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Math Help - Simplifying the cosine equation

  1. #1
    mx-
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    Question Simplifying the cosine equation

    How do i simplify the following:

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

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

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

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

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


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

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


    Therefore: . \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:

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

    Now use double angle formula:

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

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

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

    Roy
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