Question

Calculate the value of $\arctan \left( 1 \right)+\arccos \left( \dfrac{-1}{2} \right)+\arcsin \left( \dfrac{-1}{2} \right)$.

Answer 1

Hint: To solve this problem, we should know the meaning of $\arcsin ,\text{ }\arccos ,\text{ arctan}$and the range of each function. The ranges of $\arcsin ,\text{ }\arccos ,\text{ arctan}$ are $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\text{ , }\left[ 0,\pi \right]\text{ , }\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ respectively. The equation $\arcsin x=\theta $ means, in the given range of $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, the value of $\theta $ for which $\sin \theta =x$. The above definition is the same for $\arccos ,\text{ arctan}$ also. Using this concept, we can calculate the values of the terms in the required expression, we get the values of the three angles and the sum of them, will give the required answer.

Complete step-by-step solution:
The equation $\arcsin x=\theta $ means, in the given range of $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, the value of $\theta $ for which $\sin \theta =x$.
The equation $\arccos x=\theta $ means, in the given range of \[\left[ 0,\pi \right]\], the value of $\theta $ for which $\cos \theta =x$.
The equation $\arctan x=\theta $ means, in the given range of $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$, the value of $\theta $ for which $\tan \theta =x$.
Let us consider $\arctan 1$. We can infer that for the value of $\dfrac{\pi }{4}$, the value of $\tan \dfrac{\pi }{4}=1$ is valid and $\dfrac{\pi }{4}$ is in the range of $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. So, we can write that
$\arctan 1=\dfrac{\pi }{4}$
Let us consider $\arcsin \left( \dfrac{-1}{2} \right)$. We can infer that for the value of $-\dfrac{\pi }{6}$, the value of $\sin \left( -\dfrac{\pi }{6} \right)=\dfrac{-1}{2}$ is valid and $-\dfrac{\pi }{6}$ is in the range of $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. So, we can write that
$\arcsin \left( \dfrac{-1}{2} \right)=-\dfrac{\pi }{6}$
Let us consider $\arccos \left( \dfrac{-1}{2} \right)$. We can infer that for the value of $\dfrac{2\pi }{3}$, the value of $\cos \left( \dfrac{2\pi }{3} \right)=\dfrac{-1}{2}$ is valid and $\dfrac{2\pi }{3}$ is in the range of \[\left[ 0,\pi \right]\]. So, we can write that
$\arccos \left( \dfrac{-1}{2} \right)=\dfrac{2\pi }{3}$
Using the above values in the expression $\arctan \left( 1 \right)+\arccos \left( \dfrac{-1}{2} \right)+\arcsin \left( \dfrac{-1}{2} \right)$, we get
$\arctan \left( 1 \right)+\arccos \left( \dfrac{-1}{2} \right)+\arcsin \left( \dfrac{-1}{2} \right)=\dfrac{\pi }{4}+\dfrac{2\pi }{3}-\dfrac{\pi }{6}=\dfrac{\pi }{4}+\dfrac{4\pi -\pi }{6}=\dfrac{\pi }{4}+\dfrac{\pi }{2}=\dfrac{3\pi }{4}$
$\therefore $ The required value of the expression in the expression is $\dfrac{3\pi }{4}={{135}^{\circ }}$.

Note: An alternate way to do this problem is by using a property in inverse trigonometry. We can write it as $\arcsin x+\arccos x=\dfrac{\pi }{2}\text{ }\forall \text{x}\in \left[ -1,1 \right]$. Using this relation when $x=\dfrac{-1}{2}$, we get $\arcsin \left( \dfrac{-1}{2} \right)+\arccos \left( \dfrac{-1}{2} \right)=\dfrac{\pi }{2}$. Using this relation, we can directly rewrite the expression in the question as $\arctan 1+\dfrac{\pi }{2}$.