Question

Prove that ${\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}\frac{33}{65}$

Answer 1

To prove : ${\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}\frac{33}{65}$
We use $\cos (a+b)$ formula
Let $a={\cos }^{-1}\frac{4}{5}\Rightarrow \cos a=\frac{4}{5}$
$\Rightarrow \sin a=\sqrt{1-{\cos }^{2}a}$
$=\sqrt{1-{\left(\frac{4}{5}\right)}^2}=\sqrt{\frac{9}{25}}=\frac{3}{5}$

Let $b={\cos }^{-1}\frac{12}{13}\Rightarrow \cos b=\frac{12}{13}$
$\Rightarrow \sin b=\sqrt{1-{\cos }^{2}b}$
$=\sqrt{1-{\left(\frac{12}{13}\right)}^2}=\sqrt{\frac{25}{169}}=\frac{5}{13}$

$\therefore \sin a=\frac{3}{5}$ and $\sin b=\frac{5}{13}$
$\therefore \cos a=\frac{4}{5}$ and $\cos b=\frac{12}{13}$
$\cos (a+b)=\cos a\cos b-\sin a\sin b$
$\cos (a+b)=\frac{4}{5}\times \frac{12}{13}-\frac{3}{5}\times \frac{5}{13}$
$=\frac{48}{65}-\frac{3}{13}=\frac{48-15}{65}$
$\Rightarrow \cos (a+b)=\frac{33}{65}$
$\Rightarrow a+b={\cos }^{-1}\left(\frac{33}{65}\right)$
$\Rightarrow {\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}\frac{33}{65}$
Hence proved.