Question

Evaluate: ${\tan ^{ - 1}}\left( 1 \right) + {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$.

Answer 1

Here, we need to find the value of the given expression. 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 solution:
Let ${\tan ^{ - 1}}\left( 1 \right) = \theta$, ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \alpha$, and ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \beta$.
Therefore, the expression becomes
$\Rightarrow {\tan ^{ - 1}}\left( 1 \right) + {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \theta + \alpha + \beta$
First, we will simplify the equation ${\tan ^{ - 1}}\left( 1 \right) = \theta$.
Rewriting the equation ${\tan ^{ - 1}}\left( 1 \right) = \theta$, we get
$\tan \theta = 1$
The tangent of the angle measuring $\dfrac{\pi }{4}$ is equal to 1.
Thus, we get
$\tan \dfrac{\pi }{4} = 1$
From the equations $\tan \theta = 1$ and $\tan \dfrac{\pi }{4} = 1$, we get
$\Rightarrow \tan \theta = \tan \dfrac{\pi }{4}$
Therefore, we get
$\Rightarrow \theta = \dfrac{\pi }{4}$
Now, 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
$\cos \alpha = - \dfrac{1}{2}$
The trigonometric ratio cosine is negative in the second and third quadrant.
The cosine of an angle $\pi - x$, is equal to the negative of the cosine of angle $x$, where $x$ is an acute angle.
Therefore, we get
$\Rightarrow \cos \left( {\pi - x} \right) = - \cos x$
Substituting $x = \dfrac{\pi }{3}$ in the equation, we get
$\Rightarrow \cos \left( {\pi - \dfrac{\pi }{3}} \right) = - \cos \dfrac{\pi }{3}$
We know that $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$.
Therefore, we get
$\begin{array}{l} \Rightarrow \cos \left( {\pi - \dfrac{\pi }{3}} \right) = - \dfrac{1}{2}\\\ \Rightarrow \cos \dfrac{{2\pi }}{3} = - \dfrac{1}{2}\end{array}$
From the equations $\cos \alpha = - \dfrac{1}{2}$ and $\cos \dfrac{{2\pi }}{3} = - \dfrac{1}{2}$, we get
$\Rightarrow \cos \alpha = \cos \dfrac{{2\pi }}{3}$
Therefore, we get
$\Rightarrow \alpha = \dfrac{{2\pi }}{3}$
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
$\sin \beta = - \dfrac{1}{2}$
The trigonometric ratio sine is negative in the third and fourth quadrant.
The sine of an angle $- x$, is equal to the negative of the sine of angle $x$, where $x$ is an acute angle.
Therefore, we get
$\Rightarrow \sin \left( { - x} \right) = - \sin x$
Substituting $x = \dfrac{\pi }{6}$ in the equation, we get
$\Rightarrow \sin \left( { - \dfrac{\pi }{6}} \right) = - \sin \dfrac{\pi }{6}$
We know that $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$.
Therefore, we get
$\Rightarrow \sin \left( { - \dfrac{\pi }{6}} \right) = - \dfrac{1}{2}$
From the equations $\sin \beta = - \dfrac{1}{2}$ and $\sin \left( { - \dfrac{\pi }{6}} \right) = - \dfrac{1}{2}$, we get
$\Rightarrow \sin \beta = \sin \left( { - \dfrac{\pi }{6}} \right)$
Therefore, we get
$\Rightarrow \beta = - \dfrac{\pi }{6}$
Now, we will evaluate the given expression.
Substituting $\theta = \dfrac{\pi }{4}$, $\alpha = \dfrac{{2\pi }}{3}$, and $\beta = - \dfrac{\pi }{6}$ in the equation ${\tan ^{ - 1}}\left( 1 \right) + {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \theta + \alpha + \beta$, we get
$\Rightarrow {\tan ^{ - 1}}\left( 1 \right) + {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \dfrac{\pi }{4} + \dfrac{{2\pi }}{3} + \left( { - \dfrac{\pi }{6}} \right)$
Simplifying the expression, we get
$\Rightarrow {\tan ^{ - 1}}\left( 1 \right) + {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \dfrac{\pi }{4} + \dfrac{{2\pi }}{3} - \dfrac{\pi }{6}$
Taking the L.C.M., we get
$\Rightarrow {\tan ^{ - 1}}\left( 1 \right) + {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \dfrac{{3\pi + 8\pi - 2\pi }}{{12}}$
Adding and subtracting the like terms, we get
$\Rightarrow {\tan ^{ - 1}}\left( 1 \right) + {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) + {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \dfrac{{9\pi }}{{12}} = \dfrac{{3\pi }}{4}$

$\therefore$ The value of the given expression is $\dfrac{{3\pi }}{4}$.

Note:
Here, we need to remember the range of the trigonometric inverse functions. The range of ${\tan ^{ - 1}}\left( x \right)$ is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$, 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 $\cos \dfrac{{4\pi }}{3} = - \dfrac{1}{2}$, or $\sin \dfrac{{7\pi }}{6} = - \dfrac{1}{2}$, or both. This is because if $\cos \dfrac{{4\pi }}{3} = - \dfrac{1}{2}$, then ${\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \dfrac{{4\pi }}{3}$, which does not lie in the range of ${\cos ^{ - 1}}\left( x \right)$, that is $\left[ {0,\pi } \right]$. Similarly, if $\sin \dfrac{{7\pi }}{6} = - \dfrac{1}{2}$, then ${\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) = \dfrac{{7\pi }}{6}$, which does not lie in the range of ${\sin ^{ - 1}}\left( x \right)$, that is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$.