Question

Prove the given expression ${{\cos }^{-1}}\dfrac{12}{13}+{{\sin }^{-1}}\dfrac{3}{5}={{\sin }^{-1}}\dfrac{56}{65}$ .

Answer 1

Hint: 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

${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$

& \Rightarrow {{\sin }^{2}}\alpha =1-{{\cos }^{2}}\alpha \\\ & \Rightarrow \sin \alpha =\sqrt{1-{{\cos }^{2}}\alpha } \\\ \end{aligned}$$ Substituting the value of $$\cos \alpha =\dfrac{12}{13}$$ , we get $$\begin{aligned} & \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{aligned}$$ Similarly, replacing x by $$\beta $$ in equation (10), we get $${{\sin }^{2}}\beta +{{\cos }^{2}}\beta =1$$ $$\begin{aligned} & \Rightarrow {{\cos }^{2}}\beta =1-si{{n}^{2}}\beta \\\ & \Rightarrow \cos \beta =\sqrt{1-si{{n}^{2}}\beta } \\\ \end{aligned}$$ Substituting the value of $$sin\beta =\dfrac{3}{5}$$ , we get $$\begin{aligned} & \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{aligned}$$ Now, we have got all the values required to solve equation(9). $$\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{aligned} & \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{aligned}$$ Now, we have $$\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{aligned} & \alpha ={{\cos }^{-1}}\dfrac{12}{13} \\\ & \Rightarrow \cos \alpha =\dfrac{12}{13} \\\ \end{aligned}$$ Now, using a right-angled triangle, we can get the value of $$\sin \alpha $$.  Using the Pythagoras theorem, we can find the height. Height = $$\sqrt{{{\left( hypotenuse \right)}^{2}}-{{\left( base \right)}^{2}}}$$ $$\begin{aligned} & \sqrt{{{\left( 13 \right)}^{2}}-{{(12)}^{2}}} \\\ & =\sqrt{169-144} \\\ & =\sqrt{25} \\\ & =5 \\\ \end{aligned}$$ $$\begin{aligned} & \sin \alpha =\dfrac{height}{hypotenuse} \\\ & \sin \alpha =\dfrac{5}{13} \\\ \end{aligned}$$ Similarly, assume $$\begin{aligned} & \beta ={{\sin }^{-1}}\dfrac{3}{5} \\\ & \Rightarrow \sin \beta =\dfrac{3}{5} \\\ \end{aligned}$$ Now, using a right-angled triangle, we can get the value of $$\sin \alpha $$.  Using the Pythagoras theorem, we can find the base. Base = $$\sqrt{{{\left( hypotenuse \right)}^{2}}-{{\left( height \right)}^{2}}}$$ $$\begin{aligned} & \sqrt{{{\left( 5 \right)}^{2}}-{{(3)}^{2}}} \\\ & =\sqrt{25-9} \\\ & =\sqrt{16} \\\ & =4 \\\ \end{aligned}$$ $$\begin{aligned} & \cos \beta =\dfrac{base}{hypotenuse} \\\ & \cos \beta =\dfrac{4}{5} \\\ \end{aligned}$$ Now, use the formula, $$\sin (\alpha +\beta )=sin\alpha cos\beta +cos\alpha sin\beta $$, solve it further.