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Prove that $ \sin^{-1}{\frac{8}{17}}+\sin^{-1}{\frac{3}{5}}=\sin^{-1}{\frac{77}{85}}$

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  • \( sin^{-1}x+sin^{-1}y=sin^{-1} \bigg( x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg)\)
L.H.S.=
By taking \(x=\large\frac{8}{17}\:and\:y=\large\frac{3}{5},\:we\:get\)
\(\sqrt{1-y^2}=\sqrt{1-\large\frac{9}{25}}=\sqrt{\large\frac{16}{25}}=\large\frac{4}{5}\:and\:\)
\(\sqrt{1-x^2}=\sqrt{1-\large\frac{64}{289}}=\sqrt{\large\frac{225}{289}}=\large\frac{15}{17}\)
\(\Rightarrow\:x\sqrt{1-y^2}+y\sqrt{1-x^2} =\large\frac{3}{5}.\large\frac{15}{17}+\large\frac{8}{17}.\large\frac{4}{5}\)
\(=\large\frac{45}{85}+\frac{32}{85}\)
 
By using the above formula
\( sin^{-1}\large\frac{8}{17}+sin^{-1}\large\frac{3}{5}=sin^{-1}\big( \large\frac{3}{5}.\large\frac{15}{17}+\large\frac{8}{17}.\large\frac{4}{5}\big)\)
\(=sin^{-1}\large\frac{77}{85}\)
=R.H.S.

 

answered Feb 19, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 
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