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Math Help - Trig simplification (this one is tough)

  1. #1
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    Trig simplification (this one is tough)

    f(x)= (4cos (x) -cos (3x))^2 + (3sin (x) -sin (3x))^2
    f(x) has to be simplified to the form: f(x)= a cos ^2 (2x) + b cos (2x) + c where a,b,and c are real numbers.


    so far I expanded the two parentheticals to: f(x)= 16 cos ^2 (x) - 8 cos(x) cos(3x) + cos ^2 (3x) + 9 sin^2 (x) -6 sin (x) sin (3x) + sin ^2 (3x)

    after this I'm stumped.

    I tried merging the 16cos ^2 (x) + sin ^2 (3x) to a 7 cos ^2(x) + 9, and doing the same with the sin ^2 (3x) + cos ^2 (3x) = 1. But it isn't taking me anywhere.

    If someone can solve this with work, I would highly appreciate it!

    Thanks
    -kaplac
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  2. #2
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    Hello kaplac
    Quote Originally Posted by kaplac View Post
    f(x)= (4cos (x) -cos (3x))^2 + (3sin (x) -sin (3x))^2
    f(x) has to be simplified to the form: f(x)= a cos ^2 (2x) + b cos (2x) + c where a,b,and c are real numbers.


    so far I expanded the two parentheticals to: f(x)= 16 cos ^2 (x) - 8 cos(x) cos(3x) + cos ^2 (3x) + 9 sin^2 (x) -6 sin (x) sin (3x) + sin ^2 (3x)

    after this I'm stumped.

    I tried merging the 16cos ^2 (x) + sin ^2 (3x) to a 7 cos ^2(x) + 9, and doing the same with the sin ^2 (3x) + cos ^2 (3x) = 1. But it isn't taking me anywhere.

    If someone can solve this with work, I would highly appreciate it!

    Thanks
    -kaplac
    Using the identities
    \cos3x = 4\cos^3x-3\cos x; and

    \sin 3x = 3\sin x - 4\sin^3x, we get:
    f(x) = (7\cos x - 4\cos^3x)^2+(4\sin^3x)^2

    =16\cos^6x-56\cos^4x+49\cos^2x+16(1-\cos^2x)^3

    =16\cos^6x-56\cos^4x+49\cos^2x+16(1-3\cos^2x+3\cos^4x-\cos^6x)

    = -8\cos^4x+\cos^2x+16

    Now a\cos^22x+b\cos2x+c = a(2\cos^2x-1)^2 +b(2\cos^2x-1) + c
    =4a\cos^4x+(2b-4a)\cos^2x+a-b+c
    Equating coefficients gives
    a = -2, b = -\frac72, c=\frac{29}{2}
    But check my working!

    Grandad
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  3. #3
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    Thank you so much Grandad!
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