# Trigonometry problem.

• Jul 13th 2008, 07:34 PM
MatteNoob
Trigonometry problem.
Find x, when:

$\arcsin(x) = \frac \pi 3 + \arctan(-\frac 13)$

I have come this far, but here it stops:

$x = \sin\left(\frac \pi 3 + \arctan(-\frac 13)\right)$

$x = \sin(\frac \pi 3) \cdot \cos(\arctan(-\frac 13)) + \cos(\frac \pi 3) \cdot \sin(\arctan(-\frac 13))$

$x = \frac{\sqrt 3}{2} \cdot \cos(\arctan(-\frac 13)) + \frac 12 \cdot \sin(\arctan(-\frac 13))$

$x = \frac{ \sqrt 3 \cdot \cos(\arctan(-\frac 13)) + \sin(\arctan(-\frac 13))}{2}$

Firstly, is this correct?
Secondly, what should I do with $\arctan(-\frac 13)$?

• Jul 13th 2008, 07:56 PM
Jhevon
Quote:

Originally Posted by MatteNoob
Find x, when:

$\arcsin(x) = \frac \pi 3 + \arctan(-\frac 13)$

I have come this far, but here it stops:

$x = \sin\left(\frac \pi 3 + \arctan(-\frac 13)\right)$

$x = \sin(\frac \pi 3) \cdot \cos(\arctan(-\frac 13)) + \cos(\frac \pi 3) \cdot \sin(\arctan(-\frac 13))$

$x = \frac{\sqrt 3}{2} \cdot \cos(\arctan(-\frac 13)) + \frac 12 \cdot \sin(\arctan(-\frac 13))$

$x = \frac{ \sqrt 3 \cdot \cos(\arctan(-\frac 13)) + \sin(\arctan(-\frac 13))}{2}$

Firstly, is this correct?
Secondly, what should I do with $\arctan(-\frac 13)$?

Looking good!

now, let $\theta = \arctan \bigg( \frac 13 \bigg)$ (so what you are looking for is $- \sin \theta$ and $\cos \theta$ -- since $\tan (-x) = - \tan x,~ \sin (-x) = - \sin x \mbox{, and } \cos (-x) = \cos x$)

$\Rightarrow \tan \theta = \frac 13$

thus we can draw a right-triangle, with an acute angle $\theta$, where the opposite side is of length 1 and the adjacent side is of length 3, since this is how we define the tangent ratio. By Pythagoras' theorem, the hypotenuse is $\sqrt{10}$

Now, recall that $\text{sine } = \frac {\text{opposite}}{\text{hypotenuse}}$ and $\text{cosine } = \frac {\text{adjacent}}{\text{hypotenuse}}$

i leave it to you to finish up
• Jul 13th 2008, 07:59 PM
Mathstud28
Quote:

Originally Posted by MatteNoob
Find x, when:

$\arcsin(x) = \frac \pi 3 + \arctan(-\frac 13)$

I have come this far, but here it stops:

$x = \sin\left(\frac \pi 3 + \arctan(-\frac 13)\right)$

$x = \sin(\frac \pi 3) \cdot \cos(\arctan(-\frac 13)) + \cos(\frac \pi 3) \cdot \sin(\arctan(-\frac 13))$

$x = \frac{\sqrt 3}{2} \cdot \cos(\arctan(-\frac 13)) + \frac 12 \cdot \sin(\arctan(-\frac 13))$

$x = \frac{ \sqrt 3 \cdot \cos(\arctan(-\frac 13)) + \sin(\arctan(-\frac 13))}{2}$

Firstly, is this correct?
Secondly, what should I do with $\arctan(-\frac 13)$?

As you noted $x=\sin\left(\frac{\pi}{3}+\arctan\left(\frac{-1}{3}\right)\right)$

And also you noted that $\sin\left(A+B\right)=\sin\left(A\right)\cos\left(B \right)+\sin\left(B\right)\cos\left(A\right)$

So we have

$x=\sin\left(\frac{\pi}{3}\right)\cos\left(\arctan\ left(\frac{-1}{3}\right)\right)+\sin\left(\arctan\left(\frac{-1}{3}\right)\right)\cos\left(\frac{\pi}{3}\right)$

Simplifying we get

$x=\frac{\sqrt{3}}{2}\cos\left(\arctan\left(\frac{-1}{3}\right)\right)+\frac{1}{2}\sin\left(\arctan\l eft(\frac{-1}{3}\right)\right)$

Now it can be shown that $\cos\left(\arctan(x)\right)=\frac{1}{\sqrt{1+x^2}}$

and $\sin\left(\arctan(x)\right)=\frac{x}{\sqrt{x^2+1}}$

So we have that

$x=\frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{1+\left(\f rac{-1}{3}\right)^2}}+\frac{1}{2}\frac{\left(\frac{-1}{3}\right)}{\sqrt{1+\left(\frac{-1}{3}\right)^2}}$

$~=\frac{\sqrt{3}}{2}\cdot\frac{3\sqrt{10}}{10}+\fr ac{1}{2}\cdot\frac{-\sqrt{10}}{20}$

$=\frac{\left(3\sqrt{3}-1\right)\sqrt{10}}{20}\quad\blacksquare$
• Jul 13th 2008, 08:14 PM
MatteNoob
To Jhevon and Mathstud28:

Thank you very much for the good and enlightening explanations you both gave me. I really appreciate it. Glad I found this forum which has latex support and everything. You can count on seeing me post again :)
• Jul 13th 2008, 08:17 PM
Mathstud28
Quote:

Originally Posted by MatteNoob
To Jhevon and Mathstud28:

Thank you very much for the good and enlightening explanations you both gave me. I really appreciate it. Glad I found this forum which has latex support and everything. You can count on seeing me post again :)