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The trigonometric functions I must know:
(A) $\sin(-x)=-\sin x$
(B) $\cos(-x)=\cos x$
(C) $\cos(x+y)=\cos x\cos y-\sin y\sin x$
(D) $\sin(x+y)=\sin x\cos y+\cos x\sin y$

$\sin^2x+\cos^2x=1$ (Use (C) and $\cos0=1$)

Can anyone help me just understand what the first one is asking. It's giving me a property and I'm supposed to derive the identities. But I don't even know where to begin. Any help? I will post the other problems once I figure this one out.

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    Here's another approach: The identity you want can be written as $\cos 0 = 1 = \cos z \cos z + \sin z \sin z$. Now do a 'pattern match' on the identities and see what you can set $x$ and $y$ to to get this identity.2012-07-15

3 Answers 3

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Hint: Note that by (A) and (B) $\cos^2(x)+\sin^2(x)=\cos(x)\cos(-x)-\sin(-x)\sin(x)$ and apply (C).

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Take y= $cos(x)+isin(x)$

Taking derivative in either side wrt x, $\frac{dy}{dx}=-sin(x)+icos(x)=iy$

Now, integrate $\frac{dy}{y}= idx$

You will find $cos(x)+isin(x)=exp(ix)$+c (c in a undetermined constant).

Put x=0, to find c=0.

So, $cos(x)=\frac{exp(ix)+exp(-ix)}{2}$ and $sin(x)=\frac{exp(ix)-exp(-ix)}{2i}$

Use these two to make the necessary derivation(s).

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    This is not really useful to the OP.2012-07-15
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What you want is to use $(C)$.

$1=\cos 0 = \cos(x-x)=\cos x \cos (-x)-\sin x \sin(-x)$

Now you need to use $(A)$ and $(B)$ to obtain the Pythagorean identity.

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    I just got back to this problem from after a month has passed but this has been confusing me forever so sorry if my work does not make sense; $\sin^2(x)+\cos^2(x)=\alpha$ $\alpha=\cos(x)\cos(-x)-\sin(-x)\sin(x)$ $\cos(x)\cos(-x)=\cos(x-x)\Rightarrow\cos(x)\cos(-x)-\sin(-x)\sin(x)$ $\alpha=\cos(x)\cos(-x)\underbrace{-\sin(-x)\sin(x)-\sin(-x)\sin(x)}_{\text{This should equal 0}}$ $\alpha=\cos(x)\cos(-x)\Rightarrow \alpha=\cos(x-x)\Rightarrow \alpha=\cos0=1$2012-08-10