# Thread: Not sure if trig but this was given to me in trig class

1. ## Not sure if trig but this was given to me in trig class

This is an extra credit to verify two things being equal. My teacher told me that this was probably nothing we seen before so this might not be trig level.
The two things we had to verify equal was:
Sqr(2+sqr(3))
2

=

Sqr(2) + sqr(6)
4

Yes, it is square rooting 2 + sqr (3)
Also we were not allowed to square both sides cause we cannot assume that they are equal. So he said that this was extremely complicated.

2. ## Re: Not sure if trig but this was given to me in trig class

First of all this question has nothing to do with Trigonometry.
Anyways it's quite Elementary though,
$\displaystyle \frac{\sqrt{2+\sqrt{3}}}{2}=\frac{2\sqrt{2+\sqrt{3 }}}{4}=\frac{\sqrt{8+4\sqrt{3}}}{4}=\frac{\sqrt{8+ 2\sqrt{12}}}{4}=\frac{\sqrt{(\sqrt{6})^2+(\sqrt{2} )^2+2\sqrt{(2)(6)}}}{4}$
Now using $\displaystyle (a+b)^2=a^2+b^2+2ab$ we get,
$\displaystyle \frac{\sqrt{(\sqrt{6})^2+(\sqrt{2})^2+2\sqrt{(2)(6 )}}}{4}=\frac{{\sqrt{2}+\sqrt{6}}}{4}$

3. ## Re: Not sure if trig but this was given to me in trig class

Originally Posted by ignite
First of all this question has nothing to do with Trigonometry.
Anyways it's quite Elementary though,
$\displaystyle \frac{\sqrt{2+\sqrt{3}}}{2}=\frac{2\sqrt{2+\sqrt{3 }}}{4}=\frac{\sqrt{8+4\sqrt{3}}}{4}=\frac{\sqrt{8+ 2\sqrt{12}}}{4}=\frac{\sqrt{(\sqrt{6})^2+(\sqrt{2} )^2+2\sqrt{(2)(6)}}}{4}$
Now using $\displaystyle (a+b)^2=a^2+b^2+2ab$ we get,
$\displaystyle \frac{\sqrt{(\sqrt{6})^2+(\sqrt{2})^2+2\sqrt{(2)(6 )}}}{4}=\frac{{\sqrt{2}+\sqrt{6}}}{4}$
it doesn't ?

$\displaystyle \cos\left(\frac{\pi}{12}\right) = \sqrt{\frac{1 + \cos\left(\frac{\pi}{6}\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \frac{\sqrt{2+\sqrt{3}}}{2}$

$\displaystyle \cos\left(\frac{\pi}{12}\right) = \cos\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi} {4}\right) + \sin\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi} {4}\right) = \frac{\sqrt{2}+\sqrt{6}}{4}$

4. ## Re: Not sure if trig but this was given to me in trig class

Originally Posted by skeeter
it doesn't ?

$\displaystyle \cos\left(\frac{\pi}{12}\right) = \sqrt{\frac{1 + \cos\left(\frac{\pi}{6}\right)}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \frac{\sqrt{2+\sqrt{3}}}{2}$

$\displaystyle \cos\left(\frac{\pi}{12}\right) = \cos\left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{3}\right)\cos\left(\frac{\pi} {4}\right) + \sin\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi} {4}\right) = \frac{\sqrt{2}+\sqrt{6}}{4}$
Oh!I didn't think of it from that point of view