SELECT * FROM question_mgmt as q WHERE id=975 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=6 AND chapterId=89 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 12th mathematics Inverse Trigonometric Functions find the principal value of begin align sin 1

Question: Find the principal value of \begin{align} sin^{-1}\left(-\frac{1}{2}\right)\end{align}

Answer:

\begin{align} sin^{-1}\left(-\frac{1}{2}\right)=y \;\;Then\;\; sin y = -\frac{1}{2} = -sin\left(\frac{\pi}{6}\right)= sin\left(-\frac{\pi}{6}\right)\end{align} 

We know that the range of the principal value branch of sin−1 is

\begin{align} \left[-\frac{\pi}{2},\frac{\pi}{2}\right] and \;\;sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\end{align} 

Therefore, the principal value of

\begin{align} sin^{-1}\left(-\frac{1}{2}\right) is -\frac{\pi}{6}\end{align} 


SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='6' AND ex_no!=0 AND status=1 and id!=975 ORDER BY views desc, last_viewed_on desc limit 0,10
SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='3' AND subId='6' AND ex_no!=0 AND status=1 and id!=975 ORDER BY last_viewed_on desc limit 0,10

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