Question

Prove the following expression:
${{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right)={{\cos }^{-1}}\left( \dfrac{33}{65} \right)$

Answer 1

Hint: From the given expression we will take LHS and solve it by applying various trigonometric identities to prove it equal to the RHS part of the expression. We need to prove that sum of angles ${{\cos }^{-1}}\left( \dfrac{4}{5} \right)\,and\,{{\cos }^{-1}}\left( \dfrac{12}{13} \right)$ is equal to angle ${{\cos }^{-1}}\left( \dfrac{33}{65} \right)$ we know that\[\cos \left( {{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right) \right)=\cos \left( {{\cos }^{-1}}\left( \dfrac{33}{65} \right) \right)\] so we will then apply the identity $\cos \left( a+b \right)=\cos a.\cos b-\sin a.\sin b$ .

Complete step by step answer:
We know that ${{\cos }^{-1}}\theta $ is the value of angle $\theta $ in radians.
So let us suppose the angle A = ${{\cos }^{-1}}\left( \dfrac{4}{5} \right)\,$, angle B = ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)$ and angle C = ${{\cos }^{-1}}\left( \dfrac{33}{65} \right)$
Hence we get, \[\cos A=\dfrac{4}{5},\cos B=\dfrac{12}{13}\,and\,\cos C=\dfrac{33}{65}\]
Now, if we apply cosine on both the side of the given expression we get,
$\cos \left( A+B \right)=\cos \left( C \right)$
And we know that , $\cos \left( a+b \right)=\cos a.\cos b-\sin a.\sin b$ so now,
We need to find the values $\sin A\,and\,\sin B$ first, for $\sin A$
$\begin{align}
  & {{\sin }^{2}}A+{{\cos }^{2}}A=1 \\
 & {{\sin }^{2}}A=1-{{\cos }^{2}}A \\
 & \sin A=\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}} \\
 & \sin A=\dfrac{3}{5} \\
\end{align}$
Similarly for $\sin B$, we get
\[\begin{align}
  & {{\sin }^{2}}B+{{\cos }^{2}}B=1 \\
 & {{\sin }^{2}}B=1-{{\cos }^{2}}B \\
 & \sin B=\sqrt{1-{{\left( \dfrac{12}{13} \right)}^{2}}} \\
 & \sin B=\dfrac{5}{13} \\
\end{align}\]
Now, we know that
$\cos \left( A+B \right)=\cos A.\cos B-\sin A.\sin B$
Putting values $\cos A=\dfrac{4}{5},\cos B=\dfrac{3}{5},\sin A\dfrac{3}{5}\,and\,\sin B=\dfrac{5}{13}$ in the above expression we get,
$\cos \left( A+B \right)=\dfrac{4}{5}\times \dfrac{12}{13}-\dfrac{3}{5}\times \dfrac{5}{13}$
$\begin{align}
  & \cos \left( A+B \right)=\dfrac{48}{65}-\dfrac{3}{13} \\
 & \cos \left( A+B \right)=\dfrac{48-15}{65} \\
 & \cos \left( A+B \right)=\dfrac{33}{65}=\cos C \\
\end{align}$
Hence we proved that $\cos \left( A+B \right)=\cos C$ i.e. $A+B=C$ and hence, we proved
${{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right)={{\cos }^{-1}}\left( \dfrac{33}{65} \right)$

Note:
Whenever you are solving questions related to the inverse trigonometric functions try to think of ${{\cos }^{-1}}\theta ,{{\sin }^{-1}}\theta $ etc. as the value of angle $\theta $ in the radians. And always try to convert inverse trigonometric functions to simple trigonometric functions by applying sine, cosine etc. on both sides of the equation.
You also need to remember all basic trigonometric and inverse trigonometric identities in order to solve these kinds of questions and after that you will be able to easily figure out which identity to apply and when to apply it.