Trigonometry homework problems

[hrach87]

Homework Statement

I need to prove that

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

The Attempt at a Solution

I try to do it by this way

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

Please help, it is very urgent.

 

[SammyS]

Homework Statement

I need to prove that

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

The Attempt at a Solution

I try to do it by this way

Note: The cot in the first expression below should have a argument of 20°, not 10°.
$\displaystyle\frac{\frac{1}{2} \cot10^{o}-\cos10^{o}}{\frac{1}{2}+\sin10^{o}}=\frac{\cos20^{o}-2\cos10^{o}\sin20^{o}}{(1+2\sin10^{o})\sin20^{o}}= \frac{ \cos20^{o}-\frac{1}{2}-\sin10^{o}}{(1+2\sin10^{o})\sin20^{o}}$

Please help, it is very urgent.

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