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Math Help - 8cos^2 x = 4(1 + cos 2x) - how???

  1. #1
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    8cos^2 x = 4(1 + cos 2x) - how???

    Hi I have another area problem

    r = 3cos x and r = 1 + cos x

    so A i= integral sign (S) a= 0; b=pi/3 ((3cos x)^2 - (1+cos x)^2) dx

    which works out to S a=0;b= pi/3 (8cos^2 x - 2 cos x - 1) dx

    This one is actually in my answer sheet and it gives the next step as

    S a=0; b=pi/3 (4(1+cos 2x) - 2 cos x - 1) dx

    Where does the 4(1+cos 2x) from????

    Insight is appreciated

    Calculus Beginner
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  2. #2
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    Quote Originally Posted by calcbeg View Post
    Hi I have another area problem

    r = 3cos x and r = 1 + cos x

    so A i= integral sign (S) a= 0; b=pi/3 ((3cos x)^2 - (1+cos x)^2) dx

    which works out to S a=0;b= pi/3 (8cos^2 x - 2 cos x - 1) dx

    This one is actually in my answer sheet and it gives the next step as

    S a=0; b=pi/3 (4(1+cos 2x) - 2 cos x - 1) dx

    Where does the 4(1+cos 2x) from????

    Insight is appreciated

    Calculus Beginner
    It's a standard double-angle identity.

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

    which means

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