Question

How do you calculate the arctan($0$)?

Answer 1

Hint: To solve this problem we should be aware of the fact that arctan($0$) is at which at which $\tan (x)=0$ and x belongs to the range, $-{\displaystyle \frac{\pi }{2}}<x<{\displaystyle \frac{\pi }{2}}$(or $-90<x<90$ if you use degrees).
Also, $\tan (x)={\displaystyle \frac{\sin x}{\cos x}}$.

Complete step by step solution:
We need to solve arctan($0$)
We know that, arctan($0$) is at which at which $\tan (x)=0$.
Tangent of theta is defined as the ratio of sine of theta to cosine of theta, i.e., $\tan (x)={\displaystyle \frac{\sin x}{\cos x}}$
So, we get that, $\tan (x)=0$only when, $\sin x=0$
We will use the below unit circle to generalize $\sin x=0$.

We get that, $\sin x=0$in the given range of x, $-{\displaystyle \frac{\pi }{2}}<x<{\displaystyle \frac{\pi }{2}}$
$x=0+n.2.\pi$or else,
$x=\pi +2.\pi .n$
In the given case, $x=n.\pi$, where n is an integer.
Only x = 0 satisfies, $-{\displaystyle \frac{\pi }{2}}<x<{\displaystyle \frac{\pi }{2}}$.
We get, $\sin x=0$
$\tan (x)=0$and,
arctan($0$) $=0$

Note:
Arctan($0$) is at which $\tan (x)=0$. Tangent of theta is defined as the ratio of sine of theta to cosine of theta, i.e., $\tan (x)={\displaystyle \frac{\sin x}{\cos x}}$. Thus, $\tan (x)=0$ only when, $\sin x=0$.