Question

The value of ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ is equal to
(a) $\dfrac{\pi }{6}$
(b) $\dfrac{\pi }{3}$
(c) $\dfrac{{2\pi }}{3}$
(d) $\dfrac{\pi }{4}$

Answer 1

Here, we need to find the value of the given expression. We will equate each term of the given expression to a variable and form an equation. We will simplify each of these equations using trigonometric ratios of specific angles. Then, we will rewrite and simplify the given expression to get the required answer.

Complete step-by-step answer:
Let ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \alpha$, and ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \beta$.
We will simplify the two equations to find the values of $\alpha$ and $\beta$. Then, we will use these values of $\alpha$ and $\beta$ to simplify and obtain the value of the given expression.
Rewriting the given expression, we get
$\Rightarrow {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \alpha + 2\beta$
First, we will simplify the equation ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \alpha$.
Rewriting the equation ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \alpha$, we get
$\Rightarrow \cos \alpha = \dfrac{1}{2}$
The cosine of the angle measuring $\dfrac{\pi }{3}$ is $\dfrac{1}{2}$. This can be written as $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$.
From the equations $\cos \alpha = \dfrac{1}{2}$ and $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$, we get
$\Rightarrow \cos \alpha = \cos \dfrac{\pi }{3}$
Therefore, we get
$\Rightarrow \alpha = \dfrac{\pi }{3}$
Next, we will simplify the equation ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \beta$.
Rewriting the equation ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \beta$, we get
$\Rightarrow \sin \beta = \dfrac{1}{2}$
The sine of the angle measuring $\dfrac{\pi }{6}$ is $\dfrac{1}{2}$. This can be written as $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$.
From the equations $\sin \beta = \dfrac{1}{2}$ and $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$, we get
$\Rightarrow \sin \beta = \sin \dfrac{\pi }{6}$
Therefore, we get
$\Rightarrow \beta = \dfrac{\pi }{6}$
Now, we will evaluate the given expression.
Substituting $\alpha = \dfrac{\pi }{3}$ and $\beta = \dfrac{\pi }{6}$ in the equation ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \alpha + 2\beta$, we get
$\Rightarrow {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{3} + 2\left( {\dfrac{\pi }{6}} \right)$
Simplifying the expression, we get
$\Rightarrow {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{3} + \dfrac{\pi }{3}$
Taking the L.C.M., we get
$\Rightarrow {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{{\pi + \pi }}{3}$
Adding the like terms, we get
$\Rightarrow {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{{2\pi }}{3}$
$\therefore$ The value of the given expression ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) + 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ is $\dfrac{{2\pi }}{3}$.
Thus, the correct option is option (c).

Note: We need to keep in mind the range of the trigonometric inverse functions. The range of ${\cos ^{ - 1}}\left( x \right)$ is $\left[ {0,\pi } \right]$ and the range of ${\sin ^{ - 1}}\left( x \right)$ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$.
A common mistake is to use either $\cos \dfrac{{5\pi }}{3} = \dfrac{1}{2}$, or $\sin \dfrac{{5\pi }}{6} = \dfrac{1}{2}$, or both. This is because if $\cos \dfrac{{5\pi }}{3} = \dfrac{1}{2}$, then ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{{5\pi }}{3}$, which does not lie in the range of ${\cos ^{ - 1}}\left( x \right)$, that is $\left[ {0,\pi } \right]$. Similarly, if $\sin \dfrac{{5\pi }}{6} = \dfrac{1}{2}$, then ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{{5\pi }}{6}$, which does not lie in the range of ${\sin ^{ - 1}}\left( x \right)$, that is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$.