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Math Help - precalc problem help

  1. #1
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    precalc problem help

    show that cos 3(theta) = 4cos^3(theta)-3cos(theta)



    where do i even start?
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  2. #2
    GAMMA Mathematics
    colby2152's Avatar
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    Is this want you would like to prove?

    cos (3\theta) = 4cos^3(\theta)-3cos(\theta)
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  3. #3
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    yess... thanks for making it the way it should!
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  4. #4
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by ohhhh View Post
    show that cos 3(theta) = 4cos^3(theta)-3cos(theta)



    where do i even start?
    \cos{3\theta}\ =\ \cos{(2\theta+\theta)}

    Now apply the formula \cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B} as well as the double-angle formula for sine and for cosine.
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  5. #5
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    ok, so let me see.

    cos (2a+a) = cos2(a)*cos(a) - sin2(a)*sin(a)
    = cos^2a-sin^2a - 2sin(a)*cos(a)*sin(a)
    = cos^2a-sin^2a - 2sin^2(a)*cos(a)

    ...is this right so far?
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  6. #6
    Math Engineering Student
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    Quote Originally Posted by ohhhh View Post
    show that cos 3(theta) = 4cos^3(theta)-3cos(theta)
    Complex numbers also work:

    \cos 3\theta = \text{Re} \left( {\cos \theta + i\sin \theta} \right)^3 = \text{Re} \left( {\cos ^3 \theta + 3i\sin \theta\cos ^2 \theta - 3\sin ^2 \theta\cos \theta - i\sin ^3 \theta} \right).

    So \cos 3\theta = \cos ^3 \theta - 3\cos \theta(1 - \cos ^2 \theta) = 4\cos ^3 \theta - 3\cos \theta\,\blacksquare
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  7. #7
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    thanks a lot for your help guys. got it down!
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  8. #8
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    cos (2a+a) = cos2(a)*cos(a) - sin2(a)*sin(a)
    right

    = cos^2a-sin^2a - 2sin(a)*cos(a)*sin(a)
    You have lost a cos(a). It should read
    \cos (a)(\cos ^2(a)-\sin ^2(a))-2\sin (a) \cos (a) \sin (a)
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