Question

Solve the given expression and find the value of x, $\cos (2{{\sin }^{-1}}x)=\dfrac{1}{9},x>0$.

Answer 1

Hint: In this question, we have a sine inverse function in LHS. But on the other hand, we don’t have an inverse trigonometric function in RHS. So, here our main target is to remove the sine inverse function to get the numerical value as given in RHS. We consider ${{\sin }^{-1}}x=\theta$. Transform $\sin \theta$ into $\cos \theta$. And then, using the identity $\cos 2\theta =1-2{{\sin }^{2}}\theta$ , we can solve it further.

Complete step-by-step solution -
Solving LHS of the given expression, first of all, we have to remove the inverse part present in LHS of the given equation.
Let us assume, ${{\sin }^{-1}}x=\theta$……………..(1)
According to the question, the LHS part has $\cos (2{{\sin }^{-1}}x)$.
From equation (1), we can write $\cos (2{{\sin }^{-1}}x)$ as,

& \cos (2{{\sin }^{-1}}x) \\\ & =cos(2\theta ) \\\ \end{aligned}$$ We can write the equation given in the question as, $$\begin{aligned} & \cos (2{{\sin }^{-1}}x)=\dfrac{1}{9} \\\ & \Rightarrow \cos (2\theta )=\dfrac{1}{9} \\\ & \\\ \end{aligned}$$ Now, we have to expand $$\cos 2\theta$$ . We also know the identity, $$\cos 2\theta =1-2{{\sin }^{2}}\theta$$ …….(2) Using equation (2), we can write $$\begin{aligned} & \cos 2\theta =\dfrac{1}{9} \\\ & \Rightarrow 1-2{{\sin }^{2}}\theta =\dfrac{1}{9} \\\ \end{aligned}$$ Here, we need the value of $$\sin \theta$$ . For that, use equation (1). From equation(1), we have $$\theta ={{\sin }^{-1}}x$$ . Taking sin in both LHS as well as RHS in the equation (1), we get $$\sin \theta =\sin \left( {{\sin }^{-1}}x \right)$$ $$\Rightarrow \sin \theta =x$$…………….(4) We need to solve the equation, $$1-2{{\sin }^{2}}\theta =\dfrac{1}{9}$$…………….(5) Using equation(4) in equation(5), we can write equation (5) as, $$1-2{{x}^{2}}=\dfrac{1}{9}$$……………(6) As equation(6) is quadratic. Now, we can solve equation(6) for getting the values of x. $$\begin{aligned} & 1-2{{x}^{2}}=\dfrac{1}{9} \\\ & \Rightarrow 2{{x}^{2}}=1-\dfrac{1}{9} \\\ & \Rightarrow 2{{x}^{2}}=\dfrac{9-1}{9} \\\ & \Rightarrow 2{{x}^{2}}=\dfrac{8}{9} \\\ \end{aligned}$$ Dividing by 2 in both LHS and RHS, we get $$\begin{aligned} & {{x}^{2}}=\dfrac{4}{9} \\\ & \Rightarrow x=\pm \dfrac{2}{3} \\\ \end{aligned}$$ After solving the quadratic equation, we have got two values of x. One is $$+\dfrac{2}{3}$$ while the other is $$-\dfrac{2}{3}$$ . According to the question, we have $$x>0$$ . So, the value of x is $$+\dfrac{2}{3}$$ . Therefore, $$x=\dfrac{2}{3}$$ . Note: In this question, after solving the quadratic equation, we get two values of x. One can write both values of x as an answer, which is wrong. According to the information provided in the question, we have one restriction on x that is x should be greater than zero, which means x is positive. To satisfy this information, we have to take the positive value of x and ignore the negative value of x.