# Determine the missing coordinates of the points on the graph of the function. y=arctan

1. $(x,y)=(-\sqrt 3,a)$
2. $(x,y)=(b,-\dfrac{\pi}{6})$
3. $(x,y)=(c,\dfrac{\pi}{4})$

The question aims to determine the missing coordinates of the points on the graph of the function y= arctan x.

A pair of numbers that shows the exact position of a point in a cartesian plane using horizontal and vertical lines called coordinates. It is usually represented by (x, y) the value of x and the y value of the point on the graph. Each topic or paired order contains two links. The first is x coordinate or abscissa, and the second is y axis or ordinate. Point link values can be any real positive or negative number.

Part (a): For $(x,y)=(-\sqrt 3,a)$

The missing coordinate of the point on the graph pf the function $y=\arctan x$ is calculated as:

$y=\arctan (-\sqrt 3)(-\sqrt 3,y)$

$y=-\dfrac{\pi}{3}$

The output  for the missing variable $a$ for the function $y=\arctan x$ is $(x,y)=(-\sqrt 3,-\dfrac{\pi}{3})$.

Part(b): For $(x,y)=(b,-\dfrac{\pi}{6})$

The missing $x-axis$ which is represented by the variable $b$ is calculated by using following procedure.

$-\dfrac{\pi}{6}=\arctan (x)(x,-\dfrac{\pi}{6})$

$\tan(-\dfrac{\pi}{6})=x$

$x=-\dfrac{\sqrt 3}{3}$

The output of the variable $b$ for the function $y=\arctan x$ is $(x,y)=(-\dfrac{\pi}{6},-\dfrac{\sqrt 3}{3})$.

Part(c): For $(x,y)=(c,\dfrac{\pi}{4})$

The missing value of the variable $c$ which is the value of the $x-axis$ is calculated by using the following method.

$\tan\dfrac{\pi}{4}=x$

$x=1$

The output of the variable $c$ for the function $y=\arctan x$ is $(x,y)=(1,\dfrac{\pi}{4})$.

The output is (from left to right) $-\dfrac{\pi}{3},-\dfrac{\sqrt 3}{3},1$

## Numerical Result

The missing coordinates of the point for the graph of the function $y=\arctan x$ are calculated as:

Part (a)

$(x,y)=(-\sqrt 3,a)$

The missing coordinate value is $-\dfrac{\pi}{3}$.

Part(b)

-$(x,y)=(b,-\dfrac{\pi}{6})$

The missing coordinate value is $-\dfrac{\sqrt 3}{3}$.

Part(c)

-$(x,y)=(c,\dfrac{\pi}{4})$

The missing coordinate value is $1$.

$-\dfrac{\pi}{3},-\dfrac{\sqrt 3}{3},1$

## Example

Find the missing coordinates of the points on the graph of the functions: $y=cos^{-1} x$.

-$(x,y)=(-\frac{1}{2},a)$

-$(x,y)=(b,\pi)$

-$(x,y)=(c,\dfrac{\pi}{4})$

Part (a): For $(x,y)=(-\sqrt 2,a)$

The missing coordinate of the point on the graph pf the function $y=\arctan x$ is calculated as:

$y=\cos^{-1} (-\dfrac{1}{2})(-\dfrac{1}{2},y)$

$y=\dfrac{\pi}{3}$

The output of the missing variable $a$ for the function $y=\arctan x$ is $(x,y)=(-\dfrac{1}{2},\dfrac{\pi}{3})$.

Part(b): For $(x,y)=(b,\pi)$

The missing value of the variable $b$ that represents the $x-axis$ is calculated by using following procedure.

$-\pi=\cos (x)(x,\pi)$

$\cos(\pi)=x$

$x=1$

The output of the variable $b$ for the function $y=\arctan x$ is $(x,y)=(-\sqrt 3,\pi)$.

$\dfrac{\pi}{4}=\arctan(x)(x,\dfrac{\pi}{4})$

Part(c): For $(x,y)=(c,\dfrac{\pi}{4})$

The missing value of the variable $c$ that represents $x-axis$ is calculated by using the following method.

$\cos\dfrac{\pi}{4}=x$

$x=\dfrac{1}{\sqrt 2}$

The output is (from left to right) $\dfrac{\pi}{3},1,-\dfrac{1}{\sqrt 2}$