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Math Help - differential trig problem

  1. #1
    Junior Member
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    differential trig problem


    Show that d/dx (1/5 sin^5 x - 1/7 sin^7 x) = sin^4 x cos^3 x
    Also, I typed out this question in word using equation editor and tried to paste it in without luck (just thought it would be easier for helpers?). I'm not sure how to paste in the actual question, or, is the above acceptable?
    Thanks all the same.
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  2. #2
    Senior Member Spec's Avatar
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    Here's a good thread about the math code this forum uses: http://www.mathhelpforum.com/math-he...-tutorial.html

    On to your problem:

    sin^4xcos^3x = sin^4x(1-sin^2x)cos x = cosx(sin^4x-sin^6x)

    The last part is easy to integrate since \int cosxsin^{n-1}x dx = \frac{1}{n}sin^nx
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by slaypullingcat View Post

    Show that d/dx (1/5 sin^5 x - 1/7 sin^7 x) = sin^4 x cos^3 x
    Also, I typed out this question in word using equation editor and tried to paste it in without luck (just thought it would be easier for helpers?). I'm not sure how to paste in the actual question, or, is the above acceptable?
    Thanks all the same.
    Show that: \frac{d}{dx}\left[\frac{1}{5}\sin^5(x)-\frac{1}{7}\sin^7(x)\right] = \sin^4(x)\cos^3(x)

    \frac{d}{dx}\left[\frac{1}{5}\sin^5(x)\right] = 5\cdot\frac{1}{5}\sin^4(x)\cdot\cos(x) = \sin^4(x)\cos(x)

    \frac{d}{dx}\left[\frac{1}{7}\sin^7(x)\right] = 7\cdot\frac{1}{7}\sin^6(x)\cdot\cos(x) = \sin^6(x)\cos(x)

    So the derivative is \sin^4(x)\cos(x)-\sin^6(x)\cos(x) = \sin^4(x)\cos(x)\cdot(1-\sin^2(x)) = \sin^4(x)\cos(x)\cdot\cos^2(x) = \sin^4(x)\cos^3(x)
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