TOLC med 2012 trigonometry


The correct answer is sin³(x)
i know it should be a sin function because sin(-x)=-sin(x)
but i don’t know how to solve when there’s exponents

The solution is the same as solving power of negative numbers.

Sin^2(x) = sin (x) . Sin (x) =sin^2 (x)
Sin^2 (-x) = -sin (x). - sin (x) = sin^2 (x)

Sin (x^2) = 2 sin (x)
Sin (-x^2) = sin (x^2) = 2 sin (x)

Sin^3 (x) = sin (x) . Sin (x) . Sin (x) = sin^3 (x)
Sin^3 (-x) = -sin (x) . -sin (x) . - sin (x) = -sin^3 (x)

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@Juliette could u plz tell me which website u r using for this??