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Thread: Verify tanx^2 = ((1 - cos2x) / (1 + cos2x) using dbl. angle

  1. #1
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    Verify tanx^2 = ((1 - cos2x) / (1 + cos2x) using dbl. angle

    Verify tanx^2 = ((1 - cos2x) / (1 + cos2x) using double angle

    I do not know where to begin with this one. Any help would be appreciated.

  2. #2
    Elite Member stapel's Avatar
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    Quote Originally Posted by smsmith
    I do not know where to begin with this one.
    Maybe... use the double-angle identities...?

    Eliz.

  3. #3
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    First of all it has to be [tex]\left[ {\tan (x)} \right]^2 = \tan ^2 (x).[/tex]
    [tex]\begin{array}{rcl}
    \frac{{1 - \cos (2x)}}{{1 + \cos (2x)}} & = & \frac{{1 - \left[ {1 - 2\sin ^2 (x)} \right]}}{{1 + \left[ {2\cos ^2 (x) - 1} \right]}} \\
    & = & \frac{{2\sin ^2 (x)}}{{2\cos ^2 (x)}} \\
    \end{array}[/tex]
    “A professor is someone who talks in someone else’s sleep”
    W.H. Auden

  4. #4
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    Re: Verify tanx^2 = ((1 - cos2x) / (1 + cos2x) using dbl. an

    Hello, smsmith!

    Verify: [tex]\,\tan^2x \:= \:\frac{1\,-\,\cos2x}{1\,+\,\cos2x}[/tex] using double-angle identities.

    You're expected to know these identities:

    . . [tex]\sin^2x \:=\:\frac{1\,-\,\cos2x}{2}\;\;\;\;\;\cos^2x \:=\:\frac{1\,+\,\cos2x}{2}[/tex]


    We have: [tex]\:\tan^2x \:=\:\L\left(\frac{\sin x}{\cos x}\right)^2\:=\:\frac{\sin^2x}{\cos^2x} \:=\:\frac{\frac{1\,-\,\cos2x}{2}}{\frac{1\,+\,\cos2x}{2}} \:=\:\frac{1\,-\,\cos2x}{1\,+\,\cos2x}[/tex]

    I'm the other of the two guys who "do" homework.

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