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Chapter 8: Problem 60

Find the exact value of each expression. $$ \cos ^{-1}\left(\sin \frac{7 \pi}{6}\right) $$

Short Answer

Expert verified
\( \cos^{-1}\left(\sin \frac{7 \pi}{6}\right) = \frac{2 \pi}{3} \)

Step by step solution

01

Evaluate the sine

First, evaluate the sine of \(\frac{7 \pi}{6}\). Using the unit circle, \(\frac{7 \pi}{6}\) is located in the third quadrant, where the sine function is negative. The reference angle is \(\frac{\pi}{6}\), so \(\sin \frac{7 \pi}{6} = - \sin \frac{\pi}{6} = -\frac{1}{2}\).
02

Identify the range of \(\cos^{-1}\)

The range of the inverse cosine function, \( \cos^{-1} x \), is \[ 0 \leq y \leq \pi \]. This means that when we solve \(\cos^{-1}\), the answer will be an angle in this interval.
03

Solve for the inverse cosine

Find \(\cos^{-1} (-\frac{1}{2})\). This value will give an angle whose cosine is \-\frac{1}{2}\ within the range \([0, \pi]\). The angle in this range that satisfies this condition is \(\frac{2 \pi}{3}\). Thus, \(\cos^{-1} \left( \sin \frac{7 \pi}{6} \right) = \cos^{-1} (-\frac{1}{2}) \) equals to \(\frac{2 \pi}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Circle
The unit circle is a crucial tool in trigonometry that simplifies understanding angles and their trigonometric functions. On the unit circle, every point \((x, y)\) corresponds to an angle \theta\, where:
  • The x-coordinate is \cos(\theta)\
  • The y-coordinate is \sin(\theta)\
For instance, in our problem, \frac{7\pi}{6}\ lies in the third quadrant. This location means the sine function will be negative because the y-values (sine values) are negative in the third quadrant. Understanding the position of angles in the unit circle is essential for evaluating trigonometric functions and their inverses.
Sine Function
The sine function, often written as \sin(\theta)\, measures the vertical distance from the horizontal axis up to a point on the unit circle. For angle \frac{7\pi}{6}\, the reference angle is \frac{\pi}{6}\, which we use to evaluate \sin\left(\frac{7\pi}{6}\right)\. Since \frac{7\pi}{6}\ is in the third quadrant, and sine is negative there, we get:
\[ \sin\left(\frac{7\pi}{6}\right) = -\sin(\frac{\pi}{6}) = -\frac{1}{2}\]
Thus, understanding how the sine function operates on the unit circle is fundamental to solving these inverse trigonometric problems.
Inverse Cosine Function
The inverse cosine function, written as \cos^{-1}(x)\, determines what angle \y\ within the range \[0 \leq y \leq \pi\] produces a given cosine value \x\. For \cos^{-1}(-\frac{1}{2})\, we need to find which angle within \[0, \pi\] has a cosine of \-\frac{1}{2}\. Let's determine this:
The cosine function is negative in the second quadrant. The angle in the second quadrant whose cosine is \-\frac{1}{2}\ is \frac{2\pi}{3}\. Consequently, \cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}\. Mastering inverse trigonometric functions like \cos^{-1}\ helps solve for angles given specific trigonometric values within defined ranges.
Reference Angle
The reference angle is a helpful concept that simplifies evaluating trigonometric functions for any given angle. It is the smallest positive angle between the terminal side of the given angle and the x-axis. In our example, the reference angle for \frac{7\pi}{6}\ in the third quadrant is \frac{\pi}{6}\. This reference angle helps us understand the sine value:
\[ \sin\left(\frac{7\pi}{6}\right) = -\sin(\frac{\pi}{6}) = -\frac{1}{2}\]
Using reference angles makes it easier to compute trigonometric values because it reduces the problem to a simpler known angle. Incorporating reference angles into your understanding of trigonometry will greatly improve your problem-solving efficiency.

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Most popular questions from this chapter

Area under a Curve The area under the graph of \(y=\frac{1}{1+x^{2}}\) and above the \(x\) -axis between \(x=a\) and \(x=b\) is given by $$ \tan ^{-1} b-\tan ^{-1} a $$ (a) Find the exact area under the graph of \(y=\frac{1}{1+x^{2}}\) and above the \(x\) -axis between \(x=0\) and \(x=\sqrt{3}\). (b) Find the exact area under the graph of \(y=\frac{1}{1+x^{2}}\) and above the \(x\) -axis between \(x=-\frac{\sqrt{3}}{3}\) and \(x=1\)

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