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I've been trying with no success expressing this functions.

a) $\sin\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$

b) $\cos\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$

I've tried formula of the double angle ($\sin$ and $\cos$) and the ecuation $\cos^2X+\sin^2X=1$.

Can anyone point what to use or where to start?

I appreciate it, thanks.


UPDATE

I think this is achived with the half formula as njguliyev and nbubis said.

$$\sin\frac{\pi}{8} = \sin(\frac{1}{2} * \frac{\pi}{4}) = \pm\sqrt\frac{1-\cos\frac{\pi}{4}}{2}$$

$$\cos\frac{\pi}{8} = \cos(\frac{1}{2} * \frac{\pi}{4}) = \pm\sqrt\frac{1+\cos\frac{\pi}{4}}{2}$$

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3  
$\cos 2x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$. –  njguliyev Oct 20 '13 at 20:57

2 Answers 2

up vote 3 down vote accepted

Hint:

You my want to take a look at the half angle formulas.

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I mean just use the trig identities: $$\cos \frac{ PI }{4} = 1 - 2 \sin^2 \frac{ PI}{8} $$

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