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Math Help - Compound Angle Formula

  1. #1
    Junior Member
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    Question Compound Angle Formula

    Hi everyone, hope you are all ok this evening.

    Please could someone help with the working out of the following questions...

    ________________________________________
    1. Using compound angle formula show that

    Sin(x+90 º) + Cos(x-180 º) = 0

    ________________________________________
    2. Using compound angle formula show that

    Sin(A-B) + Sin(A+B) = 2Sin A Cos B

    Regards
    Paul
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  2. #2
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    Quote Originally Posted by duckegg911 View Post
    Hi everyone, hope you are all ok this evening.

    Please could someone help with the working out of the following questions...

    ________________________________________
    1. Using compound angle formula show that

    Sin(x+90 º) + Cos(x-180 º) = 0

    ________________________________________
    2. Using compound angle formula show that

    Sin(A-B) + Sin(A+B) = 2Sin A Cos B

    Regards
    Paul

    im guessing when you say "compound" you mean Double Angle forumla? I guess in Canada they call it something different.
    You have to solve the right side of the equation. 2(Sina+cosb) = (sinacosb)+(sinacosb)
    then you have to simplify using the regular formulas i think

    I took this a few months ago it is something like this, im sure someone else can help more. Its a start though
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  3. #3
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    Hello, Paul!


    You are expected to know the Compound Angle Formulas:

    . . \sin(A \pm B) \;=\;\sin A\cos B \pm \sin B\cos A

    . . \cos(A \pm B) \;=\;\cos A\cos B \mp \sin A\sin B



    (1)\;\;\sin(x+90^o)+\cos(x-180^o) \:=\: 0
    \sin(x+90^o) + \cos(x-180^o)

    . . = \;\bigg[\sin x\cos90^o + \cos x\sin90^o\bigg] + \bigg[\cos x\cos180^o + \sin x\sin180^o\bigg]

    . . = \;\bigg[\sin x\cdot 0 + \cos x\cdot 1\bigg] + \bigg[\cos x(-1) + \sin x\cdot 0\bigg]

    . . = \;0 + \cos x - \cos x + 0

    . . =\;\;0




    (2)\;\;\sin(A-B) + \sin(A+B) \:=\:2\sin A\cos B
    We have: . \begin{array}{cccc}\sin(A-B) &=& \sin A\cos B - \sin B\cos A & {\color{blue}[1]} \\ \sin(A+B) &=& \sin A\cos B + \sin B\cos A & {\color{blue}[2]} \end{array}


    Add [1] and [2]: . \sin(A-B) + \sin(A+B) \;=\;2\sin A\cos B

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  4. #4
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    Smile

    Thanks very much for question 1...

    I fully understand how you did that now!

    In question 2 can you show me how you added the two equations together?

    Best Regards

    + Many Thanks
    Paul
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  5. #5
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    Question

    Can anyone expand on the soloution for example number two please?

    Regards
    Paul
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  6. #6
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    Please may I have assistance with this problem

    Using compound angle formula show that

    cos 2A = 1 - 2sin²A

    Many thanks
    Dan
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  7. #7
    Like a stone-audioslave ADARSH's Avatar
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    2
    (A+B) =cosAcosB - sinAsinB
    --------------------------
    Here A=B

    cos(2A) =cos^2A - sin^2A
    ---------------------------
    Do you remember that

    cos^2A +sin^2A =1

    this will give

    cos^2A = 1- sin^2A

    ----------

    Hence

    cosA = cos^2A -sin^2A = 1- sin^2A -sin^2A = 1-2sin^2A
    __________________________________________________ ____________



    sin(A+B) +sin(A-B) = (sinAcosB +cosAsinB )+ (sinAcosB-cosAsinB)

    sinAcosB +sinAcossB {\color{red}+cosAsinB-cosAsinB}= 2sinAcosB

    Things in red get cancelled
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