# trigonometry functioncos5sin^-1 1/2?

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You need to substitute the corresponding angle for `sin^(-1)(1/2)` , hence `sin^(-1)(1/2) = pi/6` .

You need to evaluate `cos (5sin^(-1)(1/2) ) = cos 5pi/6`

You should consider `5pi/6 = 6pi/6 - pi/6` , hence `cos 5pi/6 = cos(6pi/6 - pi/6).`

You need to use the following formula, `cos(x-y) = cos x*cos y + sin x*sin y` , to evaluate `cos(6pi/6 - pi/6)` such that:

`cos(6pi/6 - pi/6) = cos(pi - pi/6) = cospi*cos pi/6 + sin pi*sin pi/6`

Using `cos pi = -1` and `sin pi = 0` yields:

`cos(pi - pi/6) = - cos pi/6 = -sqrt3/2`

**Hence, evaluating `cos (5sin^(-1)(1/2))` yields `cos (5sin^(-1)(1/2)) = -sqrt3/2.` **

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