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Math Help - trig derivative critical point

  1. #1
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    trig derivative critical point

    find the critical points of f(x)= 5sin^2x on [0,\pi]

    f'= 10sinxcosx

    I have a question when i get to this part.

     10sinxcosx=0

    When I set to zero, do I break into part then mutlipy the inverses of both sides?

    10sin = 0 and cosx=0
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  2. #2
    Super Member Quacky's Avatar
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    Re: trig derivative critical point

    Easier way (I think)
    Recall your double-angle formula

    We have:
    5\sin{2x}=0

    This gives us that 2x=n\pi immediately, where n is an integer.
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  3. #3
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    Re: trig derivative critical point

    Quote Originally Posted by delgeezee View Post
    find the critical points of f(x)= 5sin^2x on [0,\pi]

    f'= 10sinxcosx

    I have a question when i get to this part.

     10sinxcosx=0

    When I set to zero, do I break into part then mutlipy the inverses of both sides?

    10sin = 0 and cosx=0
    You don't MULTIPLY the inverse on both sides, you just take the inverse...

    But what you are saying is correct in essentials, so you now just need to find where sin(x) = 0 and cos(x) = 0.
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