Itsa me again T.T

**Freaky-Person** · September 9th 2007, 07:00 PM

Yeah... I'll get professional help with this tomorrow if I can...

Until then... I love you guys XD

The question said to solve the following using the double angle formulas. but I have like no clue what the double angle formula possibly has to do with any of this..

$\sin (\frac{3\pi}{8})$

**Soroban** · September 9th 2007, 07:27 PM

Hello, Freaky-Person!

Solve the following using the double-angle formulas: . $\sin\left(\frac{3\pi}{8}\right)$

One of the formulas is: . $\sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}\quad\Rightarrow\quad\sin\theta \;=\;\pm\sqrt{\frac{1-\cos2\theta}{2}}$

Let $\theta \,=\,\frac{3\pi}{8}$
Then we have: . $\sin\left(\frac{3\pi}{8}\right) \;=\;\pm\sqrt{\frac{1 - \cos\frac{3\pi}{4}}{2}} \;=\;\pm\sqrt{\frac{1 - \left(\text{-}\frac{\sqrt{2}}{2}\right)}{2}} \;=\;\pm\sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}$

. . $= \;\pm\sqrt{\frac{2+\sqrt{2}}{4}} \;=\;\pm\frac{\sqrt{2+\sqrt{2}}}{2}$

Since $\frac{3\pi}{8}$ is in Quadrant 1, where sine is positive: . $\sin\left(\frac{3\pi}{8}\right) \;=\;\frac{\sqrt{2+\sqrt{2}}}{2}$

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