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Chapter 4: Problem 101

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
\(\theta = \frac{7\pi}{4}\)

Step by step solution

01

Identification

Identify that the value \(-\frac{\sqrt{2}}{2}\) falls in the range of sine function, [-1,1], thus there exist \(\theta\) values that satisfies the equation.
02

Finding the first value

Remember the sine function value at key points, specifically at \(\frac{3\pi}{4}\), \(\frac{7\pi}{4}\), where \(\sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}\) and \(\sin \frac{7\pi}{4} = -\frac{\sqrt{2}}{2}\). Because we need \(\sin \theta = -\frac{\sqrt{2}}{2}\), the first such \(\theta\) inside the range \([0, 2\pi]\) is at \(\theta = \frac{7\pi}{4}\) (315 degrees). Note that \(\frac{7\pi}{4}\) is in the fourth quadrant where sine values are negative, which matches our requirement.
03

Finding the second value

Now to find the second value of \(\theta\), think about the periodicity of the sine function. It repeats its cycle every \(2\pi\) radians. So if \(\sin \theta = -\frac{\sqrt{2}}{2}\) at \(\theta = \frac{7\pi}{4}\), it will also hold true one cycle later at \(\theta = \frac{7\pi}{4} + 2\pi = \frac{15\pi}{4}\). However, notice \(\frac{15\pi}{4} > 2\pi\) which is out of our given range. So there indeed is only one such \(\theta\) that satisfies the equation within the range \([0, 2\pi]\), which is \(\theta = \frac{7\pi}{4}\) alone.
04

Conclusion

To summarize, we have \(\theta = \frac{7\pi}{4}\) as the only solution satisfying the equation \(\sin \theta = -\frac{\sqrt{2}}{2}\) within the range \([0, 2\pi]\).

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