Exact values using trig identities

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• Apr 29th 2009, 11:30 AM
zweevu
Exact values using trig identities
How would I find the exact value of sin(pie/16) using trig identities?
• Apr 29th 2009, 12:04 PM
TheEmptySet
You will need the the identities

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

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

Note that

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

and

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

I hope this helps
• Apr 29th 2009, 12:18 PM
Soroban
Hello, zweevu!

We need two Double angle identities:

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

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

Quote:

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

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

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

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

And I'll let you simplify it . . .

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

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

• Apr 29th 2009, 04:27 PM
zweevu
Well that was simple. Thanks!