Question

Find the values of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$.

Answer 1

Hint: To solve the question given above, we will first out the values of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ and $2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$. The value of $2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ will be calculated with the help of the formula given below:
$2{{\sin }^{-1}}\left( x \right)={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)$. After calculating the respective values of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ and $2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$, we will add both their values to get the final answer.

Complete step-by-step solution -
To start with, we will first find out the value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$. Now, we know that the value of $\cos {{60}^{\circ }}=\dfrac{1}{2}$. Thus, we have:
$\Rightarrow \cos {{60}^{\circ }}=\dfrac{1}{2}$
Now, we will convert ${{60}^{\circ }}$ to radian form. The conversion from degree to radian is achieved by the following formula:
${{x}^{\circ }}=\dfrac{\pi }{180}\times x$ radian
Thus the value of ${{60}^{\circ }}$ = $\dfrac{\pi }{180}\times {{60}^{\circ }}=\dfrac{\pi }{3}$. Thus, we will get:
$\Rightarrow \cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$
Now, we will take ${{\cos }^{-1}}$ on both sides, thus we will get the following:
$\Rightarrow {{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{3} \right) \right)={{\cos }^{-1}}\left( \dfrac{1}{2} \right)$
Now, we will use the identity shown below:
$\Rightarrow {{\cos }^{-1}}\left( \cos x \right)=x$ (if $-1\le x\le 1$)
Thus, we will get:
$\Rightarrow \dfrac{\pi }{3}={{\cos }^{-1}}\left( \dfrac{1}{2} \right)$
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3}$ ----- (1)
Now, we will find the value of $2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$. The value of $2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ is calculated by the formula given below:
$2{{\sin }^{-1}}x={{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)$
In our case the value of x is $\dfrac{1}{2}$. So, we will get:

& 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)={{\sin }^{-1}}\left[ 2\left( \dfrac{1}{2} \right)\sqrt{1-{{\left( \dfrac{1}{2} \right)}^{2}}} \right] \\\ & \Rightarrow 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)={{\sin }^{-1}}\left[ 1\times \sqrt{1-\dfrac{1}{4}} \right] \\\ \end{aligned}$$ $$\Rightarrow 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)={{\sin }^{-1}}\left[ \dfrac{\sqrt{3}}{2} \right]$$ ------ (2) Now, we know that the value of $$\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$$. Thus, we have: $$\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$$ $$\begin{aligned} & \sin \left( 60\times \dfrac{\pi }{180}radian \right)=\dfrac{\sqrt{3}}{2} \\\ & \Rightarrow \sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$$ Now, we will take $${{\sin }^{-1}}$$ on both sides. After doing this, we will get the following equation: $$\Rightarrow {{\sin }^{-1}}\left( \sin \left( \dfrac{\pi }{3} \right) \right)={{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$$ Now, we will use an inverse trigonometric identity shown below: $${{\sin }^{-1}}\left( \sin x \right)=x$$ (if $$-1\le x\le 1$$) Thus, we will get following equation: $$\dfrac{\pi }{3}={{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$$ Now, we will put this value of $${{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)$$ from above equation to (ii). Thus, we will get eh following equation: $${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3}$$ ------- (3) Now, we will add equations (1) and (3). After doing this, we will get: $$\begin{aligned} & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3}+\dfrac{\pi }{3} \\\ & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{2\pi }{3} \\\ \end{aligned}$$. Note: The above question can also be solve by the method shown below: We can also write $${{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$$ as shown below: $$\begin{aligned} & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)={{\cos }^{-1}}\left( \dfrac{1}{2} \right)+{{\sin }^{-1}}\left( \dfrac{1}{2} \right)+{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \\\ & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\left[ {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right]+{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \\\ \end{aligned}$$ Now, we will use the identity: $${{\cos }^{-1}}x+{{\sin }^{-1}}x=\dfrac{\pi }{2}$$. Thus, we will get, $$\begin{aligned} & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\left[ \dfrac{\pi }{2} \right]+{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \\\ & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{2}+\dfrac{\pi }{6} \\\ & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{2\pi }{3} \\\ \end{aligned}$$.