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Math Help - Proving Trig Equations

  1. #1
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    Proving Trig Equations

    Need help with these two, by tonight, if possible!

    1) {sin^2x - tanx}/{cos^2x - cotx} = tan^2x

    In words, sin squared x minus tan x all divided by cos squared x minus cotx equals tan squared x.

    2) {sin^3x + cos^3x}/{1 - 2cos^2x} = {secx - sinx}/{tanx - 1}

    In words, sin cubed x minus cos cubed x all divided by one minus two cos squared x equals sec x minus sin x all divided by tanx minus one.

    Thanks in advance!
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  2. #2
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    Rewrite #1 and make the cancellations.

    \frac{sin^{2}x-tanx}{cos^{2}x-cotx}

    sin^{2}x-tanx=tanx(sinxcosx-1)

    cos^{2}x-cotx=\frac{sinxcosx-1}{tanx}
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Blue Griffin View Post
    s tan squared x.

    2) {sin^3x + cos^3x}/{1 - 2cos^2x} = {secx - sinx}/{tanx - 1}

    In words, sin cubed x minus cos cubed x all divided by one minus two cos squared x equals sec x minus sin x all divided by tanx minus one.
    Hmmmm....

    How about
    \frac{sin^3(x) + cos^3(x)}{1 - 2cos^2(x)} = \frac{(sin(x) + cos(x))(sin^2(x) - sin(x)cos(x) + cos^2(x))}{1 - 2cos^2(x)}

    \frac{sin^3(x) + cos^3(x)}{1 - 2cos^2(x)} = \frac{(sin(x) + cos(x))(1 - sin(x)cos(x))}{1 - 2cos^2(x)}

    Divide both the numerator and denominator by cos^2(x). In the numerator divide each factor by cos(x):
    = \frac{(tan(x) + 1)(sec(x) - sin(x))}{sec^2(x) - 2}

    Now, sec^2(x) - 1 = tan^2(x), so
    = \frac{(tan(x) + 1)(sec(x) - sin(x))}{tan^2(x) - 1}

    = \frac{(tan(x) + 1)(sec(x) - sin(x))}{(tan(x) + 1)(tan(x) - 1)}

    = \frac{sec(x) - sin(x)}{tan(x) - 1}

    -Dan

    (Thank you. I enjoyed doing that one! )
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