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Prove that \[ cos^{-1} \frac {12}{13} +sin^{-1} \frac{3}{5}=sin^{-1} \frac{56}{65}\]

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Toolbox:
  • \( cos^{-1}x=sin^{-1}\sqrt{1-x^2}\)
  • \( sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2} +y\sqrt{1-x^2} \bigg] \)
Given L.H.S. = $cos^{-1} \large\frac {12}{13} +sin^{-1} \large\frac{3}{5}$
We know that \( cos^{-1}x=sin^{-1}\sqrt{1-x^2}\)
 
By taking $x=$\(\large\frac{12}{13},\rightarrow \sqrt{1-x^2}=\sqrt{1-\large\frac{144}{169}}=\sqrt{\large\frac{25}{169}}=\large\frac{5}{13}\)
\(\Rightarrow\:cos^{-1}\large\frac{12}{13}=sin^{-1}\large\frac{5}{13}\)
\( \Rightarrow cos^{-1}\large\frac{12}{13}+sin^{-1}\large\frac{3}{5}=sin^{-1}\large\frac{5}{13}+sin^{-1}\large\frac{3}{5}\)
 
We know that \( sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2} +y\sqrt{1-x^2} \bigg] \)
 
Given $sin^{-1}\large\frac{5}{13}+sin^{-1}\large\frac{3}{5}$, let's take $x=\large\frac{5}{13}$ and $y=\large\frac{3}{5}$
$x \sqrt {1-y^2} = \large\frac{5}{13} \sqrt {1-(\large\frac{3}{5})^2} = \large\frac{5}{13} \sqrt{1-\large\frac{9}{25}} = \large\frac{5}{13}.\large\frac{4}{5} = \large\frac{20}{65}$
$y \sqrt {1-x^2} = \large\frac{3}{5} \sqrt{1-(\large\frac{5}{13})^2} = \large\frac{3}{5} \sqrt{1-\large\frac{25}{169}} = \large\frac{3}{5}.\large\frac{12}{13} = \large\frac{36}{65}$
 
Therefore, $ sin^{-1}\large\frac{5}{13}+sin^{-1}\large\frac{3}{5} = sin^{-1} (\large \frac{20}{65}+ \large\frac{36}{65}) =\large \frac{56}{65}$ = R.H.S.

 

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