Question

Prove the following:
${{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\sin }^{-1}}\left( \dfrac{56}{65} \right)$

Answer 1

Hint: We have to prove the given equation. As you can see that R.H.S of the above equation is a ${{\sin }^{-1}}$ term so we need to convert the L.H.S of the given equation into ${{\sin }^{-1}}$. Converting the L.H.S of the given equation into ${{\sin }^{-1}}$ we can write ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)$ as $\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{12}{13} \right)$. Now, we have two terms in ${{\sin }^{-1}}$ with a negative sign so we can use the identity as ${{\sin }^{-1}}x-{{\sin }^{-1}}y={{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}} \right)$. Then put the values of x and y and solve the L.H.S of the equation.

Complete step by step answer:
We have given the following equation that we are asked to prove:
${{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\sin }^{-1}}\left( \dfrac{56}{65} \right)$
As R.H.S contains ${{\sin }^{-1}}$ term so L.H.S should contain ${{\sin }^{-1}}$ term then L.H.S becomes equal to R.H.S and we can prove the given equation.
Solving L.H.S of the given equation we get,
${{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)$
We know that,
$\begin{align}
  & {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} \\
 & \Rightarrow {{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x \\
\end{align}$
Using the above relation we can write ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)$ as $\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{12}{13} \right)$.
$\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)$
Now, we know that the expansion of ${{\sin }^{-1}}x-{{\sin }^{-1}}y$ is equal to:
${{\sin }^{-1}}x-{{\sin }^{-1}}y={{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}} \right)$
Substituting x as $\dfrac{3}{5}$ and y as $\dfrac{12}{13}$ in the above formula we get,
$\begin{align}
  & {{\sin }^{-1}}\dfrac{3}{5}-{{\sin }^{-1}}\dfrac{12}{13}={{\sin }^{-1}}\left( \dfrac{3}{5}\sqrt{1-{{\left( \dfrac{12}{13} \right)}^{2}}}-\left( \dfrac{12}{13} \right)\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}} \right) \\
 & \Rightarrow {{\sin }^{-1}}\dfrac{3}{5}-{{\sin }^{-1}}\dfrac{12}{13}={{\sin }^{-1}}\left( \dfrac{3}{5}\sqrt{1-\left( \dfrac{144}{169} \right)}-\left( \dfrac{12}{13} \right)\sqrt{1-\left( \dfrac{9}{25} \right)} \right) \\
 & \Rightarrow {{\sin }^{-1}}\dfrac{3}{5}-{{\sin }^{-1}}\dfrac{12}{13}={{\sin }^{-1}}\left( \dfrac{3}{5}\sqrt{\left( \dfrac{169-144}{169} \right)}-\left( \dfrac{12}{13} \right)\sqrt{\left( \dfrac{25-9}{25} \right)} \right) \\
 & \Rightarrow {{\sin }^{-1}}\dfrac{3}{5}-{{\sin }^{-1}}\dfrac{12}{13}={{\sin }^{-1}}\left( \dfrac{3}{5}\sqrt{\left( \dfrac{25}{169} \right)}-\left( \dfrac{12}{13} \right)\sqrt{\left( \dfrac{16}{25} \right)} \right) \\
 & \Rightarrow {{\sin }^{-1}}\dfrac{3}{5}-{{\sin }^{-1}}\dfrac{12}{13}={{\sin }^{-1}}\left( \dfrac{3}{5}\left( \dfrac{5}{13} \right)-\left( \dfrac{12}{13} \right)\left( \dfrac{4}{5} \right) \right) \\
 & \Rightarrow {{\sin }^{-1}}\dfrac{3}{5}-{{\sin }^{-1}}\dfrac{12}{13}={{\sin }^{-1}}\left( \dfrac{15-48}{65} \right) \\
 & \Rightarrow {{\sin }^{-1}}\dfrac{3}{5}-{{\sin }^{-1}}\dfrac{12}{13}={{\sin }^{-1}}\left( -\dfrac{33}{65} \right) \\
\end{align}$
Adding $\dfrac{\pi }{2}$ on both the sides we get,
${{\sin }^{-1}}\dfrac{3}{5}-{{\sin }^{-1}}\dfrac{12}{13}+\dfrac{\pi }{2}={{\sin }^{-1}}\left( -\dfrac{33}{65} \right)+\dfrac{\pi }{2}$
We know that ${{\sin }^{-1}}\left( -x \right)=-{{\sin }^{-1}}x$ so using this relation in the above problem we get,
$\begin{align}
  & {{\sin }^{-1}}\dfrac{3}{5}-{{\sin }^{-1}}\dfrac{12}{13}+\dfrac{\pi }{2}=-{{\sin }^{-1}}\left( \dfrac{33}{65} \right)+\dfrac{\pi }{2} \\
 & \Rightarrow {{\sin }^{-1}}\dfrac{3}{5}-{{\sin }^{-1}}\dfrac{12}{13}+\dfrac{\pi }{2}={{\cos }^{-1}}\left( \dfrac{33}{65} \right) \\
\end{align}$
Hence, we have resolved L.H.S to ${{\cos }^{-1}}\left( \dfrac{33}{65} \right)$ but R.H.S is equal to ${{\sin }^{-1}}\left( \dfrac{56}{65} \right)$ so we have to convert ${{\cos }^{-1}}\left( \dfrac{33}{65} \right)$ in terms of sine.
We know that:
${{\cos }^{-1}}x={{\sin }^{-1}}\left( \sqrt{1-{{x}^{2}}} \right)$
Now, using the above relation in ${{\cos }^{-1}}\left( \dfrac{33}{65} \right)$ we get,
$\begin{align}
  & {{\cos }^{-1}}\left( \dfrac{33}{65} \right)={{\sin }^{-1}}\left( \sqrt{1-{{\left( \dfrac{33}{65} \right)}^{2}}} \right) \\
 & \Rightarrow {{\cos }^{-1}}\left( \dfrac{33}{65} \right)={{\sin }^{-1}}\left( \sqrt{1-\dfrac{1089}{4225}} \right) \\
 & \Rightarrow {{\cos }^{-1}}\left( \dfrac{33}{65} \right)={{\sin }^{-1}}\left( \sqrt{\dfrac{4225-1089}{4225}} \right) \\
 & \Rightarrow {{\cos }^{-1}}\left( \dfrac{33}{65} \right)={{\sin }^{-1}}\left( \sqrt{\dfrac{3136}{4225}} \right) \\
 & \Rightarrow {{\cos }^{-1}}\left( \dfrac{33}{65} \right)={{\sin }^{-1}}\dfrac{56}{65} \\
\end{align}$

Hence, we have got the L.H.S of the given equation as ${{\sin }^{-1}}\dfrac{56}{65}$ which is equal to R.H.S of the given equation. Hence, we have proved the given equation.

Note: The other way to solve the above problem is as follows:
${{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\sin }^{-1}}\left( \dfrac{56}{65} \right)$
Subtracting ${{\sin }^{-1}}\left( \dfrac{3}{5} \right)$ on both the sides we get,
${{\cos }^{-1}}\left( \dfrac{12}{13} \right)={{\sin }^{-1}}\left( \dfrac{56}{65} \right)-{{\sin }^{-1}}\left( \dfrac{3}{5} \right)$
Solving R.H.S of the given equation and then proving the equation.
${{\sin }^{-1}}\left( \dfrac{56}{65} \right)-{{\sin }^{-1}}\left( \dfrac{3}{5} \right)$
Using the expansion of ${{\sin }^{-1}}x-{{\sin }^{-1}}y$ is equal to:
${{\sin }^{-1}}x-{{\sin }^{-1}}y={{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}} \right)$
Substituting x as $\dfrac{56}{65}$ and y as $\dfrac{3}{5}$ we get,
$\begin{align}
  & {{\sin }^{-1}}\left( \dfrac{56}{65} \right)-{{\sin }^{-1}}\left( \dfrac{3}{5} \right) \\
 & ={{\sin }^{-1}}\left( \dfrac{56}{65}\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}-\left( \dfrac{3}{5} \right)\sqrt{1-{{\left( \dfrac{56}{65} \right)}^{2}}} \right) \\
 & ={{\sin }^{-1}}\left( \dfrac{56}{65}\sqrt{1-\left( \dfrac{9}{25} \right)}-\left( \dfrac{3}{5} \right)\sqrt{1-\left( \dfrac{3136}{4225} \right)} \right) \\
 & ={{\sin }^{-1}}\left( \dfrac{56}{65}\sqrt{\left( \dfrac{25-9}{25} \right)}-\left( \dfrac{3}{5} \right)\sqrt{\left( \dfrac{4225-3136}{4225} \right)} \right) \\
 & ={{\sin }^{-1}}\left( \dfrac{56}{65}\sqrt{\left( \dfrac{16}{25} \right)}-\left( \dfrac{3}{5} \right)\sqrt{\left( \dfrac{1089}{4225} \right)} \right) \\
 & ={{\sin }^{-1}}\left( \dfrac{56}{65}\left( \dfrac{4}{5} \right)-\left( \dfrac{3}{5} \right)\left( \dfrac{33}{65} \right) \right) \\
 & ={{\sin }^{-1}}\left( \dfrac{224-99}{325} \right) \\
 & ={{\sin }^{-1}}\left( \dfrac{125}{325} \right) \\
\end{align}$
Simplifying the fraction written inside ${{\sin }^{-1}}$ we get,
${{\sin }^{-1}}\left( \dfrac{5}{13} \right)$
Now, L.H.S is equal to ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)$ so we have to convert ${{\sin }^{-1}}\left( \dfrac{5}{13} \right)$ to ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)$ which we are going to do by the following way:
${{\sin }^{-1}}x={{\cos }^{-1}}\left( \sqrt{1-{{x}^{2}}} \right)$
Substituting x as $\dfrac{5}{13}$ we get,
$\begin{align}
  & {{\sin }^{-1}}\left( \dfrac{5}{13} \right)={{\cos }^{-1}}\left( \sqrt{1-{{\left( \dfrac{5}{13} \right)}^{2}}} \right) \\
 & \Rightarrow {{\sin }^{-1}}\left( \dfrac{5}{13} \right)={{\cos }^{-1}}\left( \sqrt{\left( \dfrac{169-25}{169} \right)} \right) \\
 & \Rightarrow {{\sin }^{-1}}\left( \dfrac{5}{13} \right)={{\cos }^{-1}}\left( \sqrt{\left( \dfrac{144}{169} \right)} \right) \\
 & \Rightarrow {{\sin }^{-1}}\left( \dfrac{5}{13} \right)={{\cos }^{-1}}\left( \dfrac{12}{13} \right) \\
\end{align}$
From the above, we have converted ${{\sin }^{-1}}\left( \dfrac{5}{13} \right)$ to ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)$ which is equal to L.H.S.
Hence, L.H.S is equal to R.H.S