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Thread: Simplifying the cosine equation

  1. #1
    mx-
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    Question Simplifying the cosine equation

    How do i simplify the following:

    $\displaystyle sin(2*cos^{-1}(4x))$
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  2. #2
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    Hello, mx-!

    Simplify: .$\displaystyle \sin\left[2\cos^{-1}(4x)\right]$
    Recall that an inverse trig expression is just "some angle."

    So we have: .$\displaystyle \sin(2\theta)$, where: .$\displaystyle \theta \:=\:\cos^{-1}(4x)\quad\Rightarrow\quad \cos\theta \:=\:4x\;\;{\color{blue}[1]}$

    . . This means: .$\displaystyle \cos\theta \:=\:\frac{4x}{1} \:=\:\frac{adj}{hyp} $

    $\displaystyle \theta$ is in a right triangle with: $\displaystyle adj = 4x,\;hyp = 1$
    Using Pythagorus: .$\displaystyle opp \:=\:\sqrt{1-16x^2}$
    . . Hence: .$\displaystyle \sin\theta \:=\:\sqrt{1-16x^2}\;\;{\color{blue}[2]}$


    We want: .$\displaystyle \sin2\theta \;=\;2\sin\theta\cos\theta$

    Substitute [2] and [1]: .$\displaystyle \sin2\theta \;=\;2\left(\sqrt{1-16x^2}\right)(4x) $


    Therefore: .$\displaystyle \boxed{\sin\left[2\cos^{-1}(4x)\right] \;=\;8x\sqrt{1-16x^2}} $

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  3. #3
    Junior Member roy_zhang's Avatar
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    Quote Originally Posted by mx- View Post
    How do i simplify the following:

    $\displaystyle sin(2*cos^{-1}(4x))$
    Realize that $\displaystyle \cos^{-1}(4x)$ is an angle, whose $\displaystyle \cos$ value is $\displaystyle 4x$ providing that $\displaystyle -1\le 4x\le 1$. Mathematically, $\displaystyle \cos^{-1}(4x)=y \Leftrightarrow 4x=\cos(y)$

    Now use double angle formula:

    $\displaystyle \sin(2\cos^{-1}(4x))=2\sin(\cos^{-1}(4x))\cos(\cos^{-1}(4x))$

    where $\displaystyle \cos(\cos^{-1}(4x))=4x$ and $\displaystyle \sin(\cos^{-1}(4x))=\sqrt{1-16x^2}$

    hence $\displaystyle \sin(2\cos^{-1}(4x))=8x\sqrt{1-16x^2}$ for $\displaystyle \frac{-1}{4}\le x\le \frac{1}{4}$

    Roy
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