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True or False: The principal value of $\sin^{-1}\begin{bmatrix}\cos\bigg(\sin^{-1}\frac{1}{2}\bigg)\end{bmatrix}\;is\;\Large {\frac{\pi}{3}}$

$\begin{array}{1 1} \large {\frac{\pi}{3}} \\ \large {\frac{\pi}{6}} \\ \large {\frac{2\pi}{3}} \\ \large {\frac{\pi}{4}} \end{array} $

1 Answer

Toolbox:
  • \(sin\large\frac{\pi}{6}=\large\frac{1}{2}\)
  • \(cos\large\frac{\pi}{6}=\large\frac{\sqrt3}{2}\)
  • \(sin\large\frac{\pi}{3}=\large\frac{\sqrt3}{2}\)
Ans - True
\(sin^{-1}\large\frac{1}{2}=sin^{-1}sin\large\frac{\pi}{6}=\large\frac{\pi}{6}\)
\( cos(sin^{-1}\large\frac{1}{2})=cos\large\frac{\pi}{6}=\large\frac{\sqrt 3}{2}\)
\(sin^{-1}\bigg(cos(sin^{-1}\large\frac{1}{2})\bigg)=sin^{-1}\large\frac{\sqrt3}{2}\)
 
But \(\large\frac{\sqrt3}{2}=sin\large\frac{\pi}{3}\)
\(\Rightarrow\:sin^{-1}\large\frac{\sqrt3}{2}=sin^{-1}sin\large\frac{\pi}{3}=\large\frac{\pi}{3}\)
\(\Rightarrow\:sin^{-1}\bigg(cos(sin^{-1}\large\frac{1}{2})\bigg)=sin^{-1}sin\large\frac{\pi}{3}=\large\frac{\pi}{3}\)

 

answered Feb 18, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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