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Math Help - another question!

  1. #1
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    another question!

    sin(x)/sin(y)= 1/2 cos(x)/cos(y)= 3

    prove sin(x+y) = (7/3)sin(x)cos(x)

    I dont know what to do with the fractions, do i solve for one and then try and put it in the equation or something?

    like sin(x)= (1/2)sin(y)
    Last edited by OmegaCenturion; April 3rd 2010 at 09:54 AM.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by OmegaCenturion View Post
    sin(x)/sin(y)= 1/2 cos(x)/cos(y)= 3

    prove sin(x+y) = (7/3)sin(x)cos(x)

    I dont know what to do with the fractions, do i solve for one and then try and put it in the equation or something?
    Um, yes I think so.

    like sin(x)= (1/2)sin(y)
    Yes: Basically, the given (two) equations allow you to express sin(x) as a multiple of sin(y), and cos(x) as a multiple of cos(y). Thus, by applying the addition formula to sin(x+y), i.e. sin(x+y)=sin(x) cos(y)+cos(x) sin(y), you can reduce both sides of the equality that you have to prove to a certain multiple of sin(y) and cos(y). Therefore you only have to check whether, in fact, both sides work out to be the exact same multiple of sin(y) and cos(y).
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  3. #3
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    Hello, OmegaCenturion!

    Given: . \begin{Bmatrix}\dfrac{\sin x}{\sin y} &=& \frac{1}{2} & [1] \\ \\[-3mm]  \dfrac{\cos x}{\cos y}&=& 3 & [2] \end{Bmatrix}

    Prove: . \sin(x+y) \:=\: \tfrac{7}{3}\sin x \cos x

    From [1] we have: . \sin y \:=\:2\sin x .(a)

    From [2] we have: . \cos y \:=\:\tfrac{1}{3}\cos x .(b)


    We know that: . \sin(x+y) \;=\;\sin x\cos y + \cos x\sin y .(c)


    Substitute (a) and (b) into (c):

    . . \sin(x+y) \;=\;\sin x\left(\tfrac{1}{3}\cos x\right) + \cos x\left(2\sin x\right)

    . . . . . . . . . =\; \tfrac{1}{3}\sin x\cos x + 2\sin x\cos x

    . . . . . . . . . =\;\tfrac{7}{3}\sin x\cos x

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