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: 561335
    num_examples: 10
  download_size: 265954
  dataset_size: 561335
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"/>
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# Dataset Card for s1K-deepseek-r1-qwen-7b-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-7b-test/raw/main/pipeline.yaml"
```

or explore the configuration:

```console
distilabel pipeline info --config "https://huggingface.co/datasets/Rickxz06/s1K-deepseek-r1-qwen-7b-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 need to prove that if the function \\( f(x) = 1 - a \\cdot \\cos(x) - b \\cdot \\sin(x) - A \\cdot \\cos(2x) - B \\cdot \\sin(2x) \\) is non-negative for any real number \\( x \\), then two inequalities must hold: \\( a^2 + b^2 \\leq 2 \\) and \\( A^2 + B^2 \\leq 1 \\). \n\nHmm, let me start by understanding what the function \\( f(x) \\) looks like. It seems to be a combination of cosine and sine functions with different frequencies\u2014specifically, \\( \\cos(x) \\), \\( \\sin(x) \\), \\( \\cos(2x) \\), and \\( \\sin(2x) \\). The function is being subtracted from 1 and is supposed to be non-negative for all \\( x \\). So, my goal is to find constraints on the coefficients \\( a, b, A, B \\) such that this condition holds.\n\nI remember that when dealing with expressions involving sine and cosine, especially when they are linear combinations, it\u0027s often useful to express them as a single sine or cosine function with a phase shift. This might help in finding the maximum or minimum values of the function, which is crucial here because \\( f(x) \\geq 0 \\) for all \\( x \\) implies that the minimum value of \\( f(x) \\) is at least 0.\n\nLet me try to rewrite the terms involving \\( \\cos(x) \\) and \\( \\sin(x) \\) first. The expression \\( -a \\cos(x) - b \\sin(x) \\) can be rewritten as \\( R \\cos(x + \\phi) \\) where \\( R = \\sqrt{a^2 + b^2} \\) and \\( \\phi \\) is some phase shift. Similarly, the terms \\( -A \\cos(2x) - B \\sin(2x) \\) can be rewritten as \\( S \\cos(2x + \\theta) \\) where \\( S = \\sqrt{A^2 + B^2} \\).\n\nSo, substituting these, the function becomes:\n\\[ f(x) = 1 - R \\cos(x + \\phi) - S \\cos(2x + \\theta) \\]\n\nBut I\u0027m not sure if this substitution is helpful yet. Maybe instead of trying to combine the terms, I should think about the maximum and minimum values of the function. Since \\( f(x) \\geq 0 \\) for all \\( x \\), the minimum value must be greater than or equal to zero.\n\nAnother idea is to represent this function using complex exponentials, which might make the analysis easier. Let me recall Euler\u0027s formula: \\( e^{i\\theta} = \\cos\\theta + i \\sin\\theta \\). Maybe I can express \\( f(x) \\) as the real part of some complex expression.\n\nLet me try that. Let\u0027s consider the terms involving \\( \\cos(x) \\) and \\( \\sin(x) \\) as the real part of \\( (a - ib)e^{ix} \\), since:\n\\[ \\text{Re}[(a - ib)e^{ix}] = a \\cos x + b \\sin x \\]\nBut in our function \\( f(x) \\), we have \\( -a \\cos x - b \\sin x \\), so that would be \\( - \\text{Re}[(a - ib)e^{ix}] \\).\n\nSimilarly, the terms \\( -A \\cos(2x) - B \\sin(2x) \\) can be written as \\( - \\text{Re}[(A - iB)e^{i2x}] \\).\n\nSo, putting it all together, perhaps \\( f(x) \\) can be expressed as:\n\\[ f(x) = 1 - \\text{Re}[(a - ib)e^{ix}] - \\text{Re}[(A - iB)e^{i2x}] \\]\n\nWhich can be rewritten as:\n\\[ f(x) = 1 - \\text{Re}\\left[ (a - ib)e^{ix} + (A - iB)e^{i2x} \\right] \\]\n\nHmm, that might not directly help, but it\u0027s an interesting representation. Maybe I can think of \\( f(x) \\) as the real part of some complex function, but I\u0027m not sure how to proceed with that yet.\n\nAnother approach is to consider that \\( f(x) \\) is a continuous function and since it\u0027s non-negative everywhere, its minimum must be zero or positive. To find the minimum, I can take the derivative of \\( f(x) \\) with respect to \\( x \\) and set it to zero, then check the critical points. But that might get complicated because \\( f(x) \\) involves multiple cosine and sine terms with different frequencies.\n\nWait, maybe instead of taking derivatives, I can use the fact that the minimum of \\( f(x) \\) must be non-negative. To find the minimum, I can express \\( f(x) \\) in terms of multiple angles and then find its minimum value.\n\nAlternatively, I remember that for a function of the form \\( f(x) = C + D \\cos x + E \\cos 2x + F \\sin x + G \\sin 2x \\), the minimum can be found by considering the maximum of the negative part. But I\u0027m not sure.\n\nAlternatively, maybe I can use the method of completing the square or express \\( f(x) \\) as a sum of squares, ensuring that it\u0027s always non-negative.\n\nWait, another thought: since \\( f(x) \\) must be non-negative for all \\( x \\), perhaps I can express it as a sum of squares of sine and cosine functions, each multiplied by some coefficients, ensuring that the overall expression is always non-negative.\n\nBut I\u0027m not sure how to directly apply that here. Maybe another approach is to use Fourier series or orthogonal functions. Since the function is expressed in terms of cosines and sines of different frequencies, perhaps I can use orthogonality to find constraints on the coefficients.\n\nBut let me think differently. Since \\( f(x) \\geq 0 \\) for all \\( x \\), it must be that the function doesn\u0027t dip below zero. Therefore, the minimum value of \\( f(x) \\) is zero or positive. Maybe I can compute the minimum value of \\( f(x) \\) and set it to be greater than or equal to zero, which will give me some inequalities on \\( a, b, A, B \\).\n\nTo compute the minimum, I can take the derivative of \\( f(x) \\) with respect to \\( x \\), set it equal to zero, and solve for \\( x \\). Then, substitute back into \\( f(x) \\) to find the minimum value. But this might be complicated because \\( f(x) \\) has multiple terms with different frequencies.\n\nLet me compute the derivative:\n\\[ f\u0027(x) = a \\sin x + b \\cos x + 2A \\sin 2x + 2B \\cos 2x \\]\n\nSet this equal to zero:\n\\[ a \\sin x + b \\cos x + 2A \\sin 2x + 2B \\cos 2x = 0 \\]\n\nThis equation is going to be hard to solve analytically because of the different frequencies. Maybe instead, a better approach is to express \\( f(x) \\) in terms of multiple angles and then use the fact that the minimum is non-negative.\n\nLet me try to express \\( f(x) \\) as a combination of multiple-angle terms. Starting with \\( f(x) \\):\n\\[ f(x) = 1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x \\]\n\nI know that \\( \\cos 2x = 2 \\cos^2 x - 1 \\) and \\( \\sin 2x = 2 \\sin x \\cos x \\). Maybe substituting these into the function will help. Let\u0027s try that.\n\nSubstituting \\( \\cos 2x = 2 \\cos^2 x - 1 \\) and \\( \\sin 2x = 2 \\sin x \\cos x \\), we get:\n\\[ f(x) = 1 - a \\cos x - b \\sin x - A (2 \\cos^2 x - 1) - B (2 \\sin x \\cos x) \\]\nSimplify this:\n\\[ f(x) = 1 - a \\cos x - b \\sin x - 2A \\cos^2 x + A - 2B \\sin x \\cos x \\]\nCombine constants:\n\\[ f(x) = (1 + A) - a \\cos x - b \\sin x - 2A \\cos^2 x - 2B \\sin x \\cos x \\]\n\nHmm, this still seems complicated, but perhaps I can collect like terms. Let\u0027s group the \\( \\cos x \\) terms and the \\( \\sin x \\) terms:\n\n\\[ f(x) = (1 + A) - (a + 2B \\sin x) \\cos x - b \\sin x - 2A \\cos^2 x \\]\n\nWait, that might not be helpful. Maybe another substitution. Let me express everything in terms of \\( \\cos x \\) and \\( \\sin x \\). Let me set \\( y = \\cos x \\) and \\( z = \\sin x \\). Then, \\( y^2 + z^2 = 1 \\).\n\nSubstituting, we have:\n\\[ f(x) = 1 - a y - b z - A (2 y^2 - 1) - B (2 z y) \\]\nSimplify:\n\\[ f(x) = 1 - a y - b z - 2 A y^2 + A - 2 B z y \\]\nCombine constants:\n\\[ f(x) = (1 + A) - a y - b z - 2 A y^2 - 2 B z y \\]\n\nSo, \\( f(x) \\) is a function of \\( y \\) and \\( z \\) with the constraint \\( y^2 + z^2 = 1 \\). So, perhaps I can consider \\( f \\) as a function on the unit circle and find its minimum.\n\nLet me write \\( f \\) in terms of \\( y \\) and \\( z \\):\n\\[ f(y, z) = (1 + A) - a y - b z - 2 A y^2 - 2 B z y \\]\nwith \\( y^2 + z^2 = 1 \\).\n\nThis is a quadratic function in \\( y \\) and \\( z \\). To find its minimum, I can use the method of Lagrange multipliers, considering the constraint \\( y^2 + z^2 = 1 \\).\n\nAlternatively, maybe I can write this expression in matrix form. Let\u0027s see.\n\nLet me rewrite \\( f(y, z) \\) as:\n\\[ f(y, z) = (1 + A) - (a + 2 B z) y - b z - 2 A y^2 \\]\n\nHmm, this is a quadratic in \\( y \\) with coefficients depending on \\( z \\).\n\nAlternatively, perhaps I can arrange the terms as a quadratic form. Let me try:\n\nLet me consider the variables \\( y \\) and \\( z \\). Then:\n\\[ f(y, z) = (1 + A) - a y - b z - 2 A y^2 - 2 B y z \\]\n\nThis can be written as:\n\\[ f(y, z) = (1 + A) - \\left( a y + b z + 2 A y^2 + 2 B y z \\right) \\]\n\nHmm, maybe I can write this as a vector product. Let me think of the terms involving \\( y \\) and \\( z \\) as a vector \\( [y, z] \\) multiplied by some matrix.\n\nLet me denote the quadratic form as:\n\\[ f(y, z) = (1 + A) - \\left( \\begin{bmatrix} a \u0026 b \\end{bmatrix} \\begin{bmatrix} y \\\\ z \\end{bmatrix} + 2 A y^2 + 2 B y z \\right) \\]\n\nSo, the quadratic part is \\( 2 A y^2 + 2 B y z \\). The linear part is \\( a y + b z \\).\n\nSo, combining these, the quadratic form is:\n\\[ f(y, z) = (1 + A) - \\begin{bmatrix} a \u0026 b \\end{bmatrix} \\begin{bmatrix} y \\\\ z \\end{bmatrix} - 2 \\begin{bmatrix} A \u0026 B/2 \\\\ B/2 \u0026 0 \\end{bmatrix} \\begin{bmatrix} y \\\\ z \\end{bmatrix} \\]\n\nWait, not sure if that\u0027s correct. Let me see. The quadratic terms are \\( 2 A y^2 + 2 B y z \\). So, the corresponding matrix would have \\( A \\) on the diagonal for \\( y^2 \\) and \\( B/2 \\) for the cross term.\n\nSo, the quadratic form matrix is:\n\\[ \\begin{bmatrix} 2A \u0026 B \\\\ B \u0026 0 \\end{bmatrix} \\]\n\nBut actually, the standard quadratic form is \\( \\frac{1}{2} \\mathbf{v}^T Q \\mathbf{v} \\), where \\( Q \\) is symmetric. So, perhaps I need to adjust for that.\n\nBut maybe it\u0027s complicating things. Let me think differently. Since \\( y^2 + z^2 = 1 \\), I can parameterize \\( y \\) and \\( z \\) as \\( y = \\cos \\theta \\), \\( z = \\sin \\theta \\) for some angle \\( \\theta \\). Then, substitute into \\( f(y, z) \\):\n\nSo, substituting \\( y = \\cos \\theta \\), \\( z = \\sin \\theta \\), we have:\n\\[ f(\\theta) = (1 + A) - a \\cos \\theta - b \\sin \\theta - 2 A \\cos^2 \\theta - 2 B \\sin \\theta \\cos \\theta \\]\n\nSimplify \\( f(\\theta) \\):\nLet me write \\( \\cos^2 \\theta = \\frac{1 + \\cos 2\\theta}{2} \\) and \\( \\sin \\theta \\cos \\theta = \\frac{\\sin 2\\theta}{2} \\). So, substituting these:\n\n\\[ f(\\theta) = (1 + A) - a \\cos \\theta - b \\sin \\theta - 2 A \\left( \\frac{1 + \\cos 2\\theta}{2} \\right) - 2 B \\left( \\frac{\\sin 2\\theta}{2} \\right) \\]\nSimplify:\n\\[ f(\\theta) = (1 + A) - a \\cos \\theta - b \\sin \\theta - A (1 + \\cos 2\\theta) - B \\sin 2\\theta \\]\nExpand:\n\\[ f(\\theta) = 1 + A - a \\cos \\theta - b \\sin \\theta - A - A \\cos 2\\theta - B \\sin 2\\theta \\]\nSimplify constants:\n\\[ f(\\theta) = 1 - a \\cos \\theta - b \\sin \\theta - A \\cos 2\\theta - B \\sin 2\\theta \\]\n\nWait, that\u0027s just the original \\( f(x) \\)! So, maybe this substitution didn\u0027t help. Hmm, perhaps I need a different approach.\n\nWait, if I set \\( f(x) \\geq 0 \\) for all \\( x \\), then in particular, it must be non-negative at its minimum. So, perhaps I can compute the minimum of \\( f(x) \\) and set it to be greater than or equal to zero, which will give me some conditions on \\( a, b, A, B \\).\n\nBut as I thought earlier, computing the minimum requires taking the derivative and solving for critical points, which might be complicated.\n\nAlternatively, maybe I can express \\( f(x) \\) as a sum of squares, ensuring it\u0027s always non-negative. Let me think if that\u0027s possible.\n\nGiven that \\( f(x) = 1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x \\), perhaps I can write this as \\( 1 - (a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x) \\). So, if I can bound the maximum value of \\( a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x \\), then I can ensure that the minimum of \\( f(x) \\) is non-negative.\n\nIn other words, if \\( |a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x| \\leq 1 \\) for all \\( x \\), then \\( f(x) \\geq 0 \\). But that\u0027s not exactly the case here because the function is \\( 1 - \\) that expression, so it\u0027s non-negative if the expression is less than or equal to 1 everywhere.\n\nWait, actually, no. If \\( a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x \\leq 1 \\) for all \\( x \\), then \\( f(x) \\geq 0 \\). But the problem only states that \\( f(x) \\geq 0 \\), not necessarily that the expression is less than or equal to 1. So, perhaps it\u0027s a bit more nuanced.\n\nBut maybe the maximum of \\( a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x \\) is less than or equal to 1. Then, subtracting from 1 would make \\( f(x) \\geq 0 \\). So, if I can show that the maximum of this expression is at most 1, then \\( f(x) \\geq 0 \\) for all \\( x \\).\n\nBut the problem states that \\( f(x) \\geq 0 \\) for all \\( x \\), so perhaps that\u0027s the way to go. Let me denote \\( g(x) = a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x \\). Then, \\( f(x) = 1 - g(x) \\). So, \\( f(x) \\geq 0 \\) implies that \\( g(x) \\leq 1 \\) for all \\( x \\).\n\nSo, if \\( g(x) \\leq 1 \\) for all \\( x \\), then \\( f(x) \\geq 0 \\). Therefore, we can rephrase the problem as: if \\( g(x) \\leq 1 \\) for all \\( x \\), then prove that \\( a^2 + b^2 \\leq 2 \\) and \\( A^2 + B^2 \\leq 1 \\).\n\nBut wait, is this necessarily true? If \\( g(x) \\leq 1 \\) everywhere, then certainly the maximum of \\( g(x) \\) is 1. So, the maximum of \\( g(x) \\) is 1, which will impose some constraints on \\( a, b, A, B \\).\n\nI recall that the maximum of a function like \\( C \\cos x + D \\sin x \\) is \\( \\sqrt{C^2 + D^2} \\). So, perhaps I can bound \\( g(x) \\) by using the fact that the maximum of \\( a \\cos x + b \\sin x \\) is \\( \\sqrt{a^2 + b^2} \\) and the maximum of \\( A \\cos 2x + B \\sin 2x \\) is \\( \\sqrt{A^2 + B^2} \\). So, the maximum of \\( g(x) \\) would be at most \\( \\sqrt{a^2 + b^2} + \\sqrt{A^2 + B^2} \\).\n\nBut wait, that\u0027s not necessarily true because \\( g(x) \\) combines both frequencies. So, the maximum of the sum isn\u0027t necessarily the sum of the maxima. For example, if the two functions are in phase, their maxima add up, but if they are out of phase, they might cancel each other.\n\nBut since \\( g(x) \\leq 1 \\) for all \\( x \\), its maximum must be less than or equal to 1. Therefore, the maximum of \\( g(x) \\) is at most 1.\n\nTo find the maximum of \\( g(x) \\), perhaps I can write it as a sum of two sinusoids with different frequencies and then find its maximum.\n\nAlternatively, another method is to express \\( g(x) \\) as a single sinusoid using harmonic addition, but since it has two different frequencies, it\u0027s not straightforward.\n\nWait, maybe I can consider that \\( g(x) \\) can be written as a function with Fourier coefficients \\( a, b, A, B \\), so perhaps using the concept of the maximum modulus.\n\nAlternatively, another idea: if \\( g(x) \\leq 1 \\) for all \\( x \\), then in particular, at specific points, like \\( x = 0, \\pi/2, \\pi, 3\\pi/2 \\), etc., we can get some inequalities on \\( a, b, A, B \\).\n\nLet me compute \\( g(0) \\), \\( g(\\pi/2) \\), \\( g(\\pi) \\), \\( g(3\\pi/2) \\), and so on, and set up inequalities.\n\nCompute \\( g(0) \\):\n\\[ g(0) = a \\cos 0 + b \\sin 0 + A \\cos 0 + B \\sin 0 = a + A \\]\nSince \\( g(0) \\leq 1 \\), we have:\n\\[ a + A \\leq 1 \\]\n\nSimilarly, compute \\( g(\\pi/2) \\):\n\\[ g(\\pi/2) = a \\cos (\\pi/2) + b \\sin (\\pi/2) + A \\cos \\pi + B \\sin \\pi = 0 + b + (-A) + 0 = b - A \\]\nSince \\( g(\\pi/2) \\leq 1 \\), we have:\n\\[ b - A \\leq 1 \\]\n\nCompute \\( g(\\pi) \\):\n\\[ g(\\pi) = a \\cos \\pi + b \\sin \\pi + A \\cos 2\\pi + B \\sin 2\\pi = -a + 0 + A + 0 = -a + A \\]\nSince \\( g(\\pi) \\leq 1 \\), we have:\n\\[ -a + A \\leq 1 \\]\n\nCompute \\( g(3\\pi/2) \\):\n\\[ g(3\\pi/2) = a \\cos (3\\pi/2) + b \\sin (3\\pi/2) + A \\cos 3\\pi + B \\sin 3\\pi = 0 - b + (-A) + 0 = -b - A \\]\nSince \\( g(3\\pi/2) \\leq 1 \\), we have:\n\\[ -b - A \\leq 1 \\]\n\nSo, from these four points, we get the following inequalities:\n1. \\( a + A \\leq 1 \\)\n2. \\( b - A \\leq 1 \\)\n3. \\( -a + A \\leq 1 \\)\n4. \\( -b - A \\leq 1 \\)\n\nThese can be rewritten as:\n1. \\( a + A \\leq 1 \\)\n2. \\( b - A \\leq 1 \\)\n3. \\( -a + A \\leq 1 \\)\n4. \\( -b - A \\leq 1 \\)\n\nBut these inequalities don\u0027t directly give us \\( a^2 + b^2 \\leq 2 \\) and \\( A^2 + B^2 \\leq 1 \\). So, perhaps this approach is not sufficient.\n\nAlternatively, maybe I should use the fact that \\( f(x) \\geq 0 \\) for all \\( x \\), so the minimal value of \\( f(x) \\) is non-negative. To compute the minimal value, we can use calculus. Let\u0027s try that.\n\nFirst, compute the derivative of \\( f(x) \\):\n\\[ f\u0027(x) = -a \\sin x - b \\cos x - 2A \\sin 2x - 2B \\cos 2x \\]\n\nSet \\( f\u0027(x) = 0 \\):\n\\[ -a \\sin x - b \\cos x - 2A \\sin 2x - 2B \\cos 2x = 0 \\]\n\nThis equation is quite complex because of the different sine and cosine terms with different arguments. Solving this analytically might be difficult. Maybe instead, I can express this equation in terms of \\( \\sin x \\) and \\( \\cos x \\), and then use substitution.\n\nLet me recall that \\( \\sin 2x = 2 \\sin x \\cos x \\) and \\( \\cos 2x = 2 \\cos^2 x - 1 \\). Substituting these into the derivative equation:\n\n\\[ -a \\sin x - b \\cos x - 2A (2 \\sin x \\cos x) - 2B (2 \\cos^2 x - 1) = 0 \\]\nSimplify:\n\\[ -a \\sin x - b \\cos x - 4A \\sin x \\cos x - 4B \\cos^2 x + 2B = 0 \\]\n\nLet me rearrange terms:\n\\[ (-a - 4A \\cos x) \\sin x + (-b) \\cos x - 4B \\cos^2 x + 2B = 0 \\]\n\nHmm, this still looks complicated. Maybe another substitution. Let me set \\( t = \\cos x \\), then \\( \\sin x = \\sqrt{1 - t^2} \\) (assuming \\( \\sin x \\) is positive; we might need to consider negative as well, but let\u0027s proceed for now).\n\nSo, substituting \\( t = \\cos x \\) and \\( \\sin x = \\sqrt{1 - t^2} \\), we get:\n\\[ (-a - 4A t) \\sqrt{1 - t^2} - b t - 4B t^2 + 2B = 0 \\]\n\nThis is a complicated equation in \\( t \\). It might not be feasible to solve this analytically. Maybe another approach is needed.\n\nWait, perhaps instead of calculus, I can use the concept of positive semi-definite functions. Since \\( f(x) \\geq 0 \\) for all \\( x \\), it\u0027s a positive semi-definite function. Maybe I can express \\( f(x) \\) as a sum of squares, which would ensure it\u0027s non-negative.\n\nAlternatively, thinking back to the expression \\( f(x) = 1 - g(x) \\), since \\( f(x) \\geq 0 \\), then \\( g(x) \\leq 1 \\) for all \\( x \\). So, the maximum of \\( g(x) \\) is at most 1. Therefore, if I can bound the maximum of \\( g(x) \\) in terms of \\( a, b, A, B \\), then I can get the required inequalities.\n\nI remember that for a function like \\( g(x) = a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x \\), the maximum can be bounded using the sum of the amplitudes. But because of the different frequencies, the exact maximum isn\u0027t straightforward.\n\nAlternatively, perhaps I can use the Cauchy-Schwarz inequality. Let me consider \\( g(x) \\) as the sum of two orthogonal functions: \\( h(x) = a \\cos x + b \\sin x \\) and \\( k(x) = A \\cos 2x + B \\sin 2x \\).\n\nSince \\( h(x) \\) and \\( k(x) \\) are orthogonal over the interval \\( [0, 2\\pi] \\), the maximum of \\( g(x) \\) can be bounded by the sum of the maxima of \\( h(x) \\) and \\( k(x) \\). But actually, orthogonality doesn\u0027t directly give the maximum, but rather the integral of their product is zero.\n\nWait, but if they are orthogonal, then the maximum of their sum isn\u0027t necessarily the sum of their maxima. Hmm, maybe that approach isn\u0027t helpful.\n\nWait, another idea: using the fact that for any \\( x \\), \\( |\\cos 2x| \\leq 1 \\) and \\( |\\sin 2x| \\leq 1 \\), similarly for the single angle terms. So, \\( |g(x)| \\leq |a| + |b| + |A| + |B| \\). But this gives a rough bound, not necessarily tight.\n\nBut since \\( g(x) \\leq 1 \\) for all \\( x \\), this implies that \\( |a| + |b| + |A| + |B| \\leq 1 \\). But this is much stronger than what we need to prove, so maybe it\u0027s not useful.\n\nAlternatively, perhaps I can write \\( g(x) \\) as the sum of two functions, each with their own frequency, and then use the fact that the maximum of the sum is bounded by the sum of the maxima.\n\nBut, in reality, because of the orthogonality and the different frequencies, the maximum of the sum isn\u0027t just the sum of the individual maxima. It might be less due to possible cancellation.\n\nWait, maybe a better approach is to compute the maximum of \\( g(x) \\) by considering it as a function with multiple frequencies. For that, perhaps I can use the concept of the maximum of a trigonometric polynomial.\n\nI recall that for a trigonometric polynomial \\( P(x) = \\sum_{k=-n}^{n} c_k e^{ikx} \\), the maximum modulus can be bounded, but again, not sure how to apply that here.\n\nAlternatively, perhaps using the fact that \\( g(x) \\) is a combination of sinusoids with different frequencies, its maximum can be found by considering the envelope of these sinusoids. But that\u0027s vague.\n\nWait, maybe another approach. Let me consider specific points where \\( f(x) \\) could achieve its minimum. For example, at \\( x = 0 \\), \\( x = \\pi/2 \\), \\( x = \\pi \\), \\( x = 3\\pi/2 \\), and maybe some other points where the derivative is zero.\n\nEarlier, I found that:\n- \\( f(0) = 1 - a - A \\)\n- \\( f(\\pi/2) = 1 - b - B \\)\n- \\( f(\\pi) = 1 + a - A \\)\n- \\( f(3\\pi/2) = 1 + b - B \\)\n\nBut to get the minimum, maybe we need to evaluate \\( f(x) \\) at more points or find critical points.\n\nAlternatively, perhaps I can use the fact that \\( f(x) \\geq 0 \\) for all \\( x \\), so in particular, the integral of \\( f(x) \\) over a period must be non-negative. But actually, since \\( f(x) \\geq 0 \\), the integral is non-negative, but it\u0027s always non-negative regardless.\n\nWait, maybe integrating \\( f(x) \\) over a period and using orthogonality can give some information. Let me try that.\n\nCompute the integral of \\( f(x) \\) over \\( [0, 2\\pi] \\):\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 \\]\nCompute each integral separately:\n- \\( \\int_{0}^{2\\pi} 1 dx = 2\\pi \\)\n- \\( \\int_{0}^{2\\pi} \\cos x dx = 0 \\)\n- \\( \\int_{0}^{2\\pi} \\sin x dx = 0 \\)\n- \\( \\int_{0}^{2\\pi} \\cos 2x dx = 0 \\)\n- \\( \\int_{0}^{2\\pi} \\sin 2x dx = 0 \\)\n\nSo, the integral is \\( 2\\pi \\). Since \\( f(x) \\geq 0 \\), the average value is \\( \\pi \\), which is positive, but that doesn\u0027t give me the necessary constraints.\n\nWait, perhaps integrating \\( f(x)^2 \\) over a period would give me something. Since \\( f(x) \\geq 0 \\), \\( f(x)^2 \\) would be non-negative, and maybe I can use some inequalities.\n\nCompute the integral of \\( f(x)^2 \\):\n\\[ \\int_{0}^{2\\pi} f(x)^2 dx = \\int_{0}^{2\\pi} [1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x]^2 dx \\]\n\nExpanding the square:\n\\[ \\int_{0}^{2\\pi} [1 - 2a \\cos x - 2b \\sin x - 2A \\cos 2x - 2B \\sin 2x + a^2 \\cos^2 x + b^2 \\sin^2 x + A^2 \\cos^2 2x + B^2 \\sin^2 2x + 2ab \\cos x \\sin x + 2aA \\cos x \\cos 2x + 2aB \\cos x \\sin 2x + 2bA \\sin x \\cos 2x + 2bB \\sin x \\sin 2x + 2AB \\cos 2x \\sin 2x] dx \\]\n\nNow, integrating term by term:\n\n1. \\( \\int 1 dx = 2\\pi \\)\n2. \\( -2a \\int \\cos x dx = 0 \\)\n3. \\( -2b \\int \\sin x dx = 0 \\)\n4. \\( -2A \\int \\cos 2x dx = 0 \\)\n5. \\( -2B \\int \\sin 2x dx = 0 \\)\n6. \\( a^2 \\int \\cos^2 x dx = a^2 \\pi \\)\n7. \\( b^2 \\int \\sin^2 x dx = b^2 \\pi \\)\n8. \\( A^2 \\int \\cos^2 2x dx = A^2 \\pi \\)\n9. \\( B^2 \\int \\sin^2 2x dx = B^2 \\pi \\)\n10. \\( 2ab \\int \\cos x \\sin x dx = 0 \\) (since it\u0027s an odd function over a symmetric interval)\n11. \\( 2aA \\int \\cos x \\cos 2x dx = 0 \\) (orthogonal over the interval)\n12. \\( 2aB \\int \\cos x \\sin 2x dx = 0 \\) (orthogonal)\n13. \\( 2bA \\int \\sin x \\cos 2x dx = 0 \\) (orthogonal)\n14. \\( 2bB \\int \\sin x \\sin 2x dx = 0 \\) (orthogonal)\n15. \\( 2AB \\int \\cos 2x \\sin 2x dx = 0 \\) (orthogonal)\n\nSo, putting it all together, the integral becomes:\n\\[ 2\\pi + a^2 \\pi + b^2 \\pi + A^2 \\pi + B^2 \\pi \\]\nWhich simplifies to:\n\\[ \\pi (2 + a^2 + b^2 + A^2 + B^2) \\]\n\nSince \\( f(x) \\geq 0 \\) for all \\( x \\), \\( f(x)^2 \\geq 0 \\) as well. But this integral is just a positive quantity. How does this help me?\n\nWait, another thought. Maybe using the fact that the minimum of \\( f(x) \\) is non-negative, so the integral can also be expressed in terms of the minimum value. But I don\u0027t see an immediate connection.\n\nAlternatively, perhaps I can use the Cauchy-Schwarz inequality on the integral of \\( f(x)^2 \\). The integral is equal to \\( \\pi (2 + a^2 + b^2 + A^2 + B^2) \\). But since \\( f(x) \\geq 0 \\), the integral is at least the square of the integral of \\( f(x) \\) divided by the period, by Cauchy-Schwarz. Wait, let me recall the inequality.\n\nThe Cauchy-Schwarz inequality in integral form states that:\n\\[ \\left( \\int_{0}^{2\\pi} f(x)",
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    "generation": "\u003cthink\u003e\nOkay, so I have this problem where I need to prove that if the function \\( f(x) = 1 - a \\cdot \\cos(x) - b \\cdot \\sin(x) - A \\cdot \\cos(2x) - B \\cdot \\sin(2x) \\) is non-negative for any real number \\( x \\), then two inequalities must hold: \\( a^2 + b^2 \\leq 2 \\) and \\( A^2 + B^2 \\leq 1 \\). \n\nHmm, let me start by understanding what the function \\( f(x) \\) looks like. It seems to be a combination of cosine and sine functions with different frequencies\u2014specifically, \\( \\cos(x) \\), \\( \\sin(x) \\), \\( \\cos(2x) \\), and \\( \\sin(2x) \\). The function is being subtracted from 1 and is supposed to be non-negative for all \\( x \\). So, my goal is to find constraints on the coefficients \\( a, b, A, B \\) such that this condition holds.\n\nI remember that when dealing with expressions involving sine and cosine, especially when they are linear combinations, it\u0027s often useful to express them as a single sine or cosine function with a phase shift. This might help in finding the maximum or minimum values of the function, which is crucial here because \\( f(x) \\geq 0 \\) for all \\( x \\) implies that the minimum value of \\( f(x) \\) is at least 0.\n\nLet me try to rewrite the terms involving \\( \\cos(x) \\) and \\( \\sin(x) \\) first. The expression \\( -a \\cos(x) - b \\sin(x) \\) can be rewritten as \\( R \\cos(x + \\phi) \\) where \\( R = \\sqrt{a^2 + b^2} \\) and \\( \\phi \\) is some phase shift. Similarly, the terms \\( -A \\cos(2x) - B \\sin(2x) \\) can be rewritten as \\( S \\cos(2x + \\theta) \\) where \\( S = \\sqrt{A^2 + B^2} \\).\n\nSo, substituting these, the function becomes:\n\\[ f(x) = 1 - R \\cos(x + \\phi) - S \\cos(2x + \\theta) \\]\n\nBut I\u0027m not sure if this substitution is helpful yet. Maybe instead of trying to combine the terms, I should think about the maximum and minimum values of the function. Since \\( f(x) \\geq 0 \\) for all \\( x \\), the minimum value must be greater than or equal to zero.\n\nAnother idea is to represent this function using complex exponentials, which might make the analysis easier. Let me recall Euler\u0027s formula: \\( e^{i\\theta} = \\cos\\theta + i \\sin\\theta \\). Maybe I can express \\( f(x) \\) as the real part of some complex expression.\n\nLet me try that. Let\u0027s consider the terms involving \\( \\cos(x) \\) and \\( \\sin(x) \\) as the real part of \\( (a - ib)e^{ix} \\), since:\n\\[ \\text{Re}[(a - ib)e^{ix}] = a \\cos x + b \\sin x \\]\nBut in our function \\( f(x) \\), we have \\( -a \\cos x - b \\sin x \\), so that would be \\( - \\text{Re}[(a - ib)e^{ix}] \\).\n\nSimilarly, the terms \\( -A \\cos(2x) - B \\sin(2x) \\) can be written as \\( - \\text{Re}[(A - iB)e^{i2x}] \\).\n\nSo, putting it all together, perhaps \\( f(x) \\) can be expressed as:\n\\[ f(x) = 1 - \\text{Re}[(a - ib)e^{ix}] - \\text{Re}[(A - iB)e^{i2x}] \\]\n\nWhich can be rewritten as:\n\\[ f(x) = 1 - \\text{Re}\\left[ (a - ib)e^{ix} + (A - iB)e^{i2x} \\right] \\]\n\nHmm, that might not directly help, but it\u0027s an interesting representation. Maybe I can think of \\( f(x) \\) as the real part of some complex function, but I\u0027m not sure how to proceed with that yet.\n\nAnother approach is to consider that \\( f(x) \\) is a continuous function and since it\u0027s non-negative everywhere, its minimum must be zero or positive. To find the minimum, I can take the derivative of \\( f(x) \\) with respect to \\( x \\) and set it to zero, then check the critical points. But that might get complicated because \\( f(x) \\) involves multiple cosine and sine terms with different frequencies.\n\nWait, maybe instead of taking derivatives, I can use the fact that the minimum of \\( f(x) \\) must be non-negative. To find the minimum, I can express \\( f(x) \\) in terms of multiple angles and then find its minimum value.\n\nAlternatively, I remember that for a function of the form \\( f(x) = C + D \\cos x + E \\cos 2x + F \\sin x + G \\sin 2x \\), the minimum can be found by considering the maximum of the negative part. But I\u0027m not sure.\n\nAlternatively, maybe I can use the method of completing the square or express \\( f(x) \\) as a sum of squares, ensuring that it\u0027s always non-negative.\n\nWait, another thought: since \\( f(x) \\) must be non-negative for all \\( x \\), perhaps I can express it as a sum of squares of sine and cosine functions, each multiplied by some coefficients, ensuring that the overall expression is always non-negative.\n\nBut I\u0027m not sure how to directly apply that here. Maybe another approach is to use Fourier series or orthogonal functions. Since the function is expressed in terms of cosines and sines of different frequencies, perhaps I can use orthogonality to find constraints on the coefficients.\n\nBut let me think differently. Since \\( f(x) \\geq 0 \\) for all \\( x \\), it must be that the function doesn\u0027t dip below zero. Therefore, the minimum value of \\( f(x) \\) is zero or positive. Maybe I can compute the minimum value of \\( f(x) \\) and set it to be greater than or equal to zero, which will give me some inequalities on \\( a, b, A, B \\).\n\nTo compute the minimum, I can take the derivative of \\( f(x) \\) with respect to \\( x \\), set it equal to zero, and solve for \\( x \\). Then, substitute back into \\( f(x) \\) to find the minimum value. But this might be complicated because \\( f(x) \\) has multiple terms with different frequencies.\n\nLet me compute the derivative:\n\\[ f\u0027(x) = a \\sin x + b \\cos x + 2A \\sin 2x + 2B \\cos 2x \\]\n\nSet this equal to zero:\n\\[ a \\sin x + b \\cos x + 2A \\sin 2x + 2B \\cos 2x = 0 \\]\n\nThis equation is going to be hard to solve analytically because of the different frequencies. Maybe instead, a better approach is to express \\( f(x) \\) in terms of multiple angles and then use the fact that the minimum is non-negative.\n\nLet me try to express \\( f(x) \\) as a combination of multiple-angle terms. Starting with \\( f(x) \\):\n\\[ f(x) = 1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x \\]\n\nI know that \\( \\cos 2x = 2 \\cos^2 x - 1 \\) and \\( \\sin 2x = 2 \\sin x \\cos x \\). Maybe substituting these into the function will help. Let\u0027s try that.\n\nSubstituting \\( \\cos 2x = 2 \\cos^2 x - 1 \\) and \\( \\sin 2x = 2 \\sin x \\cos x \\), we get:\n\\[ f(x) = 1 - a \\cos x - b \\sin x - A (2 \\cos^2 x - 1) - B (2 \\sin x \\cos x) \\]\nSimplify this:\n\\[ f(x) = 1 - a \\cos x - b \\sin x - 2A \\cos^2 x + A - 2B \\sin x \\cos x \\]\nCombine constants:\n\\[ f(x) = (1 + A) - a \\cos x - b \\sin x - 2A \\cos^2 x - 2B \\sin x \\cos x \\]\n\nHmm, this still seems complicated, but perhaps I can collect like terms. Let\u0027s group the \\( \\cos x \\) terms and the \\( \\sin x \\) terms:\n\n\\[ f(x) = (1 + A) - (a + 2B \\sin x) \\cos x - b \\sin x - 2A \\cos^2 x \\]\n\nWait, that might not be helpful. Maybe another substitution. Let me express everything in terms of \\( \\cos x \\) and \\( \\sin x \\). Let me set \\( y = \\cos x \\) and \\( z = \\sin x \\). Then, \\( y^2 + z^2 = 1 \\).\n\nSubstituting, we have:\n\\[ f(x) = 1 - a y - b z - A (2 y^2 - 1) - B (2 z y) \\]\nSimplify:\n\\[ f(x) = 1 - a y - b z - 2 A y^2 + A - 2 B z y \\]\nCombine constants:\n\\[ f(x) = (1 + A) - a y - b z - 2 A y^2 - 2 B z y \\]\n\nSo, \\( f(x) \\) is a function of \\( y \\) and \\( z \\) with the constraint \\( y^2 + z^2 = 1 \\). So, perhaps I can consider \\( f \\) as a function on the unit circle and find its minimum.\n\nLet me write \\( f \\) in terms of \\( y \\) and \\( z \\):\n\\[ f(y, z) = (1 + A) - a y - b z - 2 A y^2 - 2 B z y \\]\nwith \\( y^2 + z^2 = 1 \\).\n\nThis is a quadratic function in \\( y \\) and \\( z \\). To find its minimum, I can use the method of Lagrange multipliers, considering the constraint \\( y^2 + z^2 = 1 \\).\n\nAlternatively, maybe I can write this expression in matrix form. Let\u0027s see.\n\nLet me rewrite \\( f(y, z) \\) as:\n\\[ f(y, z) = (1 + A) - (a + 2 B z) y - b z - 2 A y^2 \\]\n\nHmm, this is a quadratic in \\( y \\) with coefficients depending on \\( z \\).\n\nAlternatively, perhaps I can arrange the terms as a quadratic form. Let me try:\n\nLet me consider the variables \\( y \\) and \\( z \\). Then:\n\\[ f(y, z) = (1 + A) - a y - b z - 2 A y^2 - 2 B y z \\]\n\nThis can be written as:\n\\[ f(y, z) = (1 + A) - \\left( a y + b z + 2 A y^2 + 2 B y z \\right) \\]\n\nHmm, maybe I can write this as a vector product. Let me think of the terms involving \\( y \\) and \\( z \\) as a vector \\( [y, z] \\) multiplied by some matrix.\n\nLet me denote the quadratic form as:\n\\[ f(y, z) = (1 + A) - \\left( \\begin{bmatrix} a \u0026 b \\end{bmatrix} \\begin{bmatrix} y \\\\ z \\end{bmatrix} + 2 A y^2 + 2 B y z \\right) \\]\n\nSo, the quadratic part is \\( 2 A y^2 + 2 B y z \\). The linear part is \\( a y + b z \\).\n\nSo, combining these, the quadratic form is:\n\\[ f(y, z) = (1 + A) - \\begin{bmatrix} a \u0026 b \\end{bmatrix} \\begin{bmatrix} y \\\\ z \\end{bmatrix} - 2 \\begin{bmatrix} A \u0026 B/2 \\\\ B/2 \u0026 0 \\end{bmatrix} \\begin{bmatrix} y \\\\ z \\end{bmatrix} \\]\n\nWait, not sure if that\u0027s correct. Let me see. The quadratic terms are \\( 2 A y^2 + 2 B y z \\). So, the corresponding matrix would have \\( A \\) on the diagonal for \\( y^2 \\) and \\( B/2 \\) for the cross term.\n\nSo, the quadratic form matrix is:\n\\[ \\begin{bmatrix} 2A \u0026 B \\\\ B \u0026 0 \\end{bmatrix} \\]\n\nBut actually, the standard quadratic form is \\( \\frac{1}{2} \\mathbf{v}^T Q \\mathbf{v} \\), where \\( Q \\) is symmetric. So, perhaps I need to adjust for that.\n\nBut maybe it\u0027s complicating things. Let me think differently. Since \\( y^2 + z^2 = 1 \\), I can parameterize \\( y \\) and \\( z \\) as \\( y = \\cos \\theta \\), \\( z = \\sin \\theta \\) for some angle \\( \\theta \\). Then, substitute into \\( f(y, z) \\):\n\nSo, substituting \\( y = \\cos \\theta \\), \\( z = \\sin \\theta \\), we have:\n\\[ f(\\theta) = (1 + A) - a \\cos \\theta - b \\sin \\theta - 2 A \\cos^2 \\theta - 2 B \\sin \\theta \\cos \\theta \\]\n\nSimplify \\( f(\\theta) \\):\nLet me write \\( \\cos^2 \\theta = \\frac{1 + \\cos 2\\theta}{2} \\) and \\( \\sin \\theta \\cos \\theta = \\frac{\\sin 2\\theta}{2} \\). So, substituting these:\n\n\\[ f(\\theta) = (1 + A) - a \\cos \\theta - b \\sin \\theta - 2 A \\left( \\frac{1 + \\cos 2\\theta}{2} \\right) - 2 B \\left( \\frac{\\sin 2\\theta}{2} \\right) \\]\nSimplify:\n\\[ f(\\theta) = (1 + A) - a \\cos \\theta - b \\sin \\theta - A (1 + \\cos 2\\theta) - B \\sin 2\\theta \\]\nExpand:\n\\[ f(\\theta) = 1 + A - a \\cos \\theta - b \\sin \\theta - A - A \\cos 2\\theta - B \\sin 2\\theta \\]\nSimplify constants:\n\\[ f(\\theta) = 1 - a \\cos \\theta - b \\sin \\theta - A \\cos 2\\theta - B \\sin 2\\theta \\]\n\nWait, that\u0027s just the original \\( f(x) \\)! So, maybe this substitution didn\u0027t help. Hmm, perhaps I need a different approach.\n\nWait, if I set \\( f(x) \\geq 0 \\) for all \\( x \\), then in particular, it must be non-negative at its minimum. So, perhaps I can compute the minimum of \\( f(x) \\) and set it to be greater than or equal to zero, which will give me some conditions on \\( a, b, A, B \\).\n\nBut as I thought earlier, computing the minimum requires taking the derivative and solving for critical points, which might be complicated.\n\nAlternatively, maybe I can express \\( f(x) \\) as a sum of squares, ensuring it\u0027s always non-negative. Let me think if that\u0027s possible.\n\nGiven that \\( f(x) = 1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x \\), perhaps I can write this as \\( 1 - (a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x) \\). So, if I can bound the maximum value of \\( a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x \\), then I can ensure that the minimum of \\( f(x) \\) is non-negative.\n\nIn other words, if \\( |a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x| \\leq 1 \\) for all \\( x \\), then \\( f(x) \\geq 0 \\). But that\u0027s not exactly the case here because the function is \\( 1 - \\) that expression, so it\u0027s non-negative if the expression is less than or equal to 1 everywhere.\n\nWait, actually, no. If \\( a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x \\leq 1 \\) for all \\( x \\), then \\( f(x) \\geq 0 \\). But the problem only states that \\( f(x) \\geq 0 \\), not necessarily that the expression is less than or equal to 1. So, perhaps it\u0027s a bit more nuanced.\n\nBut maybe the maximum of \\( a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x \\) is less than or equal to 1. Then, subtracting from 1 would make \\( f(x) \\geq 0 \\). So, if I can show that the maximum of this expression is at most 1, then \\( f(x) \\geq 0 \\) for all \\( x \\).\n\nBut the problem states that \\( f(x) \\geq 0 \\) for all \\( x \\), so perhaps that\u0027s the way to go. Let me denote \\( g(x) = a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x \\). Then, \\( f(x) = 1 - g(x) \\). So, \\( f(x) \\geq 0 \\) implies that \\( g(x) \\leq 1 \\) for all \\( x \\).\n\nSo, if \\( g(x) \\leq 1 \\) for all \\( x \\), then \\( f(x) \\geq 0 \\). Therefore, we can rephrase the problem as: if \\( g(x) \\leq 1 \\) for all \\( x \\), then prove that \\( a^2 + b^2 \\leq 2 \\) and \\( A^2 + B^2 \\leq 1 \\).\n\nBut wait, is this necessarily true? If \\( g(x) \\leq 1 \\) everywhere, then certainly the maximum of \\( g(x) \\) is 1. So, the maximum of \\( g(x) \\) is 1, which will impose some constraints on \\( a, b, A, B \\).\n\nI recall that the maximum of a function like \\( C \\cos x + D \\sin x \\) is \\( \\sqrt{C^2 + D^2} \\). So, perhaps I can bound \\( g(x) \\) by using the fact that the maximum of \\( a \\cos x + b \\sin x \\) is \\( \\sqrt{a^2 + b^2} \\) and the maximum of \\( A \\cos 2x + B \\sin 2x \\) is \\( \\sqrt{A^2 + B^2} \\). So, the maximum of \\( g(x) \\) would be at most \\( \\sqrt{a^2 + b^2} + \\sqrt{A^2 + B^2} \\).\n\nBut wait, that\u0027s not necessarily true because \\( g(x) \\) combines both frequencies. So, the maximum of the sum isn\u0027t necessarily the sum of the maxima. For example, if the two functions are in phase, their maxima add up, but if they are out of phase, they might cancel each other.\n\nBut since \\( g(x) \\leq 1 \\) for all \\( x \\), its maximum must be less than or equal to 1. Therefore, the maximum of \\( g(x) \\) is at most 1.\n\nTo find the maximum of \\( g(x) \\), perhaps I can write it as a sum of two sinusoids with different frequencies and then find its maximum.\n\nAlternatively, another method is to express \\( g(x) \\) as a single sinusoid using harmonic addition, but since it has two different frequencies, it\u0027s not straightforward.\n\nWait, maybe I can consider that \\( g(x) \\) can be written as a function with Fourier coefficients \\( a, b, A, B \\), so perhaps using the concept of the maximum modulus.\n\nAlternatively, another idea: if \\( g(x) \\leq 1 \\) for all \\( x \\), then in particular, at specific points, like \\( x = 0, \\pi/2, \\pi, 3\\pi/2 \\), etc., we can get some inequalities on \\( a, b, A, B \\).\n\nLet me compute \\( g(0) \\), \\( g(\\pi/2) \\), \\( g(\\pi) \\), \\( g(3\\pi/2) \\), and so on, and set up inequalities.\n\nCompute \\( g(0) \\):\n\\[ g(0) = a \\cos 0 + b \\sin 0 + A \\cos 0 + B \\sin 0 = a + A \\]\nSince \\( g(0) \\leq 1 \\), we have:\n\\[ a + A \\leq 1 \\]\n\nSimilarly, compute \\( g(\\pi/2) \\):\n\\[ g(\\pi/2) = a \\cos (\\pi/2) + b \\sin (\\pi/2) + A \\cos \\pi + B \\sin \\pi = 0 + b + (-A) + 0 = b - A \\]\nSince \\( g(\\pi/2) \\leq 1 \\), we have:\n\\[ b - A \\leq 1 \\]\n\nCompute \\( g(\\pi) \\):\n\\[ g(\\pi) = a \\cos \\pi + b \\sin \\pi + A \\cos 2\\pi + B \\sin 2\\pi = -a + 0 + A + 0 = -a + A \\]\nSince \\( g(\\pi) \\leq 1 \\), we have:\n\\[ -a + A \\leq 1 \\]\n\nCompute \\( g(3\\pi/2) \\):\n\\[ g(3\\pi/2) = a \\cos (3\\pi/2) + b \\sin (3\\pi/2) + A \\cos 3\\pi + B \\sin 3\\pi = 0 - b + (-A) + 0 = -b - A \\]\nSince \\( g(3\\pi/2) \\leq 1 \\), we have:\n\\[ -b - A \\leq 1 \\]\n\nSo, from these four points, we get the following inequalities:\n1. \\( a + A \\leq 1 \\)\n2. \\( b - A \\leq 1 \\)\n3. \\( -a + A \\leq 1 \\)\n4. \\( -b - A \\leq 1 \\)\n\nThese can be rewritten as:\n1. \\( a + A \\leq 1 \\)\n2. \\( b - A \\leq 1 \\)\n3. \\( -a + A \\leq 1 \\)\n4. \\( -b - A \\leq 1 \\)\n\nBut these inequalities don\u0027t directly give us \\( a^2 + b^2 \\leq 2 \\) and \\( A^2 + B^2 \\leq 1 \\). So, perhaps this approach is not sufficient.\n\nAlternatively, maybe I should use the fact that \\( f(x) \\geq 0 \\) for all \\( x \\), so the minimal value of \\( f(x) \\) is non-negative. To compute the minimal value, we can use calculus. Let\u0027s try that.\n\nFirst, compute the derivative of \\( f(x) \\):\n\\[ f\u0027(x) = -a \\sin x - b \\cos x - 2A \\sin 2x - 2B \\cos 2x \\]\n\nSet \\( f\u0027(x) = 0 \\):\n\\[ -a \\sin x - b \\cos x - 2A \\sin 2x - 2B \\cos 2x = 0 \\]\n\nThis equation is quite complex because of the different sine and cosine terms with different arguments. Solving this analytically might be difficult. Maybe instead, I can express this equation in terms of \\( \\sin x \\) and \\( \\cos x \\), and then use substitution.\n\nLet me recall that \\( \\sin 2x = 2 \\sin x \\cos x \\) and \\( \\cos 2x = 2 \\cos^2 x - 1 \\). Substituting these into the derivative equation:\n\n\\[ -a \\sin x - b \\cos x - 2A (2 \\sin x \\cos x) - 2B (2 \\cos^2 x - 1) = 0 \\]\nSimplify:\n\\[ -a \\sin x - b \\cos x - 4A \\sin x \\cos x - 4B \\cos^2 x + 2B = 0 \\]\n\nLet me rearrange terms:\n\\[ (-a - 4A \\cos x) \\sin x + (-b) \\cos x - 4B \\cos^2 x + 2B = 0 \\]\n\nHmm, this still looks complicated. Maybe another substitution. Let me set \\( t = \\cos x \\), then \\( \\sin x = \\sqrt{1 - t^2} \\) (assuming \\( \\sin x \\) is positive; we might need to consider negative as well, but let\u0027s proceed for now).\n\nSo, substituting \\( t = \\cos x \\) and \\( \\sin x = \\sqrt{1 - t^2} \\), we get:\n\\[ (-a - 4A t) \\sqrt{1 - t^2} - b t - 4B t^2 + 2B = 0 \\]\n\nThis is a complicated equation in \\( t \\). It might not be feasible to solve this analytically. Maybe another approach is needed.\n\nWait, perhaps instead of calculus, I can use the concept of positive semi-definite functions. Since \\( f(x) \\geq 0 \\) for all \\( x \\), it\u0027s a positive semi-definite function. Maybe I can express \\( f(x) \\) as a sum of squares, which would ensure it\u0027s non-negative.\n\nAlternatively, thinking back to the expression \\( f(x) = 1 - g(x) \\), since \\( f(x) \\geq 0 \\), then \\( g(x) \\leq 1 \\) for all \\( x \\). So, the maximum of \\( g(x) \\) is at most 1. Therefore, if I can bound the maximum of \\( g(x) \\) in terms of \\( a, b, A, B \\), then I can get the required inequalities.\n\nI remember that for a function like \\( g(x) = a \\cos x + b \\sin x + A \\cos 2x + B \\sin 2x \\), the maximum can be bounded using the sum of the amplitudes. But because of the different frequencies, the exact maximum isn\u0027t straightforward.\n\nAlternatively, perhaps I can use the Cauchy-Schwarz inequality. Let me consider \\( g(x) \\) as the sum of two orthogonal functions: \\( h(x) = a \\cos x + b \\sin x \\) and \\( k(x) = A \\cos 2x + B \\sin 2x \\).\n\nSince \\( h(x) \\) and \\( k(x) \\) are orthogonal over the interval \\( [0, 2\\pi] \\), the maximum of \\( g(x) \\) can be bounded by the sum of the maxima of \\( h(x) \\) and \\( k(x) \\). But actually, orthogonality doesn\u0027t directly give the maximum, but rather the integral of their product is zero.\n\nWait, but if they are orthogonal, then the maximum of their sum isn\u0027t necessarily the sum of their maxima. Hmm, maybe that approach isn\u0027t helpful.\n\nWait, another idea: using the fact that for any \\( x \\), \\( |\\cos 2x| \\leq 1 \\) and \\( |\\sin 2x| \\leq 1 \\), similarly for the single angle terms. So, \\( |g(x)| \\leq |a| + |b| + |A| + |B| \\). But this gives a rough bound, not necessarily tight.\n\nBut since \\( g(x) \\leq 1 \\) for all \\( x \\), this implies that \\( |a| + |b| + |A| + |B| \\leq 1 \\). But this is much stronger than what we need to prove, so maybe it\u0027s not useful.\n\nAlternatively, perhaps I can write \\( g(x) \\) as the sum of two functions, each with their own frequency, and then use the fact that the maximum of the sum is bounded by the sum of the maxima.\n\nBut, in reality, because of the orthogonality and the different frequencies, the maximum of the sum isn\u0027t just the sum of the individual maxima. It might be less due to possible cancellation.\n\nWait, maybe a better approach is to compute the maximum of \\( g(x) \\) by considering it as a function with multiple frequencies. For that, perhaps I can use the concept of the maximum of a trigonometric polynomial.\n\nI recall that for a trigonometric polynomial \\( P(x) = \\sum_{k=-n}^{n} c_k e^{ikx} \\), the maximum modulus can be bounded, but again, not sure how to apply that here.\n\nAlternatively, perhaps using the fact that \\( g(x) \\) is a combination of sinusoids with different frequencies, its maximum can be found by considering the envelope of these sinusoids. But that\u0027s vague.\n\nWait, maybe another approach. Let me consider specific points where \\( f(x) \\) could achieve its minimum. For example, at \\( x = 0 \\), \\( x = \\pi/2 \\), \\( x = \\pi \\), \\( x = 3\\pi/2 \\), and maybe some other points where the derivative is zero.\n\nEarlier, I found that:\n- \\( f(0) = 1 - a - A \\)\n- \\( f(\\pi/2) = 1 - b - B \\)\n- \\( f(\\pi) = 1 + a - A \\)\n- \\( f(3\\pi/2) = 1 + b - B \\)\n\nBut to get the minimum, maybe we need to evaluate \\( f(x) \\) at more points or find critical points.\n\nAlternatively, perhaps I can use the fact that \\( f(x) \\geq 0 \\) for all \\( x \\), so in particular, the integral of \\( f(x) \\) over a period must be non-negative. But actually, since \\( f(x) \\geq 0 \\), the integral is non-negative, but it\u0027s always non-negative regardless.\n\nWait, maybe integrating \\( f(x) \\) over a period and using orthogonality can give some information. Let me try that.\n\nCompute the integral of \\( f(x) \\) over \\( [0, 2\\pi] \\):\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 \\]\nCompute each integral separately:\n- \\( \\int_{0}^{2\\pi} 1 dx = 2\\pi \\)\n- \\( \\int_{0}^{2\\pi} \\cos x dx = 0 \\)\n- \\( \\int_{0}^{2\\pi} \\sin x dx = 0 \\)\n- \\( \\int_{0}^{2\\pi} \\cos 2x dx = 0 \\)\n- \\( \\int_{0}^{2\\pi} \\sin 2x dx = 0 \\)\n\nSo, the integral is \\( 2\\pi \\). Since \\( f(x) \\geq 0 \\), the average value is \\( \\pi \\), which is positive, but that doesn\u0027t give me the necessary constraints.\n\nWait, perhaps integrating \\( f(x)^2 \\) over a period would give me something. Since \\( f(x) \\geq 0 \\), \\( f(x)^2 \\) would be non-negative, and maybe I can use some inequalities.\n\nCompute the integral of \\( f(x)^2 \\):\n\\[ \\int_{0}^{2\\pi} f(x)^2 dx = \\int_{0}^{2\\pi} [1 - a \\cos x - b \\sin x - A \\cos 2x - B \\sin 2x]^2 dx \\]\n\nExpanding the square:\n\\[ \\int_{0}^{2\\pi} [1 - 2a \\cos x - 2b \\sin x - 2A \\cos 2x - 2B \\sin 2x + a^2 \\cos^2 x + b^2 \\sin^2 x + A^2 \\cos^2 2x + B^2 \\sin^2 2x + 2ab \\cos x \\sin x + 2aA \\cos x \\cos 2x + 2aB \\cos x \\sin 2x + 2bA \\sin x \\cos 2x + 2bB \\sin x \\sin 2x + 2AB \\cos 2x \\sin 2x] dx \\]\n\nNow, integrating term by term:\n\n1. \\( \\int 1 dx = 2\\pi \\)\n2. \\( -2a \\int \\cos x dx = 0 \\)\n3. \\( -2b \\int \\sin x dx = 0 \\)\n4. \\( -2A \\int \\cos 2x dx = 0 \\)\n5. \\( -2B \\int \\sin 2x dx = 0 \\)\n6. \\( a^2 \\int \\cos^2 x dx = a^2 \\pi \\)\n7. \\( b^2 \\int \\sin^2 x dx = b^2 \\pi \\)\n8. \\( A^2 \\int \\cos^2 2x dx = A^2 \\pi \\)\n9. \\( B^2 \\int \\sin^2 2x dx = B^2 \\pi \\)\n10. \\( 2ab \\int \\cos x \\sin x dx = 0 \\) (since it\u0027s an odd function over a symmetric interval)\n11. \\( 2aA \\int \\cos x \\cos 2x dx = 0 \\) (orthogonal over the interval)\n12. \\( 2aB \\int \\cos x \\sin 2x dx = 0 \\) (orthogonal)\n13. \\( 2bA \\int \\sin x \\cos 2x dx = 0 \\) (orthogonal)\n14. \\( 2bB \\int \\sin x \\sin 2x dx = 0 \\) (orthogonal)\n15. \\( 2AB \\int \\cos 2x \\sin 2x dx = 0 \\) (orthogonal)\n\nSo, putting it all together, the integral becomes:\n\\[ 2\\pi + a^2 \\pi + b^2 \\pi + A^2 \\pi + B^2 \\pi \\]\nWhich simplifies to:\n\\[ \\pi (2 + a^2 + b^2 + A^2 + B^2) \\]\n\nSince \\( f(x) \\geq 0 \\) for all \\( x \\), \\( f(x)^2 \\geq 0 \\) as well. But this integral is just a positive quantity. How does this help me?\n\nWait, another thought. Maybe using the fact that the minimum of \\( f(x) \\) is non-negative, so the integral can also be expressed in terms of the minimum value. But I don\u0027t see an immediate connection.\n\nAlternatively, perhaps I can use the Cauchy-Schwarz inequality on the integral of \\( f(x)^2 \\). The integral is equal to \\( \\pi (2 + a^2 + b^2 + A^2 + B^2) \\). But since \\( f(x) \\geq 0 \\), the integral is at least the square of the integral of \\( f(x) \\) divided by the period, by Cauchy-Schwarz. Wait, let me recall the inequality.\n\nThe Cauchy-Schwarz inequality in integral form states that:\n\\[ \\left( \\int_{0}^{2\\pi} f(x)",
    "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-7B",
    "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-7b-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-7b-test")
```


</details>




