Trigonometry

• March 16th 2013, 04:54 AM
endri
Trigonometry
can u help with these 2 exercises?: 1) which are the solutions to the equation tanx=2cotx,in the interval -180<x<180?...2) Which is the value of arcsin1/2-arcsin(-1/2).. i will be glad if you could help me(Rofl)
• March 16th 2013, 05:36 AM
Prove It
Re: Trigonometry
\displaystyle \begin{align*} \tan{(x)} &= 2\cot{(x)} \\ \frac{\sin{(x)}}{\cos{(x)}} &= \frac{2\cos{(x)}}{\sin{(x)}} \\ \sin^2{(x)} &= 2\cos^2{(x)} \\ 1 - \cos^2{(x)} &= 2\cos^2{(x)} \\ 1 &= 3\cos^2{(x)} \\ \frac{1}{3} &= \cos^2{(x)} \\ \pm \frac{1}{\sqrt{3}} &= \cos{(x)} \\ x &= \left\{ \arccos{\left( \frac{1}{\sqrt{3}} \right)}, \pi - \arccos{\left( \frac{1}{\sqrt{3}} \right)}, \pi + \arccos{\left( \frac{1}{\sqrt{3}} \right)} , 2\pi - \arccos{\left( \frac{1}{\sqrt{3}} \right)} \right\} + 2\pi \, n \textrm{ where } n \in \mathbf{Z} \\ x &= \left\{ \arccos{\left( \frac{1}{\sqrt{3}} \right)} - \pi , -\arccos{\left( \frac{1}{\sqrt{3}} \right)} , \arccos{\left( \frac{1}{\sqrt{3}} \right)} , \pi - \arccos{\left( \frac{1}{\sqrt{3}} \right)} \right\} \textrm{ in the interval specified} \end{align*}

Edit: I naturally put the answer in radians when it should have been degrees. You can fix that :)
• March 16th 2013, 08:15 AM
Soroban
Re: Trigonometry
Hello, endri!

Quote:

$\text{1) What are the solutions to the equation: }\,\tan x\,=\,2\cot x,\,\text{ on the interval }(\text{-}180^o,180^o)$

We have: . $\tan x \:=\:\frac{2}{\tan x} \quad\Rightarrow\quad \tan^2x \:=\:2 \quad\Rightarrow\quad \tan x \:=\:\pm\sqrt{2}$

Therefore: . $x \;\approx\;\{\text{-}125.3^o,\:\text{-}54.7^o,\:54.7^o,\:125.3^o\}$

Quote:

$\text{2) What is the value of: }\,\arcsin(\tfrac{1}{2})-\arcsin(\text{-}\tfrac{1}{2})$

Using principal values: . $\begin{Bmatrix}\arcsin(\frac{1}{2}) &=& \frac{\pi}{6} \\ \\[-4mm] \arcsin(\text{-}\frac{1}{2}) &=& \text{-}\frac{\pi}{6} \end{Bmatrix}$

Therefore: . $\arcsin(\tfrac{1}{2}) - \arcsin(\text{-}\tfrac{1}{2}) \;=\;\frac{\pi}{6} - \left(\text{-}\frac{\pi}{6}\right) \;=\;\frac{\pi}{3}$
• March 16th 2013, 12:27 PM
endri
Re: Trigonometry
thanks for your help man ;)