Show that the formula for the distance d between two points $P_1=(r_1,{\theta }_1)$and $P_2=(r_2,{\theta }_2)$ is role="math" localid="1646753790944" $d=\sqrt{{r_1}^2+{r_2}^2-2r_1{r}_2\cos ({\theta }_2-{\theta }_1)}$.

We are given two points in polar form $P_1=(r_1,{\theta }_1)$and $P_2=(r_2,{\theta }_2)$. We need to find the distance between them.

Let the point in polar form $P_1=(r_1,{\theta }_1)$be represented in rectangular form as $(x_1,y_1)$such that

localid="1646753933006" $$\begin{gathered} x_1=r_1\cos {\theta }_1 \\ {y}_1=r_1\sin {\theta }_1 \end{gathered}$$

And the point in polar form $P_2=(r_2,{\theta }_2)$be represented in rectangular form as $\left(x_2,y_2\right)$such that

$$\begin{gathered} x_2=r_2\cos {\theta }_2 \\ {y}_2=r_2\sin {\theta }_2 \end{gathered}$$

In rectangular coordinates the distance between the points $\left(x_1,y_1\right)$and $\left(x_2,y_2\right)$is given as $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Now using the conversion it can be written in polar form as

$$\begin{gathered} d=\sqrt{{\left(r_2\cos {\theta }_2-r_1\cos {\theta }_1\right)}^2+{(r_2\sin {\theta }_2-r_1\sin {\theta }_1)}^2} \\ d=\sqrt{{r_2}^2{\cos }^2{\theta }_2+{r_1}^2{\cos }^2{\theta }_1-2\left(r_2\cos {\theta }_2\right)\left(r_1\cos {\theta }_1\right)+{r_2}^2{\sin }^2{\theta }_2+{r_1}^2{\sin }^2{\theta }_1-2\left(r_2\sin {\theta }_2\right)\left(r_1\sin {\theta }_1\right)} \\ d=\sqrt{{r_1}^2({\cos }^2{\theta }_1+{\sin }^2{\theta }_1)+{r_2}^2({\cos }^2{\theta }_2+{\sin }^2{\theta }_2)-2r_1{r}_2(\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2)} \end{gathered}$$

Using the trigonometric relations it can be simplified as

$d=\sqrt{{r_1}^2+{r_2}^2-2r_1{r}_2\cos ({\theta }_2-{\theta }_1)}$

Hence the formula is derived.