Prove the given expression \[{{\cos }^{-1}}\dfrac{12}{13}+{{\sin }^{-1}}\dfrac{3}{5}={{\sin }^{-1}}\dfrac{56}{65}\] .
VerifiedHint: In this question, we have different inverse trigonometric functions in LHS. In LHS, we have cosine inverse function and sine inverse function. On the other hand, in RHS, we have a sine inverse function. So, in LHS, our target is to make cosine inverse function into a sine inverse function. We consider \[\alpha ={{\cos }^{-1}}\dfrac{12}{13}\] and \[\beta ={{\sin }^{-1}}\dfrac{3}{5}\] . Transform \[\cos \alpha \] into \[sin\alpha \].
Then using the formula, \[\sin (\alpha +\beta )=sin\alpha .cos\beta +\sin \beta .cos\alpha \] , solve it further.
Complete step-by-step solution -
Let us assume,
\[\alpha ={{\cos }^{-1}}\dfrac{12}{13}\]……………..(1)
\[\beta ={{\sin }^{-1}}\dfrac{3}{5}\]………………..(2)
Taking cos in both LHS and RHS in equation(1), we get
\[\alpha ={{\cos }^{-1}}\dfrac{12}{13}\]
\[\Rightarrow \cos \alpha =cos\left( {{\cos }^{-1}}\dfrac{12}{13} \right)\] ……………….(3)
We know the property that, \[\cos ({{\cos }^{-1}}x)=x\] . Using this property in equation (3), we get
\[\cos \alpha =\dfrac{12}{13}\]………………..(4)
Taking sin in both LHS and RHS in equation(2), we get
\[\beta ={{\sin }^{-1}}\dfrac{3}{5}\]
\[\Rightarrow \sin \beta =\sin \left( {{\sin }^{-1}}\dfrac{3}{5} \right)\] ……………………..(5)
We know the property that, \[sin(si{{n}^{-1}}x)=x\] . Using this property in equation (5), we get
\[\sin \beta =\dfrac{3}{5}\]………………….(6)
According to the question, in LHS we have
\[{{\cos }^{-1}}\dfrac{12}{13}+{{\sin }^{-1}}\dfrac{3}{5}\]………………(7)
Using equation(1) and equation(2) in equation(7), we have
\[\alpha +\beta\]………………(8)
This means we have to find the value of \[\alpha +\beta\] .
In RHS we have inverse sine function. So, here we have to take sin in equation(8).
\[\sin (\alpha +\beta )=sin\alpha cos\beta +cos\alpha sin\beta\]……………….(9)
From equation(4) and equation(6), we have the value of \[\cos \alpha\] and \[\sin \beta\] .
But we don’t have the value of \[sin\alpha\] and \[\cos \beta\] .
Also, we know the identity, \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]………….(10)
Using equation(4) and equation(6), we can get the value of \[sin\alpha\] and \[\cos \beta\] .
Replacing x by \[\alpha \] in equation (10), we get
\[{{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1\]
\[\begin{align}
& \Rightarrow {{\sin }^{2}}\alpha =1-{{\cos }^{2}}\alpha \\
& \Rightarrow \sin \alpha =\sqrt{1-{{\cos }^{2}}\alpha } \\
\end{align}\]
Substituting the value of \[\cos \alpha =\dfrac{12}{13}\] , we get
\[\begin{align}
& \sin \alpha =\sqrt{1-{{\cos }^{2}}\alpha } \\
& \Rightarrow \sin \alpha =\sqrt{1-\dfrac{144}{169}} \\
& \Rightarrow \sin \alpha =\sqrt{\dfrac{25}{169}} \\
& \Rightarrow \sin \alpha =\dfrac{5}{13} \\
\end{align}\]
Similarly, replacing x by \[\beta \] in equation (10), we get
\[{{\sin }^{2}}\beta +{{\cos }^{2}}\beta =1\]
\[\begin{align}
& \Rightarrow {{\cos }^{2}}\beta =1-si{{n}^{2}}\beta \\
& \Rightarrow \cos \beta =\sqrt{1-si{{n}^{2}}\beta } \\
\end{align}\]
Substituting the value of \[sin\beta =\dfrac{3}{5}\] , we get
\[\begin{align}
& \cos \beta =\sqrt{1-{{\sin }^{2}}\beta } \\
& \Rightarrow \cos \beta =\sqrt{1-\dfrac{9}{25}} \\
& \Rightarrow \cos \beta =\sqrt{\dfrac{25-9}{25}} \\
& \Rightarrow \cos \beta =\sqrt{\dfrac{16}{25}} \\
& \Rightarrow \cos \beta =\dfrac{4}{5} \\
\end{align}\]
Now, we have got all the values required to solve equation(9).
\[\cos \beta =\dfrac{4}{5},\sin \beta =\dfrac{3}{5},\cos \alpha =\dfrac{12}{13},\sin \alpha =\dfrac{5}{13}\].
Putting these values in equation(9), we get
\[\begin{align}
& \sin (\alpha +\beta )=sin\alpha cos\beta +cos\alpha sin\beta \\
& =\dfrac{5}{13}.\dfrac{4}{5}+\dfrac{12}{13}.\dfrac{3}{5} \\
& =\dfrac{20+36}{65} \\
& =\dfrac{56}{65} \\
\end{align}\]
Now, we have
\[\sin (\alpha +\beta )=\dfrac{56}{65}\]
\[\Rightarrow \alpha +\beta ={{\sin }^{-1}}\left( \dfrac{56}{65} \right)\]……………….(10)
Equation(7) is equal to equation(10).
So, \[{{\cos }^{-1}}\dfrac{12}{13}+{{\sin }^{-1}}\dfrac{3}{5}={{\sin }^{-1}}\dfrac{56}{65}\].
Therefore, LHS = RHS.
Hence, proved.
Note: This question can also be solved by using the Pythagoras theorem.
Assume,
\[\begin{align}
& \alpha ={{\cos }^{-1}}\dfrac{12}{13} \\
& \Rightarrow \cos \alpha =\dfrac{12}{13} \\
\end{align}\]
Now, using a right-angled triangle, we can get the value of \[\sin \alpha \].
Using the Pythagoras theorem, we can find the height.
Height = \[\sqrt{{{\left( hypotenuse \right)}^{2}}-{{\left( base \right)}^{2}}}\]
\[\begin{align}
& \sqrt{{{\left( 13 \right)}^{2}}-{{(12)}^{2}}} \\
& =\sqrt{169-144} \\
& =\sqrt{25} \\
& =5 \\
\end{align}\]
\[\begin{align}
& \sin \alpha =\dfrac{height}{hypotenuse} \\
& \sin \alpha =\dfrac{5}{13} \\
\end{align}\]
Similarly, assume
\[\begin{align}
& \beta ={{\sin }^{-1}}\dfrac{3}{5} \\
& \Rightarrow \sin \beta =\dfrac{3}{5} \\
\end{align}\]
Now, using a right-angled triangle, we can get the value of \[\sin \alpha \].
Using the Pythagoras theorem, we can find the base.
Base = \[\sqrt{{{\left( hypotenuse \right)}^{2}}-{{\left( height \right)}^{2}}}\]
\[\begin{align}
& \sqrt{{{\left( 5 \right)}^{2}}-{{(3)}^{2}}} \\
& =\sqrt{25-9} \\
& =\sqrt{16} \\
& =4 \\
\end{align}\]
\[\begin{align}
& \cos \beta =\dfrac{base}{hypotenuse} \\
& \cos \beta =\dfrac{4}{5} \\
\end{align}\]
Now, use the formula, \[\sin (\alpha +\beta )=sin\alpha cos\beta +cos\alpha sin\beta \], solve it further.
