 2x + B sin 2x.\n\nSo, f(x) = 1 - T1 - T2\n\nSince f(x) \u2265 0 for all x, we have T1 + T2 \u2264 1 for all x.\n\nSo, T1 + T2 \u2264 1 for all x.\n\nNow, T1 and T2 are both sinusoidal functions. So, their sum is another sinusoidal function, but with potentially different amplitude.\n\nWait, but T1 has frequency x, and T2 has frequency 2x, so their sum is not a single sinusoid but a combination.\n\nHmm, perhaps I can bound T1 + T2.\n\nWait, but T1 and T2 are functions with different frequencies. So, their maximums may not align.\n\nHmm, but maybe I can use the Cauchy-Schwarz inequality or something.\n\nWait, think about T1 and T2 separately. For T1, the maximum value is \u221a(a\u00b2 + b\u00b2), and for T2, the maximum value is \u221a(A\u00b2 + B\u00b2). So, if I can bound T1 + T2, maybe I can say that the maximum of T1 + T2 is \u2264 \u221a(a\u00b2 + b\u00b2) + \u221a(A\u00b2 + B\u00b2). But if their maxima don\u0027t align, perhaps the maximum of T1 + T2 is less than that.\n\nBut to ensure that T1 + T2 \u2264 1 for all x, it\u0027s necessary that the maximum of T1 + T2 is \u2264 1.\n\nSo, the maximum of T1 + T2 is \u2264 1.\n\nBut as I thought earlier, since T1 and T2 have different frequencies, their maximums might not occur at the same x. Therefore, the maximum of their sum might be less than the sum of their individual maxima.\n\nBut how can I relate that?\n\nWait, actually, maybe using the concept of the maximum of T1 + T2.\n\nBut perhaps another idea: use the fact that for all x, T1 + T2 \u2264 1. Then, using some optimization techniques, maybe set up an optimization problem where we maximize T1 + T2 and set that maximum to be \u2264 1.\n\nBut how do I find the maximum of T1 + T2? It\u0027s a function involving cos(x) and cos(2x). Maybe I can write it in terms of multiple angles.\n\nAlternatively, perhaps I can use some trigonometric identities.\n\nWait, cos(2x) can be written as 2 cos\u00b2x - 1, and sin(2x) is 2 sinx cosx.\n\nSo, let me try that.\n\nExpress T2 in terms of cos\u00b2x and sinx cosx:\n\nT2 = A cos(2x) + B sin(2x) = A(2 cos\u00b2x - 1) + B(2 sinx cosx)\n\nSo, T2 = 2A cos\u00b2x - A + 2B sinx cosx\n\nTherefore, T1 + T2 = [a cosx + b sinx] + [2A cos\u00b2x - A + 2B sinx cosx]\n\nSimplify:\n\nT1 + T2 = 2A cos\u00b2x + [a + 2B sinx] cosx + b sinx - A\n\nHmm, that\u0027s a quadratic in cosx and sinx. Maybe I can write it as a quadratic in cosx.\n\nBut it\u0027s still a bit messy because of the sinx term.\n\nAlternatively, perhaps I can write everything in terms of cosx and sinx:\n\nLet me denote t = x, so cosx and sinx are variables, but they are dependent since cos\u00b2x + sin\u00b2x = 1.\n\nBut that might complicate things. Alternatively, maybe parametrize using t = tan(x/2), but that might not be helpful here.\n\nWait, maybe another approach: consider T1 + T2 as a function in terms of x, then find its maximum.\n\nBut as I thought earlier, it\u0027s difficult because of the different frequencies.\n\nWait, maybe I can use the Cauchy-Schwarz inequality on T1 and T2.\n\nBut T1 and T2 are different functions, so I\u0027m not sure.\n\nAlternatively, maybe I can use the fact that for any x, T1 + T2 \u2264 1, so we can write:\n\nT1 + T2 = a cosx + b sinx + A cos2x + B sin2x \u2264 1\n\nThen, square both sides:\n\n(T1 + T2)^2 \u2264 1\n\nWhich gives:\n\n(a cosx + b sinx + A cos2x + B sin2x)^2 \u2264 1\n\nExpanding the left side:\n\n= [a cosx + b sinx]^2 + [A cos2x + B sin2x]^2 + 2(a cosx + b sinx)(A cos2x + B sin2x)\n\nSo,\n\n= (a\u00b2 cos\u00b2x + 2ab cosx sinx + b\u00b2 sin\u00b2x) + (A\u00b2 cos\u00b22x + 2AB cos2x sin2x + B\u00b2 sin\u00b22x) + 2(aA cosx cos2x + aB cosx sin2x + bA sinx cos2x + bB sinx sin2x)\n\nThat\u0027s a bit messy, but maybe we can use some trigonometric identities to simplify.\n\nFirst, note that cos\u00b2x = (1 + cos2x)/2, sin\u00b2x = (1 - cos2x)/2, cos\u00b22x = (1 + cos4x)/2, sin\u00b22x = (1 - cos4x)/2.\n\nAlso, 2ab cosx sinx = ab sin2x, 2AB cos2x sin2x = AB sin4x.\n\nSimilarly, the cross terms: 2aA cosx cos2x, 2aB cosx sin2x, 2bA sinx cos2x, 2bB sinx sin2x.\n\nThese can be simplified using product-to-sum identities.\n\nFor example, cosx cos2x = [cos(x - 2x) + cos(x + 2x)] / 2 = [cos(-x) + cos3x]/2 = [cosx + cos3x]/2\n\nSimilarly, cosx sin2x = [sin(2x + x) - sin(2x - x)] / 2 = [sin3x - sinx]/2\n\nSimilarly, sinx cos2x = [sin(x + 2x) - sin(x - 2x)] / 2 = [sin3x - sin(-x)] / 2 = [sin3x + sinx]/2\n\nAnd sinx sin2x = [cos(x - 2x) - cos(x + 2x)] / 2 = [cos(-x) - cos3x]/2 = [cosx - cos3x]/2\n\nSo, substituting all these into the expanded square:\n\nFirst term:\n\na\u00b2 cos\u00b2x + 2ab cosx sinx + b\u00b2 sin\u00b2x = a\u00b2*(1 + cos2x)/2 + ab sin2x + b\u00b2*(1 - cos2x)/2\n\n= (a\u00b2 + b\u00b2)/2 + (a\u00b2 - b\u00b2)/2 cos2x + ab sin2x\n\nSecond term:\n\nA\u00b2 cos\u00b22x + 2AB cos2x sin2x + B\u00b2 sin\u00b22x = A\u00b2*(1 + cos4x)/2 + AB sin4x + B\u00b2*(1 - cos4x)/2\n\n= (A\u00b2 + B\u00b2)/2 + (A\u00b2 - B\u00b2)/2 cos4x + AB sin4x\n\nThird term:\n\n2(aA cosx cos2x + aB cosx sin2x + bA sinx cos2x + bB sinx sin2x)\n\nFirst, compute each part:\n\n2aA cosx cos2x = aA [cosx + cos3x]\n\n2aB cosx sin2x = aB [sin3x - sinx]\n\n2bA sinx cos2x = bA [sin3x + sinx]\n\n2bB sinx sin2x = bB [cosx - cos3x]\n\nSo, combining all together:\n\n= aA cosx + aA cos3x + aB sin3x - aB sinx + bA sin3x + bA sinx + bB cosx - bB cos3x\n\nNow, group like terms:\n\n= (aA + bB) cosx + (aB - bB) sinx + (aA - bB) cos3x + (aB + bA) sin3x\n\n= (aA + bB) cosx + B(a - b) sinx + (aA - bB) cos3x + (aB + bA) sin3x\n\nSo, putting everything back together, the square of (T1 + T2) is:\n\n[(a\u00b2 + b\u00b2)/2 + (a\u00b2 - b\u00b2)/2 cos2x + ab sin2x] +\n\n[(A\u00b2 + B\u00b2)/2 + (A\u00b2 - B\u00b2)/2 cos4x + AB sin4x] +\n\n[(aA + bB) cosx + B(a - b) sinx + (aA - bB) cos3x + (aB + bA) sin3x] \u2264 1\n\nSo, the entire expression is:\n\n= (a\u00b2 + b\u00b2 + A\u00b2 + B\u00b2)/2 +\n\n(a\u00b2 - b\u00b2)/2 cos2x + ab sin2x +\n\n(A\u00b2 - B\u00b2)/2 cos4x + AB sin4x +\n\n(aA + bB) cosx + B(a - b) sinx +\n\n(aA - bB) cos3x + (aB + bA) sin3x \u2264 1\n\nNow, since this inequality must hold for all x, the maximum value of the left-hand side must be \u2264 1.\n\nBut the left-hand side is a trigonometric polynomial, and for it to be \u2264 1 for all x, all the coefficients of the cosine and sine terms must be zero, except for the constant term, and the constant term must be \u2264 1.\n\nWait, is that true? Wait, actually, no. Because if you have a trigonometric polynomial that is bounded above by a constant, it doesn\u0027t necessarily mean all the coefficients must be zero except the constant term. For example, a function like cos(x) + cos(2x) can have maximum value \u221a2, but it\u0027s not just the constant term.\n\nWait, but in our case, the entire expression must be \u2264 1 for all x. So, the maximum of this trigonometric polynomial must be \u2264 1.\n\nBut trigonometric polynomials can achieve their maxima at various points, so unless all the oscillating terms cancel out, the maximum can be larger than the constant term.\n\nSo, perhaps the only way for the entire expression to be \u2264 1 for all x is that all the coefficients of cos(kx) and sin(kx) for k \u2260 0 are zero, and the constant term is \u2264 1.\n\nWait, that\u0027s a standard result in Fourier series: if a trigonometric polynomial is bounded and has zero coefficients for all non-constant terms, then it\u0027s just a constant. Otherwise, if there are non-zero coefficients, the maximum can be higher.\n\nTherefore, to ensure that the entire expression is \u2264 1 for all x, the coefficients of all cosine and sine terms must be zero, and the constant term must be \u2264 1.\n\nTherefore, setting the coefficients of cosx, sinx, cos2x, sin2x, cos3x, sin3x, cos4x, sin4x to zero.\n\nSo, let\u0027s write down the equations:\n\n1. Coefficient of cosx: (aA + bB) = 0\n\n2. Coefficient of sinx: B(a - b) = 0\n\n3. Coefficient of cos2x: (a\u00b2 - b\u00b2)/2 = 0\n\n4. Coefficient of sin2x: ab = 0\n\n5. Coefficient of cos3x: (aA - bB) = 0\n\n6. Coefficient of sin3x: (aB + bA) = 0\n\n7. Coefficient of cos4x: (A\u00b2 - B\u00b2)/2 = 0\n\n8. Coefficient of sin4x: AB = 0\n\nAnd the constant term:\n\n(a\u00b2 + b\u00b2 + A\u00b2 + B\u00b2)/2 \u2264 1\n\nSo, from equations 3 and 7:\n\n(a\u00b2 - b\u00b2)/2 = 0 =\u003e a\u00b2 = b\u00b2\n\n(A\u00b2 - B\u00b2)/2 = 0 =\u003e A\u00b2 = B\u00b2\n\nFrom equations 4 and 8:\n\nab = 0\n\nAB = 0\n\nFrom equations 1, 2, 5, 6:\n\nEquation 1: aA + bB = 0\n\nEquation 2: B(a - b) = 0\n\nEquation 5: aA - bB = 0\n\nEquation 6: aB + bA = 0\n\nLet\u0027s analyze these equations step by step.\n\nFirst, from equation 3: a\u00b2 = b\u00b2, so b = \u00b1a\n\nSimilarly, from equation 7: A\u00b2 = B\u00b2, so B = \u00b1A\n\nFrom equation 4: ab = 0\n\nIf ab = 0, either a = 0 or b = 0.\n\nSimilarly, from equation 8: AB = 0, so either A = 0 or B = 0.\n\nCase 1: a = 0 and A = 0\n\nThen, from equation 3: a\u00b2 = b\u00b2 =\u003e 0 = b\u00b2 =\u003e b = 0\n\nFrom equation 7: A\u00b2 = B\u00b2 =\u003e 0 = B\u00b2 =\u003e B = 0\n\nSo, all coefficients a, b, A, B are zero.\n\nThen, f(x) = 1, which is non-negative, so that\u0027s acceptable. But in this case, a\u00b2 + b\u00b2 = 0 \u2264 2 and A\u00b2 + B\u00b2 = 0 \u2264 1.\n\nCase 2: a = 0 and B = 0\n\nFrom equation 4: ab = 0 =\u003e since a = 0, b can be anything? Wait, no, equation 3 says a\u00b2 = b\u00b2, so a = 0 =\u003e b\u00b2 = 0 =\u003e b = 0.\n\nSimilarly, from equation 8: AB = 0. If A \u2260 0, then B = 0, but if A = 0, then B can be anything? Wait, no, from equation 7, A\u00b2 = B\u00b2. So, if A \u2260 0, then B = \u00b1A. But equation 8: AB = 0. If A \u2260 0, then B must be 0, which would require A = 0, a contradiction.\n\nTherefore, if a = 0, then b = 0, and if A = 0, then B = 0.\n\nSo, this reduces to case 1.\n\nCase 3: b = 0 and A = 0\n\nSimilarly, from equation 3: a\u00b2 = b\u00b2 =\u003e a\u00b2 = 0 =\u003e a = 0\n\nFrom equation 7: A\u00b2 = B\u00b2 =\u003e 0 = B\u00b2 =\u003e B = 0\n\nAgain, all coefficients zero.\n\nCase 4: b = 0 and B = 0\n\nFrom equation 4: ab = 0 =\u003e either a = 0 or b = 0. Here, b = 0, so a can be anything? Wait, no: equation 3: a\u00b2 = b\u00b2 = 0 =\u003e a = 0\n\nFrom equation 8: AB = 0. Since A could be non-zero, but equation 7: A\u00b2 = B\u00b2. If B = 0, then A\u00b2 = 0 =\u003e A = 0\n\nSo again, all coefficients zero.\n\nSo, the only solution when all the coefficients of the trigonometric polynomial are zero is when a = b = A = B = 0, which gives f(x) = 1.\n\nBut in the problem statement, the function is f(x) = 1 - a cosx - b sinx - A cos2x - B sin2x, so it\u0027s possible that not all coefficients are zero, but the function is still non-negative.\n\nWait, but according to our earlier reasoning, if all the coefficients of cosx, sinx, etc., are zero, then f(x) must be constant, which is 1, hence non-negative.\n\nBut in the problem statement, we have a function that is non-negative for all x, but may not be constant.\n\nWait, perhaps my earlier conclusion is incorrect because if the function f(x) is non-negative for all x, the square of f(x) is also non-negative, but the expansion leads to a trigonometric polynomial which must be \u2264 1 for all x. But that would require all the oscillating terms to cancel out, leading to f(x) being a constant function. But that\u0027s not necessarily the case because f(x) is non-negative, but not necessarily constant.\n\nWait, maybe my mistake was in squaring f(x). Because f(x) is non-negative, but squaring it doesn\u0027t necessarily give us a direct relationship. Because f(x) could be something like 1 - something that oscillates but is always \u2264 1, so f(x) is always \u2265 0.\n\nBut when I squared f(x), I got an expression that must be \u2264 1 for all x, which is a stronger condition, implying that f(x) must be constant.\n\nBut that contradicts the problem statement, since the problem allows for a, b, A, B to be non-zero, as long as f(x) is non-negative.\n\nTherefore, my approach must be flawed.\n\nWait, perhaps instead of squaring f(x), I should consider f(x) as a function that is always non-negative, and use another method, maybe optimization.\n\nLet me think: since f(x) is non-negative for all x, the minimum value of f(x) is zero.\n\nTherefore, the minimum of f(x) over all x is zero.\n\nTo find the minimum, perhaps I can find the critical points by setting the derivative to zero, as I tried earlier.\n\nBut solving f\u0027(x) = 0 seems difficult because of the different frequencies.\n\nAlternatively, maybe I can use the method of Lagrange multipliers or something similar, but I\u0027m not sure.\n\nWait, maybe another idea: express f(x) in terms of multiple angles and combine them.\n\nWait, another thought: Use the fact that for any x, f(x) \u2265 0, so for all x, 1 \u2265 a cosx + b sinx + A cos2x + B sin2x.\n\nSo, the maximum of a cosx + b sinx + A cos2x + B sin2x is \u2264 1.\n\nSo, if I can find the maximum value of this expression, it must be \u2264 1.\n\nSo, what\u0027s the maximum of a cosx + b sinx + A cos2x + B sin2x?\n\nHmm, this is a combination of sinusoidal functions with different frequencies. It might be challenging to find the maximum.\n\nBut perhaps I can use the Cauchy-Schwarz inequality in some way.\n\nWait, let me think about the expression as a sum of two functions: the first is a cosx + b sinx, which has amplitude \u221a(a\u00b2 + b\u00b2), and the second is A cos2x + B sin2x, which has amplitude \u221a(A\u00b2 + B\u00b2).\n\nSo, the maximum of their sum is \u2264 \u221a(a\u00b2 + b\u00b2) + \u221a(A\u00b2 + B\u00b2). But as I thought earlier, their maxima may not align, so the actual maximum might be less.\n\nBut to have the maximum of their sum \u2264 1, I must have \u221a(a\u00b2 + b\u00b2) + \u221a(A\u00b2 + B\u00b2) \u2264 1.\n\nWait, but that\u0027s a sufficient condition, not a necessary one.\n\nBut in the problem",
    "metadata": "{\u0027source\u0027: \u0027aops_forum\u0027, \u0027messages\u0027: [{\u0027content\u0027: \u0027Let  $a,b,A,B$  be given reals. We consider the function defined by \\\\[ f(x) = 1 - a \\\\cdot \\\\cos(x) - b \\\\cdot \\\\sin(x) - A \\\\cdot \\\\cos(2x) - B \\\\cdot \\\\sin(2x). \\\\] Prove that if for any real number  $x$  we have  $f(x) \\\\geq 0$  then  $a^2 + b^2 \\\\leq 2$  and  $A^2 + B^2 \\\\leq 1.$ \u0027, \u0027role\u0027: \u0027user\u0027}, {\u0027content\u0027: \u00271. **Rewrite the function using trigonometric identities:**\\n   \\\\[\\n   f(x) = 1 - a \\\\cos(x) - b \\\\sin(x) - A \\\\cos(2x) - B \\\\sin(2x)\\n   \\\\]\\n   We can use the angle addition formulas for sine and cosine:\\n   \\\\[\\n   \\\\cos(x + \\\\theta) = \\\\cos(x)\\\\cos(\\\\theta) - \\\\sin(x)\\\\sin(\\\\theta)\\n   \\\\]\\n   \\\\[\\n   \\\\sin(x + \\\\theta) = \\\\sin(x)\\\\cos(\\\\theta) + \\\\cos(x)\\\\sin(\\\\theta)\\n   \\\\]\\n   Let:\\n   \\\\[\\n   R_1 = \\\\sqrt{a^2 + b^2}, \\\\quad \\\\theta_1 = \\\\arctan\\\\left(\\\\frac{b}{a}\\\\right)\\n   \\\\]\\n   \\\\[\\n   R_2 = \\\\sqrt{A^2 + B^2}, \\\\quad \\\\theta_2 = \\\\arctan\\\\left(\\\\frac{B}{A}\\\\right)\\n   \\\\]\\n   Then:\\n   \\\\[\\n   a \\\\cos(x) + b \\\\sin(x) = R_1 \\\\sin(x + \\\\theta_1)\\n   \\\\]\\n   \\\\[\\n   A \\\\cos(2x) + B \\\\sin(2x) = R_2 \\\\sin(2x + \\\\theta_2)\\n   \\\\]\\n   Therefore, the function becomes:\\n   \\\\[\\n   f(x) = 1 - R_1 \\\\sin(x + \\\\theta_1) - R_2 \\\\sin(2x + \\\\theta_2)\\n   \\\\]\\n\\n2. **Analyze the function for non-negativity:**\\n   Given \\\\( f(x) \\\\geq 0 \\\\) for all \\\\( x \\\\), we need to ensure that the maximum values of the sine terms do not exceed 1:\\n   \\\\[\\n   1 - R_1 \\\\sin(x + \\\\theta_1) - R_2 \\\\sin(2x + \\\\theta_2) \\\\geq 0\\n   \\\\]\\n   The maximum value of \\\\( \\\\sin(x + \\\\theta_1) \\\\) and \\\\( \\\\sin(2x + \\\\theta_2) \\\\) is 1. Therefore:\\n   \\\\[\\n   1 - R_1 - R_2 \\\\geq 0\\n   \\\\]\\n   This implies:\\n   \\\\[\\n   R_1 + R_2 \\\\leq 1\\n   \\\\]\\n\\n3. **Consider the individual terms:**\\n   Since \\\\( R_1 = \\\\sqrt{a^2 + b^2} \\\\) and \\\\( R_2 = \\\\sqrt{A^2 + B^2} \\\\), we need to show:\\n   \\\\[\\n   \\\\sqrt{a^2 + b^2} \\\\leq 1 \\\\quad \\\\text{and} \\\\quad \\\\sqrt{A^2 + B^2} \\\\leq 1\\n   \\\\]\\n\\n4. **Prove the inequalities:**\\n   - For \\\\( \\\\sqrt{a^2 + b^2} \\\\leq 1 \\\\):\\n     \\\\[\\n     a^2 + b^2 \\\\leq 1^2 = 1\\n     \\\\]\\n   - For \\\\( \\\\sqrt{A^2 + B^2} \\\\leq 1 \\\\):\\n     \\\\[\\n     A^2 + B^2 \\\\leq 1^2 = 1\\n     \\\\]\\n\\n5. **Conclusion:**\\n   Therefore, we have shown that if \\\\( f(x) \\\\geq 0 \\\\) for all \\\\( x \\\\), then:\\n   \\\\[\\n   a^2 + b^2 \\\\leq 2 \\\\quad \\\\text{and} \\\\quad A^2 + B^2 \\\\leq 1\\n   \\\\]\\n\\n\\\\(\\\\blacksquare\\\\)\\n\\nThe final answer is \\\\( \\\\boxed{ a^2 + b^2 \\\\leq 2 } \\\\) and \\\\( A^2 + B^2 \\\\leq 1 \\\\)\u0027, \u0027role\u0027: \u0027assistant\u0027}]}",
    "model_name": "deepseek-ai/DeepSeek-R1-Distill-Qwen-14B",
    "problem": "Let  $a,b,A,B$  be given reals. We consider the function defined by \\[ f(x) = 1 - a \\cdot \\cos(x) - b \\cdot \\sin(x) - A \\cdot \\cos(2x) - B \\cdot \\sin(2x). \\] Prove that if for any real number  $x$  we have  $f(x) \\geq 0$  then  $a^2 + b^2 \\leq 2$  and  $A^2 + B^2 \\leq 1.$ ",
    "solution": "1. **Rewrite the function using trigonometric identities:**\n   \\[\n   f(x) = 1 - a \\cos(x) - b \\sin(x) - A \\cos(2x) - B \\sin(2x)\n   \\]\n   We can use the angle addition formulas for sine and cosine:\n   \\[\n   \\cos(x + \\theta) = \\cos(x)\\cos(\\theta) - \\sin(x)\\sin(\\theta)\n   \\]\n   \\[\n   \\sin(x + \\theta) = \\sin(x)\\cos(\\theta) + \\cos(x)\\sin(\\theta)\n   \\]\n   Let:\n   \\[\n   R_1 = \\sqrt{a^2 + b^2}, \\quad \\theta_1 = \\arctan\\left(\\frac{b}{a}\\right)\n   \\]\n   \\[\n   R_2 = \\sqrt{A^2 + B^2}, \\quad \\theta_2 = \\arctan\\left(\\frac{B}{A}\\right)\n   \\]\n   Then:\n   \\[\n   a \\cos(x) + b \\sin(x) = R_1 \\sin(x + \\theta_1)\n   \\]\n   \\[\n   A \\cos(2x) + B \\sin(2x) = R_2 \\sin(2x + \\theta_2)\n   \\]\n   Therefore, the function becomes:\n   \\[\n   f(x) = 1 - R_1 \\sin(x + \\theta_1) - R_2 \\sin(2x + \\theta_2)\n   \\]\n\n2. **Analyze the function for non-negativity:**\n   Given \\( f(x) \\geq 0 \\) for all \\( x \\), we need to ensure that the maximum values of the sine terms do not exceed 1:\n   \\[\n   1 - R_1 \\sin(x + \\theta_1) - R_2 \\sin(2x + \\theta_2) \\geq 0\n   \\]\n   The maximum value of \\( \\sin(x + \\theta_1) \\) and \\( \\sin(2x + \\theta_2) \\) is 1. Therefore:\n   \\[\n   1 - R_1 - R_2 \\geq 0\n   \\]\n   This implies:\n   \\[\n   R_1 + R_2 \\leq 1\n   \\]\n\n3. **Consider the individual terms:**\n   Since \\( R_1 = \\sqrt{a^2 + b^2} \\) and \\( R_2 = \\sqrt{A^2 + B^2} \\), we need to show:\n   \\[\n   \\sqrt{a^2 + b^2} \\leq 1 \\quad \\text{and} \\quad \\sqrt{A^2 + B^2} \\leq 1\n   \\]\n\n4. **Prove the inequalities:**\n   - For \\( \\sqrt{a^2 + b^2} \\leq 1 \\):\n     \\[\n     a^2 + b^2 \\leq 1^2 = 1\n     \\]\n   - For \\( \\sqrt{A^2 + B^2} \\leq 1 \\):\n     \\[\n     A^2 + B^2 \\leq 1^2 = 1\n     \\]\n\n5. **Conclusion:**\n   Therefore, we have shown that if \\( f(x) \\geq 0 \\) for all \\( x \\), then:\n   \\[\n   a^2 + b^2 \\leq 2 \\quad \\text{and} \\quad A^2 + B^2 \\leq 1\n   \\]\n\n\\(\\blacksquare\\)\n\nThe final answer is \\( \\boxed{ a^2 + b^2 \\leq 2 } \\) and \\( A^2 + B^2 \\leq 1 \\)",
    "source_type": "AI-MO/NuminaMath-CoT/aops_forum",
    "thinking_trajectories": "[Let $f(x) = 1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x \\geq 0$ for all $x \\in \\mathbb{R}$.\nWe want to show that $a^2 + b^2 \\leq 2$ and $A^2 + B^2 \\leq 1$.\n\nConsider specific values of $x$.\nFor $x=0$, $f(0) = 1 - a - A \\geq 0$, so $a+A \\leq 1$.\nFor $x=\\pi$, $f(\\pi) = 1 - a \\cos \\pi - b \\sin \\pi - A \\cos 2\\pi - B \\sin 2\\pi = 1 - a(-1) - b(0) - A(1) - B(0) = 1+a-A \\geq 0$, so $A-a \\leq 1$.\nFor $x=\\pi/2$, $f(\\pi/2) = 1 - a \\cos(\\pi/2) - b \\sin(\\pi/2) - A \\cos(\\pi) - B \\sin(\\pi) = 1 - a(0) - b(1) - A(-1) - B(0) = 1 - b + A \\geq 0$, so $b-A \\leq 1$.\nFor $x=3\\pi/2$, $f(3\\pi/2) = 1 - a \\cos(3\\pi/2) - b \\sin(3\\pi/2) - A \\cos(3\\pi) - B \\sin(3\\pi) = 1 - a(0) - b(-1) - A(-1) - B(0) = 1 + b + A \\geq 0$, so $-b-A \\leq 1$.\n\nFrom $a+A \\leq 1$ and $A-a \\leq 1$, we have $-1 \\leq a+A \\leq 1$ and $-1 \\leq a-A \\leq 1$.\nFrom $b-A \\leq 1$ and $-b-A \\leq 1$, we have $-1 \\leq b-A \\leq 1$ and $-1 \\leq b+A \\leq 1$.\n\nConsider the integral of $f(x)$ over a period.\n$\\int_0^{2\\pi} f(x) dx = \\int_0^{2\\pi} (1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x) dx \\geq 0$.\n$\\int_0^{2\\pi} 1 dx = 2\\pi$.\n$\\int_0^{2\\pi} \\cos x dx = [\\sin x]_0^{2\\pi} = 0$.\n$\\int_0^{2\\pi} \\sin x dx = [-\\cos x]_0^{2\\pi} = -1 - (-1) = 0$.\n$\\int_0^{2\\pi} \\cos 2x dx = [\\frac{1}{2} \\sin 2x]_0^{2\\pi} = 0$.\n$\\int_0^{2\\pi} \\sin 2x dx = [-\\frac{1}{2} \\cos 2x]_0^{2\\pi} = -\\frac{1}{2} - (-\\frac{1}{2}) = 0$.\nSo $\\int_0^{2\\pi} f(x) dx = 2\\pi \\geq 0$. This doesn\u0027t give any information.\n\nMultiply $f(x)$ by $\\cos x$ and integrate.\n$\\int_0^{2\\pi} f(x) \\cos x dx = \\int_0^{2\\pi} (1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x) \\cos x dx \\geq 0$.\n$\\int_0^{2\\pi} \\cos x dx = 0$.\n$\\int_0^{2\\pi} \\cos^2 x dx = \\int_0^{2\\pi} \\frac{1+\\cos 2x}{2} dx = [\\frac{x}{2} + \\frac{\\sin 2x}{4}]_0^{2\\pi} = \\pi$.\n$\\int_0^{2\\pi} \\sin x \\cos x dx = \\int_0^{2\\pi} \\frac{1}{2} \\sin 2x dx = [-\\frac{1}{4} \\cos 2x]_0^{2\\pi} = -\\frac{1}{4} - (-\\frac{1}{4}) = 0$.\n$\\int_0^{2\\pi} \\cos 2x \\cos x dx = \\int_0^{2\\pi} \\frac{1}{2} (\\cos 3x + \\cos x) dx = [\\frac{1}{6} \\sin 3x + \\frac{1}{2} \\sin x]_0^{2\\pi} = 0$.\n$\\int_0^{2\\pi} \\sin 2x \\cos x dx = \\int_0^{2\\pi} \\frac{1}{2} (\\sin 3x + \\sin x) dx = [-\\frac{1}{6} \\cos 3x - \\frac{1}{2} \\cos x]_0^{2\\pi} = (-\\frac{1}{6} - \\frac{1}{2}) - (-\\frac{1}{6} - \\frac{1}{2}) = 0$.\nSo $\\int_0^{2\\pi} f(x) \\cos x dx = 0 - a \\pi - 0 - 0 - 0 = -a\\pi \\geq 0$, so $a \\leq 0$. This is not necessarily true.\n\nConsider $f(x) \\geq 0$.\n$1 - (a \\cos x + b \\sin x) - (A \\cos 2x + B \\sin 2x) \\geq 0$.\nLet $a = r \\cos \\alpha, b = r \\sin \\alpha$. Then $a \\cos x + b \\sin x = r (\\cos \\alpha \\cos x + \\sin \\alpha \\sin x) = r \\cos(x-\\alpha)$.\nLet $A = R \\cos \\beta, B = R \\sin \\beta$. Then $A \\cos 2x + B \\sin 2x = R (\\cos \\beta \\cos 2x + \\sin \\beta \\sin 2x) = R \\cos(2x-\\beta)$.\nSo $1 - r \\cos(x-\\alpha) - R \\cos(2x-\\beta) \\geq 0$.\nLet $y = x-\\alpha$. Then $2x-\\beta = 2(y+\\alpha) - \\beta = 2y + 2\\alpha - \\beta$. Let $\\gamma = 2\\alpha - \\beta$.\n$1 - r \\cos y - R \\cos(2y+\\gamma) \\geq 0$.\n$1 - r \\cos y - R (\\cos 2y \\cos \\gamma - \\sin 2y \\sin \\gamma) \\geq 0$.\n$1 - r \\cos y - R ((2\\cos^2 y - 1) \\cos \\gamma - 2 \\sin y \\cos y \\sin \\gamma) \\geq 0$.\n$1 - r \\cos y - 2R \\cos^2 y \\cos \\gamma + R \\cos \\gamma + 2R \\sin y \\cos y \\sin \\gamma \\geq 0$.\n\nConsider $f(x) \\geq 0$.\nMultiply by $\\cos(x-\\phi)$ and integrate from $0$ to $2\\pi$.\n$\\int_0^{2\\pi} f(x) \\cos(x-\\phi) dx \\geq 0$.\n$\\int_0^{2\\pi} (1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x) (\\cos x \\cos \\phi + \\sin x \\sin \\phi) dx \\geq 0$.\n$\\cos \\phi \\int_0^{2\\pi} (\\cos x - a \\cos^2 x - b \\sin x \\cos x - A \\cos 2x \\cos x - B \\sin 2x \\cos x) dx + \\sin \\phi \\int_0^{2\\pi} (\\sin x - a \\cos x \\sin x - b \\sin^2 x - A \\cos 2x \\sin x - B \\sin 2x \\sin x) dx \\geq 0$.\n$\\cos \\phi (0 - a\\pi - 0 - 0 - 0) + \\sin \\phi (0 - 0 - b\\pi - 0 - 0) \\geq 0$.\n$-a\\pi \\cos \\phi - b\\pi \\sin \\phi \\geq 0$.\n$a \\cos \\phi + b \\sin \\phi \\leq 0$.\nThis must be true for all $\\phi$. Let $\\phi$ such that $\\tan \\phi = b/a$. Then $a \\frac{a}{\\sqrt{a^2+b^2}} + b \\frac{b}{\\sqrt{a^2+b^2}} = \\sqrt{a^2+b^2} \\leq 0$, so $a=b=0$. This is wrong.\n\nConsider $f(x) \\geq 0$.\n$1 - (a \\cos x + b \\sin x) - (A \\cos 2x + B \\sin 2x) \\geq 0$.\nThe minimum value of $a \\cos x + b \\sin x$ is $-\\sqrt{a^2+b^2}$.\nThe minimum value of $A \\cos 2x + B \\sin 2x$ is $-\\sqrt{A^2+B^2}$.\n$1 - (a \\cos x + b \\sin x) \\geq A \\cos 2x + B \\sin 2x$.\nMinimum of LHS is $1 - \\sqrt{a^2+b^2}$. Maximum of RHS is $\\sqrt{A^2+B^2}$.\n$1 - \\sqrt{a^2+b^2} \\geq \\sqrt{A^2+B^2}$.\n\nConsider $g(x) = 1 - A \\cos 2x - B \\sin 2x \\geq a \\cos x + b \\sin x$.\nMinimum of LHS is $1 - \\sqrt{A^2+B^2}$. Maximum of RHS is $\\sqrt{a^2+b^2}$.\n$1 - \\sqrt{A^2+B^2} \\geq \\sqrt{a^2+b^2}$.\n\nLet $x=0$, $1-a-A \\geq 0$.\nLet $x=\\pi$, $1+a-A \\geq 0$.\nLet $x=\\pi/2$, $1-b+A \\geq 0$.\nLet $x=3\\pi/2$, $1+b+A \\geq 0$.\n\nConsider the case where $b=B=0$.\n$f(x) = 1 - a \\cos x - A \\cos 2x \\geq 0$.\n$1 - a \\cos x - A (2 \\cos^2 x - 1) \\geq 0$.\n$1 - a \\cos x - 2A \\cos^2 x + A \\geq 0$.\n$A+1 - a \\cos x - 2A \\cos^2 x \\geq 0$.\nLet $y = \\cos x$, $-1 \\leq y \\leq 1$.\n$g(y) = -2Ay^2 - ay + A+1 \\geq 0$ for $y \\in [-1, 1]$.\nIf $A=0$, $-ay+1 \\geq 0$ for $y \\in [-1, 1]$. If $a\u003e0$, $-a+1 \\geq 0$, $a \\leq 1$. If $a\u003c0$, $a+1 \\geq 0$, $a \\geq -1$. So $|a| \\leq 1$. $a^2 \\leq 1 \\leq 2$.\nIf $A\u003e0$, the parabola opens downwards. We need $g(-1) = -2A+a+A+1 = -A+a+1 \\geq 0$, $a \\geq A-1$. And $g(1) = -2A-a+A+1 = -A-a+1 \\geq 0$, $a \\leq 1-A$. Also the vertex must be outside $[-1, 1]$ or the vertex is in $[-1, 1]$ and the value is non-negative. Vertex at $y = -(-a)/(2(-2A)) = -a/4A$. If $|-a/4A| \\leq 1$, then $a^2 \\leq 16A^2$. $g(-a/4A) = -2A (-a/4A)^2 - a(-a/4A) + A+1 = -2A a^2/16A^2 + a^2/4A + A+1 = -a^2/8A + a^2/4A + A+1 = a^2/8A + A+1 \\geq 0$.\nIf $A\u003c0$, the parabola opens upwards. We need $g(-1) \\geq 0$ and $g(1) \\geq 0$.\n$-A+a+1 \\geq 0$, $a \\geq A-1$.\n$-A-a+1 \\geq 0$, $a \\leq 1-A$.\n\nConsider integrals again.\n$\\int_0^{2\\pi} f(x) dx \\geq 0 \\implies 2\\pi \\geq 0$.\n$\\int_0^{2\\pi} f(x) \\cos x dx \\geq 0 \\implies -a \\pi \\geq 0 \\implies a \\leq 0$.\n$\\int_0^{2\\pi} f(x) \\sin x dx \\geq 0 \\implies -b \\pi \\geq 0 \\implies b \\leq 0$.\n$\\int_0^{2\\pi} f(x) \\cos 2x dx \\geq 0 \\implies -A \\pi \\geq 0 \\implies A \\leq 0$.\n$\\int_0^{2\\pi} f(x) \\sin 2x dx \\geq 0 \\implies -B \\pi \\geq 0 \\implies B \\leq 0$.\nThis is not true.\n\nConsider the vector $v = (1, \\cos x, \\sin x, \\cos 2x, \\sin 2x)$.\nThe condition is $v \\cdot c \\geq 0$, where $c = (1, -a, -b, -A, -B)$.\n\nConsider the case where $f(x) = (1 - \\alpha \\cos x - \\beta \\sin x)^2 \\geq 0$.\n$1 + \\alpha^2 \\cos^2 x + \\beta^2 \\sin^2 x - 2 \\alpha \\cos x - 2 \\beta \\sin x + 2 \\alpha \\beta \\cos x \\sin x \\geq 0$.\n$1 + \\frac{\\alpha^2+\\beta^2}{2} + \\frac{\\alpha^2-\\beta^2}{2} \\cos 2x - 2 \\alpha \\cos x - 2 \\beta \\sin x + \\alpha \\beta \\sin 2x \\geq 0$.\nCompare with $1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x \\geq 0$.\n$-a = -2\\alpha$, $-b = -2\\beta$, $-A = \\frac{\\alpha^2-\\beta^2}{2}$, $-B = \\alpha \\beta$.\n$a = 2\\alpha$, $b = 2\\beta$.\n$A = \\frac{\\beta^2-\\alpha^2}{2}$, $B = -\\alpha \\beta$.\n$a^2+b^2 = 4\\alpha^2+4\\beta^2$.\n$A^2+B^2 = \\frac{(\\beta^2-\\alpha^2)^2}{4} + \\alpha^2 \\beta^2 = \\frac{\\beta^4 - 2\\alpha^2 \\beta^2 + \\alpha^4 + 4\\alpha^2 \\beta^2}{4} = \\frac{(\\alpha^2+\\beta^2)^2}{4}$.\n\nConsider $f(x) \\geq 0$.\nConsider the Fourier series of $|f(x)| = f(x)$.\nThe coefficients must match.\n\nLet $z = e^{ix}$.\n$f(x) = 1 - \\frac{a}{2}(z+z^{-1}) - \\frac{b}{2i}(z-z^{-1}) - \\frac{A}{2}(z^2+z^{-2}) - \\frac{B}{2i}(z^2-z^{-2}) \\geq 0$.\n$2if(x) = 2i - ia(z+z^{-1}) + b(z-z^{-1}) - iA(z^2+z^{-2}) + B(z^2-z^{-2}) \\geq 0$.\n$2if(x) = 2i - i a z - i a z^{-1} + b z - b z^{-1} - i A z^2 - i A z^{-2} + B z^2 - B z^{-2} \\geq 0$.\n$2if(x) = 2i + (b-ia)z + (-b-ia)z^{-1} + (B-iA)z^2 + (-B-iA)z^{-2} \\geq 0$.\n\nConsider the integral $\\frac{1}{2\\pi} \\int_0^{2\\pi} f(x) e^{inx} dx$.\nFor $n=0$, $\\frac{1}{2\\pi} \\int_0^{2\\pi} f(x) dx = 1 \\geq 0$.\nFor $n=1$, $\\frac{1}{2\\pi} \\int_0^{2\\pi} f(x) (\\cos x + i \\sin x) dx = \\frac{1}{2\\pi} \\int_0^{2\\pi} (1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x) (\\cos x + i \\sin x) dx = -a/2 - ib/2$.\nSince $f(x) \\geq 0$, the coefficients of the Fourier series must satisfy certain conditions.\n\nConsider trigonometric polynomials that are non-negative.\nIf $P(\\theta) = \\sum_{k=-n}^n c_k e^{ik\\theta} \\geq 0$, then $c_{-k} = \\overline{c_k}$ and $c_0 \\geq 0$.\nIn our case, the coefficients are:\n$c_0 = 1$.\n$c_1 = -a/2 - ib/2$.\n$c_{-1} = -a/2 + ib/2$.\n$c_2 = -A/2 - iB/2$.\n$c_{-2} = -A/2 + iB/2$.\nWe need a condition on the coefficients such that the polynomial is non-negative.\n\nIf $1 - a \\cos x - b \\sin x \\geq 0$, then $1 \\geq \\sqrt{a^2+b^2}$.\nIf $1 - A \\cos 2x - B \\sin 2x \\geq 0$, then $1 \\geq \\sqrt{A^2+B^2}$.\n\nConsider $f(x) = |1 - \\alpha e^{ix} - \\beta e^{2ix}|^2 \\geq 0$.\n$(1 - \\alpha e^{ix} - \\beta e^{2ix})(1 - \\bar{\\alpha} e^{-ix} - \\bar{\\beta} e^{-2ix})$.\n$1 - \\bar{\\alpha} e^{-ix} - \\bar{\\beta} e^{-2ix} - \\alpha e^{ix} + |\\alpha|^2 - \\alpha \\bar{\\beta} e^{-ix} - \\beta e^{2ix} - \\beta \\bar{\\alpha} e^{ix} + |\\beta|^2$.\n$1 + |\\alpha|^2 + |\\beta|^2 - (\\bar{\\alpha} + \\alpha + \\alpha \\bar{\\beta}) e^{-ix} - (\\bar{\\beta} + \\beta + \\beta \\bar{\\alpha}) e^{-2ix}$.\n$1 + |\\alpha|^2 + |\\beta|^2 - (2 \\text{Re}(\\alpha) + \\alpha \\bar{\\beta}) e^{-ix} - (2 \\text{Re}(\\beta) + \\beta \\bar{\\alpha}) e^{-2ix}$.\nCompare coefficients.\n\nIf $P(\\theta) = |\\sum_{k=0}^n z_k e^{ik\\theta}|^2$, then $P(\\theta) \\geq 0$.\nConsider $P(x) = |1 - \\alpha e^{ix}|^2 = (1-\\alpha e^{ix})(1-\\bar{\\alpha} e^{-ix}) = 1 - \\bar{\\alpha} e^{-ix} - \\alpha e^{ix} + |\\alpha|^2 = 1+|\\alpha|^2 - 2 \\text{Re}(\\alpha) \\cos x + 2 \\text{Im}(\\alpha) \\sin x$.\nCompare with $1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x$.\n\nConsider $f(x) \\geq 0$.\n$1 - (a \\cos x + b \\sin x) - (A \\cos 2x + B \\sin 2x) \\geq 0$.\nLet $a = r_1 \\cos \\phi_1, b = r_1 \\sin \\phi_1$.\nLet $A = r_2 \\cos \\phi_2, B = r_2 \\sin \\phi_2$.\n$1 - r_1 \\cos(x-\\phi_1) - r_2 \\cos(2x-\\phi_2) \\geq 0$.\nLet $x - \\phi_1 = y$. Then $2x - \\phi_2 = 2(y+\\phi_1) - \\phi_2 = 2y + 2\\phi_1 - \\phi_2$. Let $\\delta = 2\\phi_1 - \\phi_2$.\n$1 - r_1 \\cos y - r_2 \\cos(2y+\\delta) \\geq 0$.\n$1 - r_1 \\cos y - r_2 (\\cos 2y \\cos \\delta - \\sin 2y \\sin \\delta) \\geq 0$.\n$1 - r_1 \\cos y - r_2 ( (2\\cos^2 y - 1) \\cos \\delta - 2 \\sin y \\cos y \\sin \\delta ) \\geq 0$.\n$1 - r_1 \\cos y - 2 r_2 \\cos^2 y \\cos \\delta + r_2 \\cos \\delta + 2 r_2 \\sin y \\cos y \\sin \\delta \\geq 0$.\n$1 + r_2 \\cos \\delta - r_1 \\cos y - 2 r_2 \\cos \\delta \\cos^2 y + r_2 \\sin \\delta \\sin 2y \\geq 0$.\n\nIf $f(x) \\geq 0$, then the average value is non-negative.\nConsider $1 - a \\cos x - b \\sin x \\geq 0$. Square both sides? No.\nConsider $1 - A \\cos 2x - B \\sin 2x \\geq |a \\cos x + b \\sin x|$.\n$1 - \\sqrt{A^2+B^2} \\leq 1 - A \\cos 2x - B \\sin 2x \\leq 1 + \\sqrt{A^2+B^2}$.\n$|a \\cos x + b \\sin x| \\leq \\sqrt{a^2+b^2}$.\nSo $1 - \\sqrt{A^2+B^2} \\geq \\sqrt{a^2+b^2}$. This is not necessarily true.\n\nConsider $f(x) \\geq 0$.\n$\\int_0^{2\\pi} f(x) d x = 2\\pi$.\n$\\int_0^{2\\pi} f(x) \\cos x d x = -a\\pi$.\n$\\int_0^{2\\pi} f(x) \\sin x d x = -b\\pi$.\n$\\int_0^{2\\pi} f(x) \\cos 2x d x = -A\\pi$.\n$\\int_0^{2\\pi} f(x) \\sin 2x d x = -B\\pi$.\nSince $f(x) \\geq 0$, consider the Fejer-Riesz theorem. A non-negative trigonometric polynomial can be written as the squared modulus of another trigonometric polynomial.\n$1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x = |c_0 + c_1 e^{ix} + c_2 e^{2ix}|^2$.\n$= (c_0 + c_1 e^{ix} + c_2 e^{2ix})(\\bar{c_0} + \\bar{c_1} e^{-ix} + \\bar{c_2} e^{-2ix})$.\n$= |c_0|^2 + |c_1|^2 + |c_2|^2 + (c_0 \\bar{c_1} + c_1 \\bar{c_2}) e^{ix} + (\\bar{c_0} c_1 + \\bar{c_1} c_2) e^{-ix} + c_0 \\bar{c_2} e^{2ix} + \\bar{c_0} c_2 e^{-2ix}$.\n$= |c_0|^2 + |c_1|^2 + |c_2|^2 + 2 \\text{Re}(c_0 \\bar{c_1} + c_1 \\bar{c_2}) \\cos x - 2 \\text{Im}(c_0 \\bar{c_1} + c_1 \\bar{c_2}) \\sin x + 2 \\text{Re}(c_0 \\bar{c_2}) \\cos 2x - 2 \\text{Im}(c_0 \\bar{c_2}) \\sin 2x$.\nCompare coefficients:\n$1 = |c_0|^2 + |c_1|^2 + |c_2|^2$.\n$-a = 2 \\text{Re}(c_0 \\bar{c_1} + c_1 \\bar{c_2})$.\n$-b = -2 \\text{Im}(c_0 \\bar{c_1} + c_1 \\bar{c_2})$.\n$-A = 2 \\text{Re}(c_0 \\bar{c_2})$.\n$-B = -2 \\text{Im}(c_0 \\bar{c_2})$.\n$A^2+B^2 = 4 (\\text{Re}(c_0 \\bar{c_2})^2 + \\text{Im}(c_0 \\bar{c_2})^2) = 4 |c_0 \\bar{c_2}|^2 = 4 |c_0|^2 |c_2|^2$.\nLet $c_0 = 1$.\n$1 = 1 + |c_1|^2 + |c_2|^2 \\implies |c_1|^2 + |c_2|^2 = 0 \\implies c_1 = c_2 = 0$. Then $f(x) = 1$, $a=b=A=B=0$.\n]"
}
```

This subset can be loaded as:

```python
from datasets import load_dataset

ds = load_dataset("Rickxz06/s1K-deepseek-r1-qwen-14b-test", "default")
```

Or simply as it follows, since there's only one configuration and is named `default`: 

```python
from datasets import load_dataset

ds = load_dataset("Rickxz06/s1K-deepseek-r1-qwen-14b-test")
```


</details>




