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Thread: Need help simplifying a trigonometric expression

  1. #1
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    Need help simplifying a trigonometric expression

    Cot(2arcsinx)

    Thanks in advance
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Need help simplifying a trigonometric expression

    Note that $\displaystyle \cot(2\arcsin x) = \frac{1}{\tan(2\arcsin x)}$

    From this point you can use the formula $\displaystyle \tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$ and $\displaystyle \tan(\arcsin x) = \frac{\sin(\arcsin x)}{\cos(\arcsin x)} = \frac{x}{\sqrt{1-x^2}}$
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  3. #3
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    Re: Need help simplifying a trigonometric expression

    Hello, nubshat!

    $\displaystyle \text{Simplify: }\cot(2\arcsin x)$

    Let $\displaystyle \theta \,=\,\arcsin x\quad\Rightarrow\quad \sin\theta \,=\,x$

    That is: .$\displaystyle \sin\theta \:=\:\frac{x}{1}\:=\:\frac{opp}{hyp}$

    $\displaystyle \theta$ is in a right triangle with: $\displaystyle opp = x$ and $\displaystyle hyp = 1.$

    Pythagorus says: .$\displaystyle adj \,=\,\sqrt{1-x^2}$

    Hence: .$\displaystyle \cot\thet \,=\,\frac{adj}{opp} \,=\,\frac{\sqrt{1-x^2}}{x}$


    Identity: .$\displaystyle \cos2\theta \:=\:\frac{\cot^2\theta - 1}{2\cot\theta}$

    We have: .$\displaystyle \cot2\theta \:=\:\frac{\left(\frac{\sqrt{1-x^2}}{x}\right)^2-1}{2\left(\frac{\sqrt{1-x^2}}{x}\right)} \;=\; \frac{\frac{1-x^2}{x^2} - 1}{\frac{2\sqrt{1-x^2}}{x}}$

    Multiply by $\displaystyle \tfrac{x^2}{x^2}\!:\;\;\frac{x^2}{x^2}\cdot\frac{ \frac{1-x^2}{x^2} - 1}{\frac{2\sqrt{1-x^2}}{x}} \;=\;\frac{1-x^2-x^2}{2x\sqrt{1-x^2}} $

    Therefore: .$\displaystyle \cos2\theta \;=\;\frac{1-2x^2}{2x\sqrt{1-x^2}}$

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  4. #4
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    Re: Need help simplifying a trigonometric expression

    Hi Soroban,

    a little attention error in the last row:

    cot (2arcsin(x)) = ....
    and not
    cos(2arcsin(x))=...
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