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    Hi.
    https://983c9f06eb1f75af6e83364e092b...20C4%20OCR.pdf
    How is question 7(i) done.
    I cant see how to get two equations.
    Thanks
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    (Original post by SamuelN98)
    Hi.
    https://983c9f06eb1f75af6e83364e092b...20C4%20OCR.pdf
    How is question 7(i) done.
    I cant see how to get two equations.
    Thanks
    Expand your brackets to (in gonna use x to replace theta)
    Asinx + Acosx+Bcosx-Bsinx=4sinx
    So you have 2 equations here, one in terms of cosx and one in terms of sinx
    Ainx - Bsinx=4sinx
    Acosx+Bcosx=0

    Therefore you get A-B=4, A+B=0
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    IF you expand it out you get  (A-B)\sin \theta + (A+b)\cos \theta .
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    You could also note that
    B(\cos\theta - \sin\theta) = 0 when \theta = 45^{\circ}

    Therefore
    A(\sin{45} + \cos{45})=4\sin{45}
    As such A = 2

    You can then get:
    B(\cos\theta - \sin\theta)=2\sin\theta - 2\cos\theta

    Therefore B = -2

    Thats just another way if you get stuck with the equations method (or expanding is a pain). Usually a longer way though
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    In the same vein as above, using x=0 immediately gives you A+B= 0 and using x = \frac{\pi}{2} gets you A-B=4.
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    (Original post by Zacken)
    In the same vein as above, using x=0 immediately gives you A+B= 0 and using x = \frac{\pi}{2} gets you A-B=4.
    How did you chose suitable values of x to substitute in?
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    (Original post by SamuelN98)
    How did you chose suitable values of x to substitute in?
    I wanted to choose values that made one of sine and cosine 0 in turn so that I could have A(one thing only) + B(one thing only) instead of A(sqrt(3)/2 + (1/2)) + B(blah) or the likes. So x=0, \frac{\pi}{2} fit perfectly.
 
 
 
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