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

1. ## 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

2. Originally Posted by calcbeg
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.

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

which means

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

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# 8cos^2x 8sec^x=65

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