Prove the equation ${{\sin }^{-1}}\left( \dfrac{8}{17} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\cos }^{-1}}\left( \dfrac{36}{65} \right)$.
Answer
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Hint: To solve this question, we should know the relation between the trigonometric ratios sine and cosine. We know the relation ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ which implies that $\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$. We should assume the terms in L.H.S as two different angles, that is, $\alpha ={{\sin }^{-1}}\left( \dfrac{8}{17} \right),\beta ={{\sin }^{-1}}\left( \dfrac{3}{5} \right)$. From these equations, we can write the values of $\sin \alpha ,\sin \beta ,\cos \alpha ,\cos \beta $. We have to evaluate the value of $\cos \left( \alpha +\beta \right)$ and by applying ${{\cos }^{-1}}$ to the equation, we get the required result.
Complete step by step answer:
Let us consider the two inverse sine terms in the L.H.S as two different angles.
$\alpha ={{\sin }^{-1}}\left( \dfrac{8}{17} \right)$
$\beta ={{\sin }^{-1}}\left( \dfrac{3}{5} \right)$
Applying sine on both sides in the above two equations, we get
$\begin{align}
& \sin \alpha =\dfrac{8}{17} \\
& \sin \beta =\dfrac{3}{5} \\
\end{align}$
We know the relation between sine and cosine functions as
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Subtracting ${{\sin }^{2}}\theta $ on both sides, we get
${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $
Applying square root on both sides, we get
$\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$
Using the above relation to get the values of $\cos \alpha ,\cos \beta $, we get
$\begin{align}
& \cos \alpha =\sqrt{1-{{\sin }^{2}}\alpha } \\
& \cos \alpha =\sqrt{1-{{\left( \dfrac{8}{17} \right)}^{2}}} \\
& \cos \alpha =\sqrt{1-\dfrac{64}{289}}=\sqrt{\dfrac{289-64}{289}}=\sqrt{\dfrac{225}{289}}=\dfrac{15}{17} \\
\end{align}$
$\begin{align}
& \cos \beta =\sqrt{1-{{\sin }^{2}}\beta } \\
& \cos \beta =\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}} \\
& \cos \beta =\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{25-9}{25}}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5} \\
\end{align}$
In the question, the L.H.S is in the form of $\alpha +\beta $ and the R.H.S contains ${{\cos }^{-1}}$ term in it. So, let us consider $\cos \left( \alpha +\beta \right)$
We know the relation $\cos (A+B)=\cos A\cos B-\sin A\sin B$
Using this relation, we get
$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta $
Substituting the values of $\sin \alpha ,\sin \beta ,\cos \alpha ,\cos \beta $ in the above equation, we get
$\cos \left( \alpha +\beta \right)=\dfrac{15}{17}\times \dfrac{4}{5}-\dfrac{8}{17}\times \dfrac{3}{5}=\dfrac{60}{85}-\dfrac{24}{85}$
As the denominator is same, we can simplify the fraction as
$\cos \left( \alpha +\beta \right)=\dfrac{60-24}{85}=\dfrac{36}{85}$
By applying the function of ${{\cos }^{-1}}$ on both sides, we get
${{\cos }^{-1}}\left( \cos \left( \alpha +\beta \right) \right)={{\cos }^{-1}}\left( \dfrac{36}{85} \right)\to \left( 1 \right)$
We know that $0<\alpha ,\beta <\dfrac{\pi }{2}$ from the inference that both the ${{\sin }^{-1}}$ values are positive,
$0<\alpha +\beta <\pi $.
So we can write that ${{\cos }^{-1}}\left( \cos \left( \alpha +\beta \right) \right)=\alpha +\beta $,
Rewriting the equation-1, we get
$\alpha +\beta ={{\cos }^{-1}}\left( \dfrac{36}{85} \right)$
Substituting the values of $\alpha ,\beta $, we get
${{\sin }^{-1}}\left( \dfrac{8}{17} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\cos }^{-1}}\left( \dfrac{36}{65} \right)$
$\therefore $The statement ${{\sin }^{-1}}\left( \dfrac{8}{17} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\cos }^{-1}}\left( \dfrac{36}{65} \right)$ is proved.
Note:
The alternate way to do this problem is by applying cosine function in the first step of the solution. Applying cosine function and using $\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta $, we get
$\begin{align}
& \cos \left( {{\sin }^{-1}}\left( \dfrac{8}{17} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)=\cos \left( {{\sin }^{-1}}\left( \dfrac{8}{17} \right) \right)\cos \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)-\sin \left( {{\sin }^{-1}}\left( \dfrac{8}{17} \right) \right)\sin \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right) \\
& \Rightarrow \cos \left( {{\sin }^{-1}}\left( \dfrac{8}{17} \right) \right)\cos \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)-\dfrac{8}{17}\times \dfrac{3}{5} \\
\end{align}$
We know the formula ${{\sin }^{-1}}\dfrac{x}{y}={{\cos }^{-1}}\dfrac{\sqrt{{{y}^{2}}-{{x}^{2}}}}{y}$.
Using this relation, we get
$\cos \left( {{\cos }^{-1}}\left( \dfrac{15}{17} \right) \right)\cos \left( {{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right)-\dfrac{8}{17}\times \dfrac{3}{5}=\dfrac{15}{17}\times \dfrac{4}{5}-\dfrac{8}{17}\times \dfrac{3}{5}=\dfrac{36}{65}$
$\begin{align}
& \cos \left( {{\sin }^{-1}}\left( \dfrac{8}{17} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)=\dfrac{36}{65} \\
& {{\sin }^{-1}}\left( \dfrac{8}{17} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\cos }^{-1}}\left( \dfrac{36}{65} \right) \\
\end{align}$
Complete step by step answer:
Let us consider the two inverse sine terms in the L.H.S as two different angles.
$\alpha ={{\sin }^{-1}}\left( \dfrac{8}{17} \right)$
$\beta ={{\sin }^{-1}}\left( \dfrac{3}{5} \right)$
Applying sine on both sides in the above two equations, we get
$\begin{align}
& \sin \alpha =\dfrac{8}{17} \\
& \sin \beta =\dfrac{3}{5} \\
\end{align}$
We know the relation between sine and cosine functions as
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Subtracting ${{\sin }^{2}}\theta $ on both sides, we get
${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $
Applying square root on both sides, we get
$\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$
Using the above relation to get the values of $\cos \alpha ,\cos \beta $, we get
$\begin{align}
& \cos \alpha =\sqrt{1-{{\sin }^{2}}\alpha } \\
& \cos \alpha =\sqrt{1-{{\left( \dfrac{8}{17} \right)}^{2}}} \\
& \cos \alpha =\sqrt{1-\dfrac{64}{289}}=\sqrt{\dfrac{289-64}{289}}=\sqrt{\dfrac{225}{289}}=\dfrac{15}{17} \\
\end{align}$
$\begin{align}
& \cos \beta =\sqrt{1-{{\sin }^{2}}\beta } \\
& \cos \beta =\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}} \\
& \cos \beta =\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{25-9}{25}}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5} \\
\end{align}$
In the question, the L.H.S is in the form of $\alpha +\beta $ and the R.H.S contains ${{\cos }^{-1}}$ term in it. So, let us consider $\cos \left( \alpha +\beta \right)$
We know the relation $\cos (A+B)=\cos A\cos B-\sin A\sin B$
Using this relation, we get
$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta $
Substituting the values of $\sin \alpha ,\sin \beta ,\cos \alpha ,\cos \beta $ in the above equation, we get
$\cos \left( \alpha +\beta \right)=\dfrac{15}{17}\times \dfrac{4}{5}-\dfrac{8}{17}\times \dfrac{3}{5}=\dfrac{60}{85}-\dfrac{24}{85}$
As the denominator is same, we can simplify the fraction as
$\cos \left( \alpha +\beta \right)=\dfrac{60-24}{85}=\dfrac{36}{85}$
By applying the function of ${{\cos }^{-1}}$ on both sides, we get
${{\cos }^{-1}}\left( \cos \left( \alpha +\beta \right) \right)={{\cos }^{-1}}\left( \dfrac{36}{85} \right)\to \left( 1 \right)$
We know that $0<\alpha ,\beta <\dfrac{\pi }{2}$ from the inference that both the ${{\sin }^{-1}}$ values are positive,
$0<\alpha +\beta <\pi $.
So we can write that ${{\cos }^{-1}}\left( \cos \left( \alpha +\beta \right) \right)=\alpha +\beta $,
Rewriting the equation-1, we get
$\alpha +\beta ={{\cos }^{-1}}\left( \dfrac{36}{85} \right)$
Substituting the values of $\alpha ,\beta $, we get
${{\sin }^{-1}}\left( \dfrac{8}{17} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\cos }^{-1}}\left( \dfrac{36}{65} \right)$
$\therefore $The statement ${{\sin }^{-1}}\left( \dfrac{8}{17} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\cos }^{-1}}\left( \dfrac{36}{65} \right)$ is proved.
Note:
The alternate way to do this problem is by applying cosine function in the first step of the solution. Applying cosine function and using $\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta $, we get
$\begin{align}
& \cos \left( {{\sin }^{-1}}\left( \dfrac{8}{17} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)=\cos \left( {{\sin }^{-1}}\left( \dfrac{8}{17} \right) \right)\cos \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)-\sin \left( {{\sin }^{-1}}\left( \dfrac{8}{17} \right) \right)\sin \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right) \\
& \Rightarrow \cos \left( {{\sin }^{-1}}\left( \dfrac{8}{17} \right) \right)\cos \left( {{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)-\dfrac{8}{17}\times \dfrac{3}{5} \\
\end{align}$
We know the formula ${{\sin }^{-1}}\dfrac{x}{y}={{\cos }^{-1}}\dfrac{\sqrt{{{y}^{2}}-{{x}^{2}}}}{y}$.
Using this relation, we get
$\cos \left( {{\cos }^{-1}}\left( \dfrac{15}{17} \right) \right)\cos \left( {{\cos }^{-1}}\left( \dfrac{4}{5} \right) \right)-\dfrac{8}{17}\times \dfrac{3}{5}=\dfrac{15}{17}\times \dfrac{4}{5}-\dfrac{8}{17}\times \dfrac{3}{5}=\dfrac{36}{65}$
$\begin{align}
& \cos \left( {{\sin }^{-1}}\left( \dfrac{8}{17} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right) \right)=\dfrac{36}{65} \\
& {{\sin }^{-1}}\left( \dfrac{8}{17} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\cos }^{-1}}\left( \dfrac{36}{65} \right) \\
\end{align}$
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