Chapter 7.2, Problem 6QR

Solution

To calculate: The coordinates of the point $P$ in terms of $\theta$ by using the following figure.

  

The obtained point is $\left(r\cos \theta ,r\sin \theta \right)$ .

Given Information:

The figure is,

  

Formula Used:

The sine and cosine ratios are defined as,

  $\sin \theta =\frac{\text{hyp}}{\text{opp}\text{. side}}$

And,

  $\cos \theta =\frac{\text{base}}{\text{opp}\text{. side}}$

Calculation:

Consider the given matrix,

Rewrite the given figure as following,

  

There is given the hypotenuse so in order to figure out the $x$ and $y$ . There must be use SOH CAH, or TOA, it can cross out tangent because here do not have opposite or adjacent, opposite from theta we have the $y$ and adjacent to theta, have $x$ .

Now, use the sine ratio to find the $\sin \theta$ in the triangle $\angle PAB$ .

  $\begin{array}{c}\sin \theta =\frac{PB}{PA}\\ \sin \theta =\frac{y}{r}\\ y=r\sin \theta \end{array}$

Use the cosine ratio to find the $\cos \theta$ in the triangle $\angle PAB$ .

  $\begin{array}{c}\cos \theta =\frac{AB}{AP}\\ \cos \theta =\frac{x}{r}\\ x=r\cos \theta \end{array}$

Substitute the obtained values in the point $P$ .

  $P\left(x,y\right)=\left(r\cos \theta ,r\sin \theta \right)$

Therefore, the obtained point is $\left(r\cos \theta ,r\sin \theta \right)$ .