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(A) $\sin(-x)=-\sin x$
(B) $\cos(-x)=\cos x$
(C) $\cos(x+y)=\cos x\cos y-\sin x\sin y$
(D) $\sin(x+y)=\sin x\cos y+\cos x\sin y$
Use these equalities to derive the following important trigonometric functions:
(b) $\sin 2x=2\sin(x)\cos(x)$

I know I've asked this same question about a month ago, but I just got back to this problem in my packet and I have yet to come up with anything. For (a), I may have gotten it. So please check here. I moved on to the second one and I have no idea how to start this one off. It's just a confusing question I think. Can someone at least tell me which equation(s) to use and why? Thanks a lot.

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I believe from the wording that you are not expected to prove these facts. In particular, (C) and (D) are harder than other things you have been asking about. I think you are just supposed to use facts chosen from (A) to (D) to prove (b). This is easy. – André Nicolas Aug 10 '12 at 6:55
Right, so for the next one, (e) $\cos2x=1-2\sin^2x$ (using d then a & $y=x$); $\sin(x+y)=\sin x\cos y+\sin y\cos x\Rightarrow \sin2x=-(-\sin x\cos x)+\cos x\sin x\Rightarrow \sin2x=\sin(-x)\cos x\sin x \cdots$. I believe I messed this one up horribly... – Austin Broussard Aug 10 '12 at 7:30
Whoa, what? I never found such problem in (e). Oh, wait. You want me to start off using (C) instead of (D)? – Austin Broussard Aug 10 '12 at 7:34
You already showed $\cos 2x=\cos^2 x-\sin^2 x$ using (C). Now it's just replacing $\cos^2 x$ by $1-\sin^2 x$. – André Nicolas Aug 10 '12 at 7:45
Yes, anything that takes space in typing is too hard to do in comment. One doesn't get feedback on the LaTeX, a little LaTeX error can make an awful mess. – André Nicolas Aug 10 '12 at 7:53

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