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:
\[\begin{align}
  & 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)={{\sin }^{-1}}\left[ 2\left( \dfrac{1}{2} \right)\sqrt{1-{{\left( \dfrac{1}{2} \right)}^{2}}} \right] \\
 & \Rightarrow 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)={{\sin }^{-1}}\left[ 1\times \sqrt{1-\dfrac{1}{4}} \right] \\
\end{align}\]
\[\Rightarrow 2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)={{\sin }^{-1}}\left[ \dfrac{\sqrt{3}}{2} \right]\] ------ (2)
Now, we know that the value of \[\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\]. Thus, we have:
\[\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\]
\[\begin{align}
  & \sin \left( 60\times \dfrac{\pi }{180}radian \right)=\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow \sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2} \\
\end{align}\]
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:
\[\begin{align}
  & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3}+\dfrac{\pi }{3} \\
 & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{2\pi }{3} \\
\end{align}\].

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:
\[\begin{align}
  & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)={{\cos }^{-1}}\left( \dfrac{1}{2} \right)+{{\sin }^{-1}}\left( \dfrac{1}{2} \right)+{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \\
 & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\left[ {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \right]+{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \\
\end{align}\]
Now, we will use the identity: \[{{\cos }^{-1}}x+{{\sin }^{-1}}x=\dfrac{\pi }{2}\]. Thus, we will get,
\[\begin{align}
  & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\left[ \dfrac{\pi }{2} \right]+{{\sin }^{-1}}\left( \dfrac{1}{2} \right) \\
 & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{2}+\dfrac{\pi }{6} \\
 & \Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)+2{{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{2\pi }{3} \\
\end{align}\].