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Question

Mathematics Question on Inverse Trigonometric Functions

Find the values of sin⁡−1(sin⁡2π3)\sin^{-1}(\sin^2\frac{\pi}{3})

Answer

sin⁡−1(sin⁡2π3)\sin^{-1}(\sin^2\frac{\pi}{3})
We know that sin⁡−1\sin^{-1}(sin x) = x if x∈ [π2,π2][\frac{\pi}{2},\frac{\pi}{2}], which is the principal value branch of sin⁡−1\sin^{-1} x.

Here, 2π32 \frac{\pi}{3} ∉ [−π2,π2-\frac{\pi}{2},\frac{\pi}{2} ]

Now,sin⁡−1(sin⁡2π3)\sin^{-1}(\sin^2\frac{\pi}{3})can be written as:

sin⁡−1(sin⁡2π3)=sin⁡−1[(sin⁡π−2π3)]=sin⁡−1(sinπ3)\sin^{-1}(\sin^2\frac{\pi}{3})=\sin^{-1}[(\sin\frac{\pi-2\pi}{3})]=\sin^{-1}(sin\frac{\pi}{3}) where π3∈[−π2,π2]\frac{\pi}{3}\in[-\frac{\pi}{2},\frac{\pi}{2}]

∴sin⁡−1(sin⁡2π3=sin⁡−1(sin⁡π3)=π3\therefore \sin^{-1}(\sin^2\frac{\pi}{3}=\sin^{-1}(\sin\frac{\pi}{3})=\frac{\pi}{3}