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    hi guys, could someone help me with this question.

    Solve: sinƟ = 2sin (60-Ɵ) for 0≤Ɵ≤180

    thanks
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    Use the identity for \sin (A-B) on the RHS, collect like terms and divide through to put the equation in terms of \tan \theta.
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    beat me to it
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    You have to use  \sin(60-x) = \sin60 \cos x - \cos60 \sin x
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    (Original post by nuodai)
    Use the identity for \sin (A-B) on the RHS, collect like terms and divide through to put the equation in terms of \tan \theta.

    (Original post by Dr Grip)
    You have to use  \sin(60-x) = \sin60 \cos x - \cos60 \sin x

    Thanks for that, got it now

    I've just got one more question if you don't mind.

    how would i prove

    (sinA + cosA)(sin2A-cos2A) == sinA-cos3A
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    (Original post by tehonlyone)
    Thanks for that, got it now

    I've just got one more question if you don't mind.

    how would i prove

    (sinA + cosA)(sin2A-cos2A) == sinA-cos3A
    Expand brackets on the LHS. Then, on the RHS, write A=2A-A and 3A=2A+A and expand using the addition/subtraction formulae.
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    (Original post by nuodai)
    Expand brackets on the LHS. Then, on the RHS, write A=2A-A and 3A=2A+A and expand using the addition/subtraction formulae.
    I don't understand why you have to use the RHS. Surely you just have to use the LHS to get the RHS.
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    (Original post by tehonlyone)
    I don't understand why you have to use the RHS. Surely you just have to use the LHS to get the RHS.
    Surely what is RHS or LHS is irrelevant to proving the said conjecture?

    tanx = sinx/cosx just as equally as sinx/cosx = tanx.

    What I mean to say is RHS and LHS doesn't matter, it is merely a way of referencing an expression when proving equality. RHS = LHS just as much as LHS = RHS and you don't have to always start from the LHS to prove the RHS.
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    (Original post by tehonlyone)
    I don't understand why you have to use the RHS. Surely you just have to use the LHS to get the RHS.
    If you can show that the LHS and the RHS are both equal to the same thing, then they must be equal to each other. That's the idea with my method. You could expand the RHS, compare it with the expressions for \sin (A \pm 2A) and \cos (A \pm 2A), but that would require a bit of extra work which is ultimately unnecessary.

    For example, suppose you were asked to show that 1^2=\dfrac{2}{2}. Well 1^2=1 and \dfrac{2}{2}=1, so 1^2=\dfrac{2}{2}. You're doing exactly the same thing here.
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    Ok, finally figured it out, thanks guys
 
 
 
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