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Math Help - Deceptively difficult trig problem

  1. #1
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    Deceptively difficult trig problem

    I'm not sure what I'm missing here:
    Solve a\sin(2\theta)-b\cos(2\theta)=\sin\theta for θ in terms of a and b.
    Every attempt I've made has gotten rather messy.

    --Kevin C.
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  2. #2
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    Quote Originally Posted by TwistedOne151 View Post
    I'm not sure what I'm missing here:
    Solve a\sin(2\theta)-b\cos(2\theta)=\sin\theta for θ in terms of a and b.
    Every attempt I've made has gotten rather messy.

    --Kevin C.
    Hi Kevin,

    Tell us your answer. I have a nagging suspicion that you have the right answer
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  3. #3
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    I get to 4(a^2+b^2)\cos^4\theta-4a\cos^3\theta+(1-4(a^2+b^2))\cos^2\theta+4a\cos\theta+(b^2-1)=0, a quartic equation for \cos\theta. Attempting to solve that via the standard "double completing the square" method led to a horrible cubic.

    --Kevin C.
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  4. #4
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    Quote Originally Posted by TwistedOne151 View Post
    I'm not sure what I'm missing here:
    Solve a\sin(2\theta)-b\cos(2\theta)=\sin\theta for θ in terms of a and b.
    Every attempt I've made has gotten rather messy.

    --Kevin C.
    Do you have reason to believe a 'nice' general solution exists .....?
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  5. #5
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    I think a lot of times as math students we want a nice or "pretty" if you will solution to problems. Sometimes though, we get solutions that look messy and automatically think they must be wrong.

    I do this all the time.

    Eventually, you will learn to trust and not doubt yourself so much!
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