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Prove that \[\frac{9 \pi}{8} - \frac{9}{4} sin ^{-1}\frac{1}{3}= \frac{9}{4} sin^{-1}\frac{2\sqrt 2}{3}\]

1 Answer

Toolbox:
  • \( sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]\)
  • \( sin^{-1}1=\large\frac{\pi}{2}\)
Given $\frac{9 \pi}{8} - \large\frac{9}{4} sin ^{-1}\large\frac{1}{3}= \large\frac{9}{4} sin^{-1}\large\frac{2\sqrt 2}{3}$
 
Rearrranging the terms, we need to prove that \( \large\frac{\pi}{8}=\large\frac{1}{4} \bigg[ sin^{-1}\large\frac{2\sqrt 2}{3}+sin^{-1}\large\frac{1}{3} \bigg]\)
 
R.H.S. = \( \large\frac{1}{4} \bigg[ sin^{-1}\large\frac{2\sqrt 2}{3}+sin^{-1}\large\frac{1}{3} \bigg]\)
We know that \( sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]\)
 
By taking \(x=\large\frac{2\sqrt 2}{3}\:and\:y=\large\frac{1}{3}\)
\(\sqrt{1-y^2}=\sqrt{1-\large\frac{1}{9}}=\sqrt{\large\frac{8}{9}}=\large\frac{2\sqrt{2}}{3}\)
\(\sqrt{1-x^2}=\sqrt{1-\large\frac{8}{9}}=\large\frac{1}{3}\)
 
\( \Rightarrow\:\) R.H.S. = \( \large\frac{1}{4} \bigg[ sin^{-1} \bigg(\large \frac{2\sqrt 2}{3} \sqrt{1-\large\frac{1}{9}} +\large\frac{1}{3} \sqrt{1-\frac{8}{9}} \bigg) \bigg]\)
\(=\large\frac{1}{4}sin^{-1}\bigg(\large\frac{2\sqrt{2}}{3}.\large\frac{2\sqrt{2}}{3}+\large\frac{1}{3}.\large\frac{1}{3}\bigg)\)
\( = \large\frac{1}{4}sin^{-1} \bigg(\large\frac{8}{9}+\large\frac{1}{9}\bigg) \)
 
\( \large\frac{1}{4}sin^{-1}1=\large\frac{1}{4}.\large\frac{\pi}{2}=\large\frac{\pi}{8}\) =L.H.S
 

 

answered Feb 23, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 

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