Evaluate each of the following:

sin^{–1} (sin12)

sin^{–1}(sin x) = x

Provided x ϵ ≈ [–1.57,1.57]

And in our equation x is 4 which does not lie in the above range.

We know sin[2nπ – x] = sin[–x]

∴ sin(2nπ – 12) = sin(–12)

Here n = 2

Also 2π–12 belongs in

∴ sin^{–1}(sin12) = 2π – 12

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