How do I solve $$3\sin t=18(1-\sin ^2t)$$  Because, $$3\sin t=18(1-\sin ^2t)$$ is like quadratic, but using ABC I get not special angles.
This is trigonometry from double angles and compound angles. Thus far I've simplified it up to this point. $$2 \sin^2θ – 3\sinθ +1 = 0$$  I'm trying to factorize this to solve for General Solution. I tried taking out a common but that +1 is giving me a problem. I used the quadratic equation to get 1 and 0.5 as an answer, but i don't know how to write it in terms of 2sin(squared)θ and -3sinθ.  Any help will be appreciated.
Find the last two digits of $9^{{9}^{9}}$ I have to find the last two decimal digits of the number $9^{{9}^{9}}$.  That's what I did:  $$m=100 , \phi(m)=40, a=9$$ $$(100,9)=1, \text{ so from Euler's theorem,we have :} 9^{40} \equiv 1 \pmod{100}$$  $$9^{{9}^{9}} \equiv 9^{9^{2+2+2+2+1}} \equiv 9^{9^2 \cdot 9^2 \cdot 9^2 \cdot 9^2 \cdot 9} \equiv 9^{81 \cdot 81 \cdot 81 \cdot 81 \cdot 9} \equiv 9^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv (9^{(40+40+1)})^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv 9^9 \equiv 9^4 \cdot 9^4 \cdot 9 \equiv 6561 \cdot 6561 \cdot 9 \equiv 3721 \cdot 9 \\ \equiv 21 \cdot 9 \equiv 89 \pmod{100}$$  So,the last two digits are $8 \text{ and } 9$. $$$$But,is there also an other way two calculate the last two digits of $9^{{9}^{9}}$ or is the above the only one? Examples proving why the tensor product does not distribute over direct products. I recently read about the result that the tensor product distributes over direct sums. I was curious if it also distributes over direct products, but google tells me it doesn't.   What are some simple counterexamples to why this property isn't true? I know that there is a natural homomorphism  $$ \left(\prod M_i\right)\otimes N\to \prod (M_i\otimes N) $$ given by $(\prod m_i)\otimes n\mapsto \prod (m_i\otimes n)$ when $M$ and $N$ are modules over some commutative ring $R$. Are there standard examples where this homomorphism is not injective/surjective and hence not an isomorphism? Why can't you get awarded the bounty for answering your own question? Consider this question on .  I put out a 200 bounty on the question, but did not get an answer from anyone other than myself that worked.    In fact, my answer is also the highest up-voted on it at this point.  I know that allowing OP's to retract a bounty can be a slippery slope; in fact I think that once the bounty is put up it should stay up.  But why not allow the OP to answer their own questions with a bounty and reward it to them if they receive the same up-votes as others?  Yes, I am fully aware that bounties are a risk.  People will use them for numerous reasons such as trying to draw extra attention to their issues.  However, in my specific question, the only valid answer is the one that I gave.  Had it been anyone else's question, I would have received the bounty, but since it was my own I did not. How to test for minimum amount of items in inventory and execute a command if true Functions of Independent Random Variables Solving cubic equations with another method. I want to solve cubic equations with some pattern or formula. My maths teacher told me that if we want to solve cubic equations we have to find any one value for the equation which satisfy it and then divide the main equation by it. But I find that method little useless . Is there any other simplier or good method to solve it. You can explain it by an example. How to prove that orthochronous Lorentz transformations $O^+(1,3)$ form a group? Orthochronous Lorentz transform are Lorentz transforms that satisfy the conditions (sign convention of Minkowskian metric $+---$) $$ \Lambda^0{}_0 \geq +1.$$ How to prove they form a subgroup of Lorentz group? All books I read only give this result, but no derivation.  Why is this condition $ \Lambda^0{}_0 \geq +1$ enough for a Lorentz transform to be orthochronous?  The temporal component of a transformed vector is  $$x'^0=\Lambda^0{}_0 x^0+\Lambda^0{}_1 x^1+\Lambda^0{}_2 x^2+\Lambda^0{}_3 x^3,$$ the positivity of $\Lambda^0{}_0$ alone does not seem at first glance sufficient for the preservation of the sign of temporal component.  And how to prove that all Lorentz transform satisfying such simple conditions can be generated from $J_i,\ K_i$?    For those who think that closure and invertibility are obvious, keep in mind that $$\left(\bar{\Lambda}\Lambda \right)^0{}_0\neq \bar{\Lambda}^0{}_0\Lambda^0{}_0,$$ but instead $$\left(\bar{\Lambda}\Lambda \right)^0{}_0= \bar{\Lambda}^0{}_0\Lambda^0{}_0+\bar{\Lambda}^0{}_1\Lambda^1{}_0+\bar{\Lambda}^0{}_2\Lambda^2{}_0+\bar{\Lambda}^0{}_3\Lambda^3{}_0.$$  And I'm looking for a rigorous proof, not physical "intuition". How can I chain implicits in Scala? Let $f: \mathbb{R} \to \mathbb{R}$ be continuous. Then $G = \{ (x, f(x) ) : x \in \mathbb{R} \} $ is a closed set.  My try: Suppose $(z_n) =  (x_n, f(x_n) ) $ is sequence in $G$ with limit $(x,y)$. We must show $(x,y) \in G$. Since $x_n \to x$ and since $f$ is continuous, then we must have that $f(x_n) \to f(x) $. Since limits are unique. Then $y = f(x) $. Is this enough to conlude that $(x,y) \in G $ ? Hence showing $G$ is closed ?  thanks Does there exist a surjective homomorphism from $(\mathbb R,+)$ to $(\mathbb Q,+)$ ? Does there exist a surjective homomorphism from $(\mathbb R,+)$ to $(\mathbb Q,+)$ ?  ( I know that there 'is' a   'surjection' , but I don't know whether any surjective homomrophism from $\mathbb R$ to $\mathbb Q$ exist or not . Please help . Thanks in advance ) Is there a way to ignore flow errors? I have an auto launched flow that I would like to just ignore the errors of if they occur. I currently have several fault connectors going to a dummy assignment. However I keep receiving error emails which is what I am trying to avoid. What is the logical form of the definition of validity? My text book gives the definition of validity as "An argument is valid IFF if the premises are true, then the conclusion must be true". Using a conditional elimination on the RHS of the IFF yields "An argument is valid IFF the premises are false or the conclusion is true".   Yet, this seems strange to me. Am I doing something incorrect with my manipulation or is this definition actually true? Surely the criteria for a valid argument goes beyond a false set of premises or a true conclusion. Understanding the length extension attack I have been trying to understand exactly how a length extension attack works on SHA-1. I'll detail below what I've understood so far, so that I can convey my understanding of the same and hopefully get advice on where I'm going wrong.  Lets assume that for a message $m$, a secret $s$ is appended to it and the SHA-1 hash is calculated. If $\textrm{len}(m) + \textrm{len}(s)$ is not a multiple of the block size, a certain amount of padding is added to it to make it a multiple, and the hash is generated using what is essentially $s || m || \textrm{padding}$.  Now SHA-1 internally uses 5 registers — it takes in an block, generates new register values, turns the crank, takes in another block, generates new register values, turns the crank ... and so on till the end. When all blocks are exhausted, the register values are concatenated and spit out.  Assume that an attacker knows a certain value of $m$, for which he also knows the hashed key. In such a case, he knows the state of the 5 registers when $(s || m || \textrm{padding}$ was processed. Lets also assume that the attacker knows the length of $s$, so that he can predict how much padding was added.  Here is where I get confused. From what I've read so far, the attack involves taking the known hash, adding a certain amount of padding and then the value that we want. However, isn't the padding already used up in creating the hash? Why are we using padding in our attack? Is commitment necessary to participate in beta (as of June 2015)? There's this proposal for a new S.E. site which is about to enter the beta phase. It reads  This proposal is in: Commitment A successful site needs people to commit to use it. This proposal is 100% complete. Committed users will be invited to the private beta soon.  I'd like to participate in the beta but I don't want to commit (due to a full name being required). Will I be unable to participate in the beta if I don't commit? How to derive $[x_i, F(\vec p)] = i \hbar \frac {\partial F(\vec p)}{\partial p_i}$ indicates that the following relation is &quot;easily shown&quot;: $[x_i, F(\vec p)] = i \hbar \frac {\partial F(\vec p)}{\partial p_i}$, however I'm having some trouble showing it.  I think I'm just messing up the multivariable Taylor expansion (of $F(\vec p)$).  Can one of you walk me through it or link me to site that will?  Thanks.  Edit: Here's what I get (without using $x=i\hbar \frac \partial {\partial p}$ which I haven't proven yet): $$F(\vec p) = F(\vec 0) + \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}p_j + \frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} p_j p_k + \dots$$ so $$(x_iF(\vec p) - F(\vec p)x_i)\psi$$ $$= x_i[F(\vec 0)\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\nabla \psi)_j + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\nabla \psi)_j (\nabla \psi)_k + \dots]$$ $$- [F(\vec 0)x_i\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\nabla x_i\psi)_j + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\nabla x_i\psi)_j (\nabla x_i\psi)_k + \dots]$$ where $$(\nabla x_i\psi)_j=\frac {\partial x_i}{\partial x_j}\psi + x_i\frac {\partial \psi}{\partial x_j} = \delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j}$$ and $$(\nabla x_i\psi)_j(\nabla x_i\psi)_k=(\delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j})(\delta_{ik}\psi + x_i\frac {\partial \psi}{\partial x_k}) = \delta_{jk}\psi^2 + x_j \psi \frac {\partial \psi}{\partial x_k} + x_k \frac {\partial \psi}{\partial x_j} \psi + x_i^2 \frac {\partial^2 \psi}{\partial x_j \partial x_k}$$ Thus: $$(x_iF(\vec p) - F(\vec p)x_i)\psi$$ $$=  x_i[F(\vec 0)\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}\frac {\partial \psi}{\partial x_j} + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} \frac {\partial \psi}{\partial x_j} \frac {\partial \psi}{\partial x_k} + \dots]$$ $$- [F(\vec 0)x_i\psi -i\hbar \sum^3_{j=1} \frac {\partial F(\vec 0)}{\partial p_j}(\delta_{ij}\psi + x_i\frac {\partial \psi}{\partial x_j}) + \hbar^2\frac 12 \sum^3_{k=1} \sum^3_{j=1} \frac {\partial^2 F(\vec 0)}{\partial p_k \partial p_j} (\delta_{jk}\psi^2 + x_j \psi \frac {\partial \psi}{\partial x_k} + x_k \frac {\partial \psi}{\partial x_j} \psi + x_i^2 \frac {\partial^2 \psi}{\partial x_j \partial x_k}) + \dots]$$ From here it doesn't look like those higher order terms are all going to cancel out. Looking for a short story about an insect-like alien captured by humans on its own planet This poor creature is taken aboard the spaceship and the humans do not know that it is intelligent.  The alien can repeat what it hears, e,g, "pass the sugar".  It is able to escape at night and it communicates messages to its fellow aliens.  It directs that they turn up the heat of the atmosphere (somehow they can do this) so as to force the humans to leave.  The alien protagonist states that it will die w/ the humans or leave with them. Will metering from a DSLR apply to a film SLR? I still like to shoot with film (I use a Nikon F3) but I like the immediate feedback of digital.  I'm thinking of buying a used DSLR to help me get the framing and lighting right when I shoot static subjects.  Then I'd take the final shot with my F3. But would the metering be transferable?  If the shot looks perfect at 1/125 with f/2 on a DSLR, will the same settings work on a film camera? Given the functional dependencies of a relation is there more then one way to normalize it? Can I upload an entire folder using FTP? I need to upload a full folder using FTP.  Is there is any option for transferring a folder and all of its contents at once? I want to know how EXACTLY the WPA password gets encrypted  I have been searching through and I was able to catch:   The actual password is converted to some hash Salting is applied to the hash The operation is not reversible The hash comes with the 4-way handshake   So what kind of hash is it converted? like, SHA1? MD5? and how the salting process is done? like, is the SSID appended to the password before the hashing operation?  Or maybe I have some misunderstanding here?