Gujarat BoardEnglish MediumSTD 12 ScienceMathsINVERSE TRIGNOMETRIC FUNCTIONS2 Marks
Question
Find the values: $\sin^{-1}\bigg(\sin\frac{2\pi}{3}\bigg)$
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Answer
For $\sin^{-1}\left(\sin x\right)$ type of problem we have to always check whether the angle is in the principal range or not. This angle must be in the principal range $\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg].$ $=\sin^{-1}\bigg(\sin\frac{3\pi-\pi}{3}\bigg)$ $=\sin^{-1}\bigg[\sin\bigg(\pi-\frac{\pi}{3}\bigg)\bigg]$ $=\sin^{-1}\sin\frac{\pi}{3}=\frac{\pi}{3}$
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