Question

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

Answer 1

Let $a={\cos }^{-1}\frac{4}{5}$

$\Rightarrow$ $\cos a=\frac{4}{5}$ ---- ( 1 )

We know that,

$\Rightarrow$ ${\sin }^{2}a=1-{\cos }^{2}a$

$\Rightarrow$ $\sin a=\sqrt{1-{\cos }^{2}a}$

$=\sqrt{1-{\left(\frac{4}{5}\right)}^2}$

$=\sqrt{1-\frac{16}{25}}$

$=\sqrt{\frac{25-16}{25}}$

$=\sqrt{\frac{9}{25}}$

$=\frac{3}{5}$

$\Rightarrow$ $\sin a=\frac{3}{5}$ ---- ( 2 )

Let $b={\cos }^{-1}\frac{12}{13}$

$\Rightarrow$ $\cos b=\frac{12}{13}$ ---- ( 3 )

We know that,

$\Rightarrow$ ${\sin }^{2}b=1-{\cos }^{2}b$

$\Rightarrow$ $\sin b=\sqrt{1-{\cos }^{2}b}$

$=\sqrt{1-{\left(\frac{12}{13}\right)}^2}$

$=\sqrt{1-\frac{144}{169}}$

$=\sqrt{\frac{169-144}{169}}$

$=\sqrt{\frac{25}{169}}$

$=\frac{5}{13}$

$\Rightarrow$ $\sin b=\frac{5}{13}$ ----- ( 4 )

$\cos (a+b)=\cos a\cos b-\sin a\sin b$

From ( 1 ), ( 2 ), ( 3 ) and ( 4 ) we get

$\cos (a+b)=\frac{4}{5}\times \frac{12}{13}-\frac{3}{5}\times \frac{5}{13}$

$=\frac{48}{65}-\frac{3}{13}$

$=\frac{48-15}{65}$

$=\frac{33}{65}$

$\therefore$ $\cos (a+b)=\frac{33}{65}$

$\Rightarrow$ $a+b={\cos }^{-1}\left(\frac{33}{65}\right)$

$\Rightarrow$ ${\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{12}{15}={\cos }^{-1}\frac{33}{65}$

Hence, $L.H.S=R.H.S$