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Evaluate $\sin(\frac{\pi}{8})$ and $\cos(\frac{\pi}{8})$

I was just wondering what I am doing wrong, as I don't seem to be arriving at the correct answer for $\sin(\frac{\pi}{8})$

What I did:

Let $\theta = \frac{\pi}{8}$

$\cos(2\theta) = 2\cos^2(\theta) - 1$

$\therefore \cos(\theta) = \sqrt{\frac{\cos(2\theta) + 1}{2}} = \frac{\sqrt{\sqrt{2} + 2}}{2}$

Now, $\sin(2\theta) = 2\cos(\theta)\sin(\theta)$

$\therefore \sin(\theta) = \frac{\sin(2\theta)}{2\cos(\theta)}$

Solving for $\sin(2\theta)$ and substituting in my answer for $\cos(\theta)$, I get:

$\frac{\sqrt{2}}{2\sqrt{\sqrt{2} + 2}}$ but I have an answer saying that $\sin(\frac{\pi}{8}) = \frac{\sqrt{2 - \sqrt{2}}}{2}$ and I couldn't seem to arrive at that.

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Found a related post here math.stackexchange.com/questions/66100/… –  ja72 Jul 30 '12 at 17:25
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1 Answer

up vote 8 down vote accepted

The trick is to rationalize the denominator by multiplying by the conjugate, along the following lines:

$$\begin{align}\frac{\sqrt{2}}{2\sqrt{\sqrt{2} + 2}}&=\frac{\sqrt{2}}{2\sqrt{\sqrt{2} + 2}}\cdot\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2-\sqrt{2}}}\\\&=\dfrac{\sqrt{2}\sqrt{2-\sqrt{2}}}{2 \sqrt {4-2}}=\dfrac{\sqrt{\not 2}\sqrt{2-\sqrt 2}}{2 \sqrt {\not 2} }\\\ &=\frac{\sqrt{2-\sqrt{2}}}{2}\end{align}$$

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I have beautified the post; hope it looks good now. +1. –  user21436 Apr 2 '12 at 1:56
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