Thread: Itsa me again T.T

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..

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

Hello, Freaky-Person!

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

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


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

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


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