# what is sin(3sin^-1 angle)?

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You need to remember that the domain of inverse trigonometric function contains real numbers, not angles, hence you need to provide a value instead of angle for the problem to be solved.

I'll select a value and I'll show how to solve this problem.

`sin(3sin^(-1) (1/2)) = sin(3 alpha)` (notice that the value of inverse trigonometric function is an angle)

You need to remember the formula of sine of triple angle such that:

`sin 3 alpha = 3 sin alpha - 4 sin^3 alpha`

Plugging `alpha = sin^(-1) (1/2)` yields:

`sin 3(sin^(-1) (1/2)) = 3 sin(sin^(-1) (1/2)) - 4 sin^3 (sin^(-1) (1/2))`

`sin 3(sin^(-1) (1/2)) = 3*(1/2) - 4*(1/2)^3 = 3/2 - 4/8 = 3/2 - 1/2 = 2/2 = 1`

**Hence, evaluating the value of `sin(3(sin^(-1) (1/2))) ` yields 1.**