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Math Help - trig identity

  1. #1
    Senior Member polymerase's Avatar
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    trig identity

    prove that sin(3x)=3sinx-4sin^3x
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Use the fact \sin3x=\sin(x+2x)

    Then use \sin(\alpha+\beta) formula.

    Of course remember \sin2x and \cos2x
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  3. #3
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    Hello, polymerase!

    Here it is . . . in baby-steps.


    Prove that: . \sin(3x)\:=\:3\sin x - 4\sin^3x
    We're expected to know these identities:

    . . \begin{array}{ccc}\sin(A + B) & = & \sin A\cos B + \sin B\cos A \\ \sin2x & = & 2\sin x\cos x \\ \cos2x & = & 2\cos^2x - 1 \\ \cos^2\!x & = & 1 - \sin^2\!x\end{array}


    \sin3x \:=\:\sin(2x+x)

    . . . . =\;\sin2x\cos x + \sin x\cos2x

    . . . . = \;(2\sin x\cos x)\!\cdot\cos x + \sin x(2\cos^2\!x-1)

    . . . . = \;2\sin x\cos^2\!x + 2\sin x\cos^2\!x - \sin x

    . . . . = \;4\sin x\cos^2\!x - \sin x

    . . . . = \;4\sin x(1-\sin^2\!x) - \sin x

    . . . . = \;4\sin x - 4\sin^3\!x - \sin x

    . . . . = \;3\sin x - 4\sin^3\!x

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  4. #4
    MHF Contributor red_dog's Avatar
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    Another approach:

    \cos 3x+i\sin 3x=(\cos x+i\sin x)^3=
    =\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x
    Identifying the real parts and the imaginary parts we have
    \cos 3x=\cos^3x-3\cos x\sin^2x=\cos^3x-3\cos x(1-\cos^2x)=4\cos^3x-3\cos x
    \sin 3x=3\cos^2x\sin x-\sin^3x=3(1-\sin^2x)\sin x-\sin^3x=3\sin x-4\sin^3x
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by red_dog View Post
    Another approach:

    \cos 3x+i\sin 3x=(\cos x+i\sin x)^3=
    =\cos^3x+3i\cos^2x\sin x-3\cos x\sin^2x-i\sin^3x
    Identifying the real parts and the imaginary parts we have
    \cos 3x=\cos^3x-3\cos x\sin^2x=\cos^3x-3\cos x(1-\cos^2x)=4\cos^3x-3\cos x
    \sin 3x=3\cos^2x\sin x-\sin^3x=3(1-\sin^2x)\sin x-\sin^3x=3\sin x-4\sin^3x
    nice! how do you come up with this stuff?!
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    nice! how do you come up with this stuff?!
    This is actually a well known trick.

    To find \sin nx \mbox{ and }\cos nx.
    Use de Moiver's Theorem.

    So, (\cos x + i \sin x)^n = \cos nx + i \sin nx.
    Apply Binomial Expansion on RHS.
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