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Math Help - Identity Proofs - Simple Question Help

  1. #1
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    Identity Proofs - Simple Question Help

    Hi everyone!

    I just have an identity proof that I am stuggling with.

    Identity Proofs - Simple Question Help-lesson3hw_q2k.gif

    Could anyone give me a hand?

    Thanks!
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Identity Proofs - Simple Question Help

    That's not an identity.
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  3. #3
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    Re: Identity Proofs - Simple Question Help

    No wonder it was so hard. Didn't even copy the question correctly haha. Thanks!
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  4. #4
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    Re: Identity Proofs - Simple Question Help

    Quote Originally Posted by Higg View Post
    Hi everyone!

    I just have an identity proof that I am stuggling with.

    Click image for larger version. 

Name:	lesson3hw_Q2k.gif 
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ID:	25724

    Could anyone give me a hand?

    Thanks!
    It's not an identity, and there are not even any real solutions for x.

    \displaystyle \begin{align*} \tan^4{x} - \sec^4{x} &= 2\sin^2{x} \\ \left( \tan^2{x} - \sec^2{x} \right)\left( \tan^2{x} + \sec^2{x} \right) &= 2\sin^2{x} \\ -1\left( \tan^2{x} + \sec^2{x} \right) &= 2\sin^2{x} \\ -\left( \frac{\sin^2{x}}{\cos^2{x}} + \frac{1}{\cos^2{x}} \right) &= 2\sin^2{x} \\ -\left( \frac{\sin^2{x} + 1 }{\cos^2{x}} \right) &= 2\sin^2{x} \\ \frac{-\left( \sin^2{x} + 1 \right) }{1 - \sin^2{x}} &= 2\sin^2{x} \\ -\left( \sin^2{x} + 1 \right) &= 2\sin^2{x} \left( 1 - \sin^2{x} \right) \\ -\sin^2{x} - 1 &= 2\sin^2{x} - 2\sin^4{x} \\ 2\sin^4{x} - 3\sin^2{x} - 1 &= 0 \\ \sin^2{x} &= \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-1)}}{2(2)} \\ \sin^2{x} &= \frac{3 \pm \sqrt{13}}{4} \\ \sin^2{x} &= \frac{3 + \sqrt{13}}{4} \textrm{ if we are assuming } x \textrm{ is real} \\ \sin{x} &= \pm \frac{\sqrt{{3 + \sqrt{13}} }}{2} \end{align*}

    Since \displaystyle \begin{align*} \left| \frac{\sqrt{3 + \sqrt{13}}}{2} \right| > 1 \end{align*}, there are not any values of x which satisfy this equation.

    Of course, complex numbers are another story...
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