Show sin^2xcos^2x=(1-cos4x)/9

1 answer:

8 0

Did you mean sin^2xcos^2x=(1-cos4x)/8?

Because I have my solution here to that problem:

sin²(x)*cos²(x) = 1/8*(1-cos(4x))

[sin(2x)/2]² = 1/8*(1-cos(4x)) 

sin²(2x)/4 = 1/8*(1-cos(4x)) 

Recall that sin²(x) = (1-cos(2x))/2 

From this, we can tell that sin²(2x) = 1-cos(4x)/2 

[(1-cos(4x)/2]/4 = 1/8*(1-cos(4x)) 

(1-cos(4x))/8 = (1-cos(4x))/8

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