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Mathematics 7 Online
OpenStudy (anonymous):

(4cos^3x/sin^2x) * (sinx/4cosx)2

OpenStudy (solomonzelman):

Expand the second fraction, you will get sin^2(x) / 16cos^2(x) Do you see which terms can cancel out?

OpenStudy (anonymous):

Yes, I do. I was thinking that the two meant multipy by two, not to the power of two for some reason

OpenStudy (anonymous):

so then would you leave it as 4cos^2x(cosx/4), or would you put hem ezual to zero and get cosx=0 ?

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