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Write an algebraic expression for sin(arccos4x)?
the answer is the sq. root of 1-16x^2 ....how? hahah thankss
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- A square ELv 61 decade ago
take arccos4x = y
4x = cosy
sin(y) = sqrt(1-cos^2(x)) = sqrt(1 - 16x^2)
y = arcsin(sqrt(1-16x^2))
sin(arccos(4x)) = sin(y) = sin(arcsin(sqrt(1-16x^2))) = sqrt(1-16x^2)
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