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Thread: Trig Proof

  1. #1
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    Trig Proof

    Required to Prove the following trig identity:

    (cos2x)^2 + (sin2x)^2 = 1
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    Quote Originally Posted by skeske1234 View Post
    Required to Prove the following trig identity:

    (cos2x)^2 + (sin2x)^2 = 1

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

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


    add'em up ...

    $\displaystyle \cos^4{x} + 2\sin^2{x}\cos^2{x} + \sin^4{x} = (\cos^2{x} + \sin^2{x})^2 = (1)^2 = 1$
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  3. #3
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    (I don't know if you can prove it this way.)

    Draw a unit circle. Plot a point P(x, y) on the unit circle. Drop a perpendicular from point P to the x-axis. You formed a right triangle. Then
    $\displaystyle \cos \theta = \frac{\text{adj}}{\text{hyp}} = \frac{x}{1} = x$
    and
    $\displaystyle \sin \theta = \frac{\text{opp}}{\text{hyp}} = \frac{y}{1} = y$

    According to the Pythagorean theorem,
    $\displaystyle x^2 + y^2 = 1$,
    so plug in $\displaystyle \cos \theta$ for x and $\displaystyle \sin \theta$ for y and you get
    $\displaystyle \cos^2 \theta + \sin^2 \theta = 1$.


    01

    EDIT: Wow, I completely misread the problem. Didn't see the 2 in front of the x.
    Last edited by yeongil; Jun 18th 2009 at 08:41 PM.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by skeske1234 View Post
    Required to Prove the following trig identity:

    (cos2x)^2 + (sin2x)^2 = 1
    I don't know why everybody is making this problem complicated.

    Set $\displaystyle y=2x$. Then

    $\displaystyle (\cos2x)^2 + (\sin2x)^2 = \cos^2y+\sin^2y =1$

    !!
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