Find the principal value of \(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\).\(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\) का principal value ज्ञात कीजिए।
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(\frac{\pi}{3}\)\(\frac{\pi}{3}\)
Let \(\theta=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\). Then \(\sin\theta=\frac{\sqrt{3}}{2}\) with \(\theta\) in the principal range \( \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\). We know \(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\), and \(\frac{\pi}{3}\) lies in the principal range. So \(\theta=\frac{\pi}{3}\). Therefore, \(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}\).मान लीजिए \(\theta=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\). तब \(\sin\theta=\frac{\sqrt{3}}{2}\) और \(\theta\) principal range \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) में होता है। हमें पता है कि \(\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}\) और \(\frac{\pi}{3}\) इसी principal range में आता है। इसलिए \(\theta=\frac{\pi}{3}\)। अतः \(\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}\).
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