# Use equalities to derive important trigonometric functions

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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You have an expression that starts off with $\cos x\cos y$ and you really want an expression involving $\cos^2x$. So you might try setting $y = x$. This doesn't quite work, because in (C) you'll get $\cos 2x$ and not know what to do. But maybe you can tweak this slightly, using (A) and (B)? – Dylan Moreland Jul 15 '12 at 3:57
The paper says I have to use C for number one. Which I honestly do not understand considering there is no $y$ in the problem. – Austin Broussard Jul 15 '12 at 3:59
You will use (C). In (C) you are allowed to plug in different things for $y$. Try setting $y = x$. This will not work, but it will be close to what you want and you should try it for yourself and see why. – Dylan Moreland Jul 15 '12 at 4:02
I will thanks a lot! – Austin Broussard Jul 15 '12 at 4:03
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. – copper.hat Jul 15 '12 at 4:34

## 3 Answers

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. – Pedro Tamaroff Jul 15 '12 at 5:16

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; $$\sin^2(x)+\cos^2(x)=\alpha$$ $$\alpha=\cos(x)\cos(-x)-\sin(-x)\sin(x)$$ $$\cos(x)\cos(-x)=\cos(x-x)\Rightarrow\cos(x)\cos(-x)-\sin(-x)\sin(x)$$ $$\alpha=\cos(x)\cos(-x)\underbrace{-\sin(-x)\sin(x)-\sin(-x)\sin(x)}_{\text{Thi‌​s should equal 0}}$$ $$\alpha=\cos(x)\cos(-x)\Rightarrow \alpha=\cos(x-x)\Rightarrow \alpha=\cos0=1$$ – Austin Broussard Aug 10 '12 at 6:28