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Math Help - Simplify trig equation using identities

  1. #1
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    Simplify trig equation using identities

    So I need to simplify a problem down to ten characters, and I got this far and now I'm stuck:

    2cos3t(3sint-4sin3t) +1

    I don't really know what to do with the 2cos3t... I feel like distributing it in would just make the problem even messier...

    Here's the original problem:
    2cos(3t)(sin(2t)cos(t)+cos2t* sqr(1-cos2t))+1

    Here's what I was able to do:
    I turned the (sin2tcost)
    into 2sintcost(cost)
    then into 2sintcos2t,
    then into 2sint(1-sin2t),
    then into 2sint-2sin3t.

    Then I turned cos2t*sqr(1-cos2t)
    into 1-2sin2t*sqrt(sin2t),
    then into 1-2sin2t(sint),
    then into sint-2sin3t.

    Then I added the two sides together to get 3sint-4sin3t

    Right now we're using reciprocal id, quotient id, pythagorean id, cofunction id, even/odd id, sum&difference formulas, half-angle formulas, double angle formulas, sum-to-product formulas


    Any help would be fantastic! Thanks!
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Simplify trig equation using identities

    Hi dftba28!

    I think you should have gone the other way.

    In your initial expression you have something that looks a lot like: sin2t cost + cos2t sint.
    This happens to be part of the sum formula: sin(2t + t)=sin2t cos t + cos2t sint.

    So instead of making the angles smaller, I think you should make them bigger.
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  3. #3
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    Re: Simplify trig equation using identities

    Hello, dftba28!

    ILikeSerena has great advice . . . did you follow it?


    \text{Simplify: }\:2\cos 3t(\sin2t\cos t+\cos2t\sqrt{1-\cos^2t})+1

    \text{We have: }\:2\cos 3t(\sin2t\cos t+\cos2t\underbrace{\sqrt{1-\cos^2t}}_{\text{This is }\sin t})+1

    . . . . . . . =\;2\cos3t\underbrace{(\sin2t\cos t + \cos2t\sin t)}_{\text{This is }\sin3t} + 1

    . . . . . . . =\;\underbrace{2\cos3t\sin3t}_{\text{This is }\sin6t} + 1

    . . . . . . . =\;\sin6t + 1
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  4. #4
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    Re: Simplify trig equation using identities

    Why all the trouble. we know that sin 3A = 3sinA-4sin^3A
    Thus we get 2cos3t(3sint-4sin3t) +1 = 2 cos 3t sin 3t + 1 = sin 6t + 1 [Because sin 2A = 2 sinA cosA]
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