Chapter 4.4, Problem 92E

(a)

To determine

Two solutions of $\cos \theta =\frac{\sqrt{2}}{2}$ in radians $0\le \theta <2\pi$ and in degrees $0^0\le \theta <{360}^0$ .

Answer to Problem 92E

  $\theta =\frac{\pi }{4}={45}^0\text{ and }\theta =\frac{7\pi }{4}={315}^0$

Explanation of Solution

Given information:

  $\cos \theta =\frac{\sqrt{2}}{2}$

Formula used:

In quadrant I all trigonometric functions are positive.

In quadrant II sine functions are positive.

In quadrant III tangent functions are positive.

In quadrant IV cosine functions are positive.

Calculation:

Given that $\cos \theta =\frac{\sqrt{2}}{2}$ , Now,

  $\begin{array}{l}\cos \theta =\frac{\sqrt{2}}{2}\\ \theta =\frac{\pi }{4}={45}^0\text{ and }\theta =2\pi -\frac{\pi }{4}=\frac{7\pi }{4}={315}^0\end{array}$

Conclusion:

Thus,the solutions are $\theta =\frac{\pi }{4}={45}^0\text{ and }\theta =\frac{7\pi }{4}={315}^0$

(b)

To determine

Two solutions of $\cos \theta =-\frac{\sqrt{2}}{2}$ in radians $0\le \theta <2\pi$ and in degrees $0^0\le \theta <{360}^0$ .

Answer to Problem 92E

  $\theta =\frac{3\pi }{4}={135}^0\text{ and }\theta =\frac{5\pi }{4}={225}^0$

Explanation of Solution

Given info:

  $\cos \theta =-\frac{\sqrt{2}}{2}$

Formula used:

In quadrant I all trigonometric functions are positive.

In quadrant II sine functions are positive.

In quadrant III tangent functions are positive.

In quadrant IV cosine functions are positive.

Calculation:

Given that $\cos \theta =-\frac{\sqrt{2}}{2}$ , Now,

  $\begin{array}{l}\cos \theta =-\frac{\sqrt{2}}{2}\\ \theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}={135}^0\text{ and }\theta =\pi +\frac{\pi }{4}=\frac{5\pi }{4}={225}^0\end{array}$

Conclusion:

Thus,the solutions are $\theta =\frac{3\pi }{4}={135}^0\text{ and }\theta =\frac{5\pi }{4}={225}^0$