Question

Prove the given inverse trigonometric expression,$\dfrac{9\pi }{8}-\dfrac{9}{4}{{\sin }^{-1}}\dfrac{1}{3}=\dfrac{9}{4}{{\sin }^{-1}}\dfrac{2\sqrt{2}}{3}$.

Answer 1

Hint: In this question, we can see that $\dfrac{9}{4}$ is common in both of the terms present in LHS of the expression. So, we can take $\dfrac{9}{4}$ as common. Using the property ${{\sin }^{-1}}\theta +{{\cos }^{-1}}\theta =\dfrac{\pi }{2}$ , we can simplify the expression and then transform the inverse cosine function into inverse sine function.

Complete step-by-step solution -
Let us first consider the LHS of the given expression. Now taking $\dfrac{9}{4}$ as common, we get

& \dfrac{9\pi }{8}-\dfrac{9}{4}{{\sin }^{-1}}\dfrac{1}{3} \\\ & =\dfrac{9}{4}\left( \dfrac{\pi }{2}-{{\sin }^{-1}}\dfrac{1}{3} \right) \\\ \end{aligned}$$ In the above expression, we have $$\left( \dfrac{\pi }{2}-{{\sin }^{-1}}\dfrac{1}{3} \right)$$ . So, we need to convert this into a simpler form. And we also know that the property that, $${{\sin }^{-1}}\theta +{{\cos }^{-1}}\theta =\dfrac{\pi }{2}$$…………(1) In the above equation (1), taking $${{\sin }^{-1}}\theta$$ to the RHS, we get $${{\cos }^{-1}}\theta =\dfrac{\pi }{2}-{{\sin }^{-1}}\theta$$………….(2) We are replacing $$\theta$$ by $$\dfrac{1}{3}$$ in the equation (2), we get $${{\cos }^{-1}}\dfrac{1}{3}=\dfrac{\pi }{2}-{{\sin }^{-1}}\dfrac{1}{3}$$……………..(3) Now, according to the question we have, $$\dfrac{9}{4}\left( \dfrac{\pi }{2}-{{\sin }^{-1}}\dfrac{1}{3} \right)$$…………….(4) Using equation (3), and putting the value of $$\dfrac{\pi }{2}-{{\sin }^{-1}}\dfrac{1}{3}$$ in equation (4). We get, $$\begin{aligned} & \dfrac{9}{4}\left( \dfrac{\pi }{2}-{{\sin }^{-1}}\dfrac{1}{3} \right) \\\ & \\\ \end{aligned}$$ $$=\dfrac{9}{4}{{\cos }^{-1}}\dfrac{1}{3}$$………………..(5) But according to the question, we have the inverse of sine in RHS. So, here we have to convert it into the inverse of sine. Converting inverse cosine function to inverse sine function. Let us assume, $$x={{\cos }^{-1}}\dfrac{1}{3}$$……………..(6) Taking cosine in both of LHS as well as RHS in the equation (6), we get $$\begin{aligned} & x={{\cos }^{-1}}\dfrac{1}{3} \\\ & \Rightarrow \cos x=\dfrac{1}{3} \\\ \end{aligned}$$ Now, we have to find $$\sin x$$ . For that, we also know the identity $$\begin{aligned} & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\\ & \Rightarrow si{{n}^{2}}x=1-{{\cos }^{2}}x \\\ & \Rightarrow \sin x=\sqrt{1-{{\cos }^{2}}x} \\\ \end{aligned}$$ Using this formula, we can get the value of $$\sin x$$ . $$\begin{aligned} & \sin x=\sqrt{1-{{\cos }^{2}}x} \\\ & \Rightarrow \sin x=\sqrt{1-\dfrac{1}{9}} \\\ & \Rightarrow \sin x=\sqrt{\dfrac{9-1}{9}} \\\ & \Rightarrow \sin x=\sqrt{\dfrac{8}{9}} \\\ & \Rightarrow \sin x=\dfrac{2\sqrt{2}}{3} \\\ \end{aligned}$$ Now we have, $$\sin x=\dfrac{2\sqrt{2}}{3}$$……………(7) Taking the inverse of sine in both LHS as well as RHS in equation (7), we get $$\begin{aligned} & \sin x=\dfrac{2\sqrt{2}}{3} \\\ & \\\ \end{aligned}$$ $$\Rightarrow x={{\sin }^{-1}}\dfrac{2\sqrt{2}}{3}$$…………….(8) From equation (5), we have $$\dfrac{9}{4}{{\cos }^{-1}}\dfrac{1}{3}$$ . Using equation (6), we can write the equation (5) as $$\dfrac{9}{4}x$$ . And by using equation (8) we can write $$\dfrac{9}{4}x$$ as, $$\dfrac{9}{4}{{\sin }^{-1}}\dfrac{2\sqrt{2}}{3}$$………(9) So, $$\dfrac{9\pi }{8}-\dfrac{9}{4}{{\sin }^{-1}}\dfrac{1}{3}=\dfrac{9}{4}{{\sin }^{-1}}\dfrac{2\sqrt{2}}{3}$$ Therefore, LHS = RHS. Hence, proved. Note: In this question, one can think to find the principal value of inverse sine functions which are provided in LHS and RHS of the given equation, that is $${{\sin }^{-1}}\dfrac{1}{3}$$ and $$\dfrac{9}{4}{{\sin }^{-1}}\dfrac{2\sqrt{2}}{3}$$ . If we do so, then only we are increasing the complexity and here, we don’t need the principal value of inverse sine function.