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

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.

edited Mar 15, 2013