How can I use the result of sin(PI/8) to find the cos and sin of other angles?

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I have (from a complex equation problem) found the following angles in the answers:

5PI/8, 9PI/8, 13PI/8, 17PI/8

In the same assignement I found sin(PI/8) = .5*sqrt(2-sqrt(2)) and I found the cos(PI/8) value by using standard trig. identities.

Is there an easy way to use this result to find the cos and sin of the angles stated? (The assignement tells us to use the answer we found from sin(PI/8))

I know I can find formulas for cos(5x) etc by using this method: http://library.thinkquest.org/C0110248/trigonometry/form3.htm

But that is going to take ages up to cos(17x) !

I think I've missed something here.

Thanks
 
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Hmm.. 5/8-1/8=1/2, I think. And 17=16+1, if I'm not mistaken
 
I'm not getting it... :P
 
\frac{17\pi}{8}=\frac{16\pi}{8}+\frac{1\pi}{8}=2\pi+\frac{\pi}{8}
For example
 
ah, thanks I get it!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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