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Solve for $x$: $\sin\bigg[\sin^{-1}\frac{1}{3}+\cos^{-1}x\bigg]=1$

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  • \( sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2}\)
  • \( sin\large\frac{\pi}{2}=1\: or \: sin^{-1}1=\large\frac{\pi}{2}\)
Since sin\(\large\frac{\pi}{2}=1,\:sin^{-1}\large\frac{1}{3}+cos^{-1}x\:should\:be\:=\large\frac{\pi}{2}\)
\( sin^{-1}\large\frac{1}{3}+cos^{-1}x=\large\frac{\pi}{2}\)
From the formula sin\(^{-1}x+cos^{-1}x=\large\frac{\pi}{2}\) we get
\( \Rightarrow x=\large\frac{1}{3}\)

 

answered Feb 20, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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