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    Need help with this question. Its on the attachment.
    I have done parts i)=PROVE, ii)= (root3)-(root2) and iii)=10.7, 79.3

    I can not do part iv). Hope you can help me.

    PS. Sorry its a bit small.
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    Use the identity

     cos(A) + cos(B) = 2 cos \frac{A+B}{2} cos \frac{A-B}{2}

    where

     \frac{A+B}{2} = \theta + 60
     \frac{A-B}{2} = \theta + 30

    Solve these simultaneously to find the value of A and B, then use this identitiy

     cos(u+v) = cosu cosv + sinu sinv

    You should get the result  2\theta = arcsin(k - cos(30))

    So make k - cos(30) outside the range of arcsin, ie k - cos(30) > 1 or k - cos(30) < 1

    sorry for rushed responce, gotta go - will check back later!
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    sqrt(3) - 2sin(2x) = k

    2sin(2x) = sqrt(3) - k

    sin(2x) = [sqrt(3) - k] / 2

    2x = arcsin [sqrt(3) - k] / 2

    When [sqrt(3) - k] / 2 is greater than 1 or less than -1, the RHS goes tits up and x (or theta) does not exist. Simply write:

    [sqrt(3) - k] / 2 > 1

    and

    [sqrt(3) - k] / 2 < - 1



    Solving for k gives:

    k < sqrt(3) - 2

    or

    k > sqrt(3) + 2
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    (Original post by spread_logic_not_hate)

     cos(u+v) = cosu cosv + sinu sinv
    Should it not be

     cos(u+v) = cosu cosv - sinu sinv

    For cos double angle formulae, sign inside bracket is OPPOSITE to the sign in the expansion.


    N'est-ce pas?
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    Should it not be

     cos(u+v) = cosu cosv - sinu sinv

    For cos double angle formulae, sign inside bracket is OPPOSITE to the sign in the expansion.
    Indeed it should - thank you!
 
 
 
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