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Math Help - [SOLVED] (7cosx+5sinx)²=Acos2x+Bsin2x+C

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    [SOLVED] (7cosx+5sinx)²=Acos2x+Bsin2x+C

    Hello there

    I need help with the following trig question...

    (7cosx+5sinx)²=Acos2x+Bsin2x+C

    I've come about this far...

    (7cosx+5sinx)²=24cos²x+35sin2x+25

    I need to do something about the 24cos²x but im unsure how to proceed... I was thinking i could use cos²(x/2)=(1+cosx)/2

    but im not sure...

    Thanks in advance
    Janne the Swede
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    Quote Originally Posted by JannetheSwede View Post
    Hello there

    I need help with the following trig question...

    (7cosx+5sinx)²=Acos2x+Bsin2x+C

    I've come about this far...

    (7cosx+5sinx)²=24cos²x+35sin2x+25

    I need to do something about the 24cos²x but im unsure how to proceed... I was thinking i could use cos²(x/2)=(1+cosx)/2

    but im not sure...

    Thanks in advance
    Janne the Swede
    (7\cos{x} + 5\sin{x})^2 = 49\cos^2{x} + 70\cos{x}\sin{x} + 25\sin^2{x}

     = 24\cos^2{x} + 25\cos^2{x} + 35\cdot 2\cos{x}\sin{x} + 25\sin^2{x}

     = 24\cos^2{x} + 35\sin{(2x)} + 25(\cos^2{x} + \sin^2{x})

     = 24\cos^2{x} + 35\sin{(2x)} + 25\cdot 1

     = 24\cos^2{x} + 35\sin{(2x)} + 25.
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  3. #3
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    mmh, thats about as far as I got, but my awnser have to be A*cos (2x)+B*sin (2x)+C where A, B and C are integers...

    somehow i have to get rid of 24*cos²x but i don't know how
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    Quote Originally Posted by JannetheSwede View Post
    mmh, thats about as far as I got, but my awnser have to be A*cos (2x)+B*sin (2x)+C where A, B and C are integers...

    somehow i have to get rid of 24*cos²x but i don't know how
    Oh yeah, you need to remember

    \cos^2{x} - \sin^2{x} = \cos{(2x)}.


    And since \sin^2{x} = 1 - \cos^2{x} we get


    \cos^2{x} - (1 - \cos^2{x}) = \cos{(2x)}

    2\cos^2{x} - 1 = \cos{(2x)}

    \cos^2{x} = \frac{1}{2} + \frac{1}{2}\cos{(2x)}.


    So substitute this back into the equation we found:

    24\cos^2{x} + 35\sin{(2x)} + 25 = 24\left(\frac{1}{2} + \frac{1}{2}\cos{(2x)}\right) + 35\sin{(2x)} + 25

     = 12 + 12\cos{(2x)} + 35\sin{(2x)} + 25

     = 12\cos{(2x)} + 35\sin{(2x)} + 37.


    So A = 12, B = 35, C = 37.
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