# Thread: Exact values using trig identities

1. ## Exact values using trig identities

How would I find the exact value of sin(pie/16) using trig identities?

2. You will need the the identities

$\displaystyle \sin\left( \frac{\theta}{2}\right)=\pm \sqrt{\frac{1-\cos(\theta)}{2}}$

$\displaystyle \cos\left( \frac{\theta}{2}\right)=\pm \sqrt{\frac{1+\cos(\theta)}{2}}$

Note that

$\displaystyle \sin\left( \frac{\pi}{16}\right)=\sqrt{\frac{1-\cos\left( \frac{\pi}{8}\right)}{2}}$

and

$\displaystyle \cos\left( \frac{\pi}{8}\right)= \sqrt{\frac{1+\cos(\frac{\pi}{4})}{2}}$

I hope this helps

3. Hello, zweevu!

We need two Double angle identities:

. . $\displaystyle \sin\theta \:=\:\sqrt{\frac{1-\cos2\theta}{2}}$

. . $\displaystyle \cos\theta \:=\:\sqrt{\frac{1+\cos2\theta}{2}}$

Find the exact value of $\displaystyle \sin\left(\frac{\pi}{16}\right)$ using trig identities.

$\displaystyle \sin\left(\frac{\pi}{16}\right) \;=\;\sqrt{\frac{1 - \cos(\frac{\pi}{8})}{2}}$

. . . $\displaystyle =\;\sqrt{\frac{1-\sqrt{\dfrac{1 + \cos(\frac{\pi}{4})}{2}}}{2}}$

. . . . $\displaystyle =\;\sqrt{\frac{1-\sqrt{\dfrac{1 + \frac{\sqrt{2}}{2} }{2}}}{2}}$

And I'll let you simplify it . . .

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Well, okay . . . I got: .$\displaystyle \frac{\sqrt{2- \sqrt{2 + \sqrt{2}}}}{2}$

4. Well that was simple. Thanks!