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Thread: Trigonometry problem.

  1. #1
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    Trigonometry problem.

    Find x, when:

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

    I have come this far, but here it stops:

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

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

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

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

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

    In advance: Thank you for your assistance. :]
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by MatteNoob View Post
    Find x, when:

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

    I have come this far, but here it stops:

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

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

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

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

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

    In advance: Thank you for your assistance. :]
    Looking good!

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

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

    thus we can draw a right-triangle, with an acute angle $\displaystyle \theta$, 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 $\displaystyle \sqrt{10}$

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

    i leave it to you to finish up
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by MatteNoob View Post
    Find x, when:

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

    I have come this far, but here it stops:

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

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

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

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

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

    In advance: Thank you for your assistance. :]
    As you noted $\displaystyle x=\sin\left(\frac{\pi}{3}+\arctan\left(\frac{-1}{3}\right)\right)$

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

    So we have

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

    Simplifying we get

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

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

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

    So we have that

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

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

    $\displaystyle =\frac{\left(3\sqrt{3}-1\right)\sqrt{10}}{20}\quad\blacksquare$
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  4. #4
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    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
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by MatteNoob View Post
    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
    Glad to hear it!
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