Question

How do you graph $r=1-\sin \left(\theta \right)$ ?

Answer 1

Hint:The given expression is $r=1-\sin \left(\theta \right)$ which produces a cardioid. In the given expression $r=1-\sin \left(\theta \right)$ try to substitute different values for $\theta$ and find the corresponding values of $r$ and plot the graph for the same values.

Complete step by step answer:
The given expression that is $r=1-\sin \left(\theta \right)$ which is a polar coordinate produces the cardioid. Cardioid is nothing but a curve or a graph that somewhat looks like a heart-shaped curve.
The graph of a cardioid looks as shown below.

Now, to draw a graph for $r=1-\sin \left(\theta \right)$ , try to substitute different values for $\theta$ which varies from $0$ to $2\pi$.
The below table gives us the values of sine function for different values:
$\begin{array}{ccccccccc}\theta & 0^{\circ }& {30}^{\circ }& {45}^{\circ }& {60}^{\circ }& {90}^{\circ }& {180}^{\circ }& {270}^{\circ }& {360}^{\circ }\\ \sin \theta & 0& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{\sqrt{2}}{2}}& {\displaystyle \frac{1}{2}}& 1& 0& -1& 0\end{array}$
Now we consider different values for $\theta$ to which we need to find the corresponding values of $r$ .
So let $\theta =0$ now to find the corresponding value of $r$ we can write as below,
$\Rightarrow r=1-\sin \left(0^{\circ }\right)=1-0=1$
At $\theta ={30}^{\circ }$ the value of $r$ is
$\Rightarrow r=1-\sin \left({30}^{\circ }\right)=1-{\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{2}}$
At $\theta ={60}^{\circ }$ the value of $r$ is
$\Rightarrow r=1-\sin \left({30}^{\circ }\right)=1-{\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{2}}$
At $\theta ={90}^{\circ }$ the value of $r$ we get as
$\Rightarrow r=1-\sin \left({90}^{\circ }\right)=1-1=0$
In the same way the values of $r$ can be listed as below for different values of $\theta$ .
$$\begin{array}{cccccccc}\theta & 0^{\circ }& {30}^{\circ }& {60}^{\circ }& {90}^{\circ }& {180}^{\circ }& {270}^{\circ }& {360}^{\circ }\\ r& 1& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{2}}& 0& 1& 2& 1\end{array}$$
Now, plot the graph for the above values. Which is shown as in the below figure.

Therefore, the graph for the given expression $r=1-\sin \left(\theta \right)$ is as shown in the above figure.

Note: Whenever they ask us to draw a graph by giving an equation, then just take some values for one unknown that is for $\theta$ in the given equation and find the corresponding values of another unknown that is $r$ in this problem. Plot the same on a graph sheet as we did above.