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Math Help - hey one more trig question im stuck on

  1. #1
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    Question hey one more trig question im stuck on

    hey its tri compund angle formulae proof.

    Prove that 4sin(x+1/6Pi)sin(x-1/6pi) is identical to 3-4cos^2x

    just tried to solve it but really confused.


    thanks
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  2. #2
    Super Member Deadstar's Avatar
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    Haven't tried it myself but try using the identity

    \sin(u)\sin(v) = \frac{1}{2}[\cos(u-v) - \cos(u+v)]

    EDIT - Having re-read the question I dont think it'll work.
    Last edited by Deadstar; October 7th 2009 at 09:03 AM.
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  3. #3
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    Quote Originally Posted by Deadstar View Post
    Haven't tried it myself but try using the identity

    \sin(u)\sin(v) = \frac{1}{2}[\cos(u-v) - \cos(u+v)]

    EDIT - Having re-read the question I dont think it'll work.
    yep lol havent seen that idenity in my life
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  4. #4
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    Quote Originally Posted by elliegurl View Post
    yep lol havent seen that idenity in my life
    have you seen these two idenities
    \cos(c-d)=\cos c \cos d+\sin c \sin d \ \ .............(1)
    \cos(c+d)=\cos c \cos d-\sin c \sin d \ \ .............(2)
    on subtracting (2) from (1), you get
    \cos(c-d)-\cos(c+d)=2\sin c \sin d

    now LHS
    4\sin(x+\frac{1}{6}  \pi) \sin(x-\frac{1}{6}  \pi) = 2\{2\sin(x+\frac{1}{6}  \pi) \sin(x-\frac{1}{6}  \pi)\}
    = 2 \{ \cos \{(x+\frac{1}{6} \pi)-(x-\frac{1}{6}  \pi)\} - \cos \{(x+\frac{1}{6}  \pi)+(x-\frac{1}{6}  \pi)\} \}
    = 2 \{\cos \frac{\pi}{3} - \cos {2x}\}= 2 \{\frac{1}{2} - \cos {2x}\}
    = 1- 2 \cos {2x}  =1-2(2\cos^2x-1) =1-4\cos^2x+2
    =3-4\cos^2x=RHS
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