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Chapter 4: Problem 99

In Exercises \(99-104,\) find two values of \(\theta, 0 \leq \theta<2 \pi,\) that satisfy each equation. $$\sin \theta=\frac{\sqrt{2}}{2}$$

Short Answer

Expert verified
The two values of \(\theta\) that satisfy \(\sin \theta = \frac{\sqrt{2}}{2}\) and lies within the range \(0 \leq \theta < 2 \pi\) are \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{5\pi}{4}\)

Step by step solution

01

Recall the Definition of Sine

The sine of an angle in the unit circle is defined as the y-coordinate of the point where the terminal side of the angle intersects the circle. We know that \(\sin \theta = \frac{\sqrt{2}}{2}\) in the first quadrant when \(\theta = \frac{\pi}{4}\).
02

Identify the Values of Theta

Since sine is positive in both first and second quadrants, and the range here is \(0 \leq \theta < 2 \pi\), we can determine that when \(\sin \theta = \frac{\sqrt{2}}{2}\), \(\theta = \frac{\pi}{4}\) or \(\theta = \frac{5\pi}{4}\).
03

Check the results

You can check these results by inserting them into the original equation: \(\sin \theta = \frac{\sqrt{2}}{2}\). For both values of \(\theta\), the equation holds true. Therefore, the two values of \(\theta\) are valid solutions.

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