# determining Sin x as Cos x = sqrt 3/2

• Sep 6th 2008, 04:35 AM
Meowet
determining Sin x as Cos x = sqrt 3/2
Hello I need help in determining Sin x as Cos x = sqrt 3/2.

What should I know to be able to do this. and how?
I would like a fully detailed explanation if possible.
• Sep 6th 2008, 06:54 AM
Soroban
Hello, Meowet!

Quote:

$\displaystyle \text{If }\cos x = \frac{\sqrt{3}}{2}\text{, find }\sin x$

We have: .$\displaystyle \cos x \:=\:\frac{\sqrt{3}}{2} \:=\:\frac{adj}{hyp}$

$\displaystyle x$ is a in a right triangle with: .$\displaystyle adj = \sqrt{3},\;hyp = 2$

The triangle looks like this:
Code:

                        *                     *  *               2  *    *               *        *             *          *         * x            *       *  *  *  *  *  *  *               √3

Pythagorus says: .$\displaystyle \text{(opp)}^2 + \text{(adj)}^2 \:=\:\text{(hyp)}^2$

So we have: .$\displaystyle \text{(opp)}^2 + (\sqrt{3})^2 \:=\:2^2 \quad\Rightarrow\quad \text{(opp)}^2 + 3 \:=\:4$

. . Hence: .$\displaystyle \text{(opp)}^2 \:=\:1 \quad\Rightarrow\quad \text{opp} \:=\:1$

Therefore: .$\displaystyle \sin x \;=\;\frac{opp}{hyp} \;=\;\frac{1}{2}$