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Math Help - Proving/Solving trig equations

  1. #1
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    Post Proving/Solving trig equations

    1) Solve
    (sinx)/(cot^2x) - (sinx)/(cos^2x)

    2) Prove
    tan^2x - sin^2x = tan^2x*sin^2x

    3) Prove
    sin(x - π/2) = -cosx

    Thanks for any help!
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  2. #2
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    Quote Originally Posted by live_laugh_luv27 View Post
    1) Solve
    (sinx)/(cot^2x) - (sinx)/(cos^2x)

    2) Prove
    tan^2x - sin^2x = tan^2x*sin^2x

    3) Prove
    sin(x - π/2) = -cosx

    Thanks for any help!
    #1 can only be simplified ... it's not an equation.

    \sin{x}\left(\frac{1}{\cot^2{x}} - \frac{1}{\cos^2{x}}\right)

    \sin{x}(\tan^2{x} - \sec^2{x})

    \sin{x}[\tan^2{x} - (1 + \tan^2{x})] = -\sin{x}

    #2 change the right side to \tan^2{x}(1 - \cos^2{x}) ... work from there to get the left side.

    #3 use the difference identity for sine on the left side ...

    \sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b)<br />
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  3. #3
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    Smile

    Thanks so much..you've been really helpful
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  4. #4
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    Post one question..

    sorry, i just have one question.
    Could you please explain how you did #2? I can't get past the second step.
    How did you get tan^2x(1-cos2x) to equal tan^2x - sin^2x? After I distribute the tan^2x, I'm not sure what to do.
    Thanks!
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  5. #5
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    Quote Originally Posted by live_laugh_luv27 View Post
    sorry, i just have one question.
    Could you please explain how you did #2? I can't get past the second step.
    How did you get tan^2x(1-cos2x) to equal tan^2x - sin^2x? After I distribute the tan^2x, I'm not sure what to do.
    Thanks!
    change tan^2{x} into \frac{\sin^2{x}}{\cos^2{x}} , then distribute.
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  6. #6
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    great, thanks!
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