The value of sin^{-1}(sin 10) is | vedclass

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(C) We know that $\sin^{-1}(\sin x) = x$ only when $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.Since $10$ radians is not in the interval $[-\frac{\pi}{2},

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$3\pi - 10$

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(C) We know that $\sin^{-1}(\sin x) = x$ only when $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.Since $10$ radians is not in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we need to find an angle $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ such that $\sin(y) = \sin(10)$.We know that $\sin(x) = \sin(n\pi + (-1)^n x)$.For $n = 3$,$\sin(10) = \sin(3\pi - 10)$.Since $3\pi \approx 9.42$,we have $3\pi Specifically,$3\pi Subtracting $3\pi$ from all sides,we get $0 Multiplying by $-1$,we get $-\frac{\pi}{2} Since $3\pi - 10$ lies in the principal value branch $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we have $\sin^{-1}(\sin 10) = \sin^{-1}(\sin(3\pi - 10)) = 3\pi - 10$.

The value of $\sin^{-1}(\sin 10)$ is

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