If sin^{-1} theta = sin^{-1}(sin 5),then thet | vedclass

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(C) We are given the equation $\sin^{-1} 	heta = \sin^{-1}(\sin 5)$.By the definition of the inverse sine function,$\sin^{-1}(\sin x) = x$ only if $x

Vedclass · Accepted Answer

$\sin(5 - 2\pi)$

Vedclass · Answer

(C) We are given the equation $\sin^{-1} 	heta = \sin^{-1}(\sin 5)$.By the definition of the inverse sine function,$\sin^{-1}(\sin x) = x$ only if $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.Here,$x = 5$ radians. Since $\pi \approx 3.14$,$2\pi \approx 6.28$. Thus,$5$ is not in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}] \approx [-1.57, 1.57]$.We know that $\sin(x) = \sin(2\pi - x)$ is not correct,but $\sin(x) = \sin(x - 2\pi)$ is true because the period of the sine function is $2\pi$.Specifically,$\sin(5) = \sin(5 - 2\pi)$.Since $5 - 2\pi \approx 5 - 6.28 = -1.28$,which lies within the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we have $\sin^{-1}(\sin 5) = \sin^{-1}(\sin(5 - 2\pi)) = 5 - 2\pi$.Therefore,$	heta = \sin(5 - 2\pi)$.

If $\sin^{-1} \theta = \sin^{-1}(\sin 5)$,then $\theta$ is

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