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Math Help - Trig Derivative

  1. #1
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    Question Trig Derivative

    I have the problem
    Find the derivative
    r=sin^2 3*pi*theta

    First off, since there are no parentheses in the problem, I don't know if this equals sin(3*pi*theta)^2. I tried that and got a very long answer.
    I know the sin will change to cos, but how many of these terms are being operated on by the function of sin?

    The book answer was 3pi sin 6pi theta

    Thanks.
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  2. #2
    Senior Member
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    So I'm assuming your book writes it like this:

    r=\sin^2 \, 3 \pi \theta

    In general it should be interpreted as

    r=\sin^2 (3 \pi \theta) = \left[ \sin (3 \pi \theta) \right] ^2

    but I definitely agree that the notation is confusing and should be avoided without parentheses.

    Would you mind showing us what answer you got? I should point out that the book is using a trig identity to get to the final answer.

    In particular, notice that

    2 \sin u \cos u = \sin (2u)
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by dwatkins741 View Post
    I have the problem
    Find the derivative
    r=sin^2 3*pi*theta

    First off, since there are no parentheses in the problem, I don't know if this equals sin(3*pi*theta)^2. I tried that and got a very long answer.
    I know the sin will change to cos, but how many of these terms are being operated on by the function of sin?

    The book answer was 3pi sin 6pi theta

    Thanks.
    It almost certainly means find:

    \frac{d}{d\theta} [\sin(3\pi\theta)]^2

    which should not be much of a problem, you need the chain rule.

    The less likely alternative (thought it is what the notation means in other contexts) is:

    \frac{d}{d\theta} \sin[\sin(3\pi\theta)]

    which again is a chain rule problem.

    CB
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