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Show that $\tan({\frac{1}{2}}\sin^{-1}{\frac{3}{4}})=\frac{4-\sqrt 7}{3}$

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Toolbox:
  • Take \( sin^{-1}\frac{3}{4}=x\) and proceed
  • \( cosx=2cos^2\frac{x}{2}-1\)
  • \(cosx=\sqrt{1-sin^2x}\)
L.H.S.
Let \( sin^{-1}\frac{3}{4}=x \Rightarrow sinx=\frac{3}{4}\)
L.H.S=\(tan\frac{x}{2}\)
\( cos x=\sqrt{1-sin^2x}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt 7}{4}\)
\( \Rightarrow\:cosx= 2cos^2\frac{x}{2}-1=\frac{\sqrt 7}{4}\)
\( \Rightarrow 2cos^2\frac{x}{2}=\frac{\sqrt{7}}{4}+1=\frac{4+\sqrt 7}{4}\)
\( \Rightarrow cos\frac{x}{2}=\sqrt{\frac{4+\sqrt 7}{8}}\)
\(sin\frac{x}{2}=\sqrt{1-cos^2\frac{x}{2}}=\sqrt{1-\big(\frac{4+\sqrt7}{8}\big)}=\sqrt{\frac{4-\sqrt 7}{8}}\)
\(\Rightarrow\: tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\frac{4-\sqrt 7}{4+\sqrt 7}\)
\(=\frac{4-\sqrt 7}{3}\) (by rationalising the denominator)
\( tan\frac{x}{2}=\frac{4-\sqrt 7}{3}\)
\(\Rightarrow\:L.H.S=R.H.S.\)
answered Jun 12, 2014 by balaji.thirumalai
 

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