# Trigonometry homework problems

1. Feb 26, 2012

### hrach87

1. The problem statement, all variables and given/known data
I need to prove that

$\frac{\frac{1}{2} cot20^{o}-cos10^{o}}{\frac{1}{2}+sin10^{o}}=\frac{\sqrt{3}}{3}$

3. The attempt at a solution

I try to do it by this way

$\frac{\frac{1}{2} cot10^{o}-cos10^{o}}{\frac{1}{2}+sin10^{o}}=\frac{cos20^{o}-2cos10^{o}sin20^{o}}{(1+2sin10^{o})sin20^{o}}= \frac{ cos20^{o}-\frac{1}{2}-sin10^{o}}{(1+2sin10^{o})sin20^{o}}$

Last edited: Feb 26, 2012
2. Feb 26, 2012

### SammyS

Staff Emeritus
Re: Trigonometry

After taking some time, I figured out how to get that $\displaystyle 2\cos10^{o}\sin20^{o}=\frac{1}{2}+\sin10^{o}\,,$ using a product to sum identity.

Do a similar thing to your denominator:
$\displaystyle (1+2\sin10^{o})\sin20^{o}=\sin20^{o}+\sin10^{o}+ \sin30^{o}=\sin10^{o}+\sin20^{o}+\sin30^{o}$​
That's also the same as cos60° + cos70° + cos80° .

Your numerator can be written as -cos60° + sin70° - cos80° or equivalently, -sin10° + cos20° - sin30° .

I don't know if any of that helps.

Also remember that $\displaystyle \tan30^{o}=\cot60^{o}=\frac{\sqrt{3}}{3}$

Last edited: Feb 26, 2012