Chapter 11.1, Problem 10QR

To determine

The equation relating $x$ and $y$ , when $x=\cos \theta$ and $y=\sin \theta$ for $\left(\pi \le \theta \le 2\pi \right)$ .

Answer to Problem 10QR

The relation between $x$ and $y$ is $y=-\sqrt{1-x^2}$ .

Explanation of Solution

Given information:

  $x=\cos \theta$ , $y=\sin \theta$

Such that $\pi \le \theta \le 2\pi$

Formula used:

Trigonometric formula for angle $\theta$ of the trigonometric identity:

  ${\cos }^2\theta +{\sin }^2\theta =1$

Calculation:

Substituting the value $x$ for $\cos \theta$ and $y$ for $\sin \theta$ in the formula ${\cos }^2\theta +{\sin }^2\theta =1$ , we get

  $\begin{array}{c}{\cos }^2\theta +{\sin }^2\theta =1\\ x^2+y^2=1\\ y^2=1-x^2\\ y=\pm \sqrt{1-x^2}\end{array}$

Since the range $\theta$ is from $\pi$ to $2\pi$

Thus, we have to consider the lower half of the circle.

Conclusion:

Hence, the required relation between $x$ and $y$ is $y=-\sqrt{1-x^2}$ .