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Chapter 8: Problem 21

Find the exact value of each expression. \(\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)\)

Short Answer

Expert verified
\( \cos^{-1}\big(-\frac{\root{3}}{2}\big) = \frac{5\pi}{6} \)

Step by step solution

01

- Understand the Problem

We need to find the angle whose cosine is \(-\frac{\root 3 //2}\). This is often denoted by \( \theta = \cos^{-1}\big(-\frac{\root{3}}{2}\big)\).
02

- Recall the Range of \( \theta \) for \( \cos^{-1}\)

The range of the inverse cosine function, \( \cos^{-1}\) is \[0, \pi\]. This means that \( \theta \) must be within this interval.
03

- Identify the Cosine Value

We know the angle whose cosine gives us \(-\frac{\root 3 //2}\) in the range \[0, \pi\] is \( \theta = \frac{5\pi}{6}\). This since \(\cos\big(\frac{5\pi}{6}\big) = -\frac{\root{3}}{2}\).

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Key Concepts

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

cosine inverse
The inverse cosine function, denoted as \(\cos^{-1}(x)\), is used to find the angle whose cosine value is \(x\). For example, \(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)\) gives us the angle whose cosine is \(-\frac{\sqrt{3}}{2}\). This function is very helpful in trigonometry when we need to solve for angles rather than side lengths. The notation \(\cos^{-1}\) is called 'arc cosine' or 'inverse cosine'. It's important to note that \(\cos^{-1}(x)\) does not mean \(\frac{1}{\cos(x)}\); it specifically refers to the angle calculation.
exact values
In trigonometry, 'exact values' mean specific results that do not rely on decimal approximations but rather on known values. These are usually derived from special angles such as 0, 30, 45, 60, and 90 degrees (or their radian equivalents). For example, we know that \(\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\). Hence, when finding the inverse cosine of \(-\frac{\sqrt{3}}{2}\), the 'exact value' of the angle is \(\frac{5\pi}{6}\). These exact values are derived from the unit circle and fundamental trigonometric identities.
trigonometric ranges
Trigonometric ranges are important because they tell us the values that a trigonometric function can take. For the inverse cosine function, the range is limited to \[0, \pi\]. This means any angle \(\theta = \cos^{-1}(x)\theta\) must be between 0 and \(\pi\) radians (or 0 to 180 degrees). This restriction exists because the cosine function is not one-to-one over its entire set of possible values, but it is one-to-one from 0 to \(\pi\). Therefore, this range ensures that each value of \(\cos^{-1}(x)\) corresponds to exactly one angle within this interval.

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Most popular questions from this chapter

Convert \(6^{x}=y\) to an equivalent statement involving a logarithm.

Calculus Show that the difference quotient for \(f(x)=\cos x\) is given by $$ \begin{aligned} \frac{f(x+h)-f(x)}{h} &=\frac{\cos (x+h)-\cos x}{h} \\ &=-\sin x \cdot \frac{\sin h}{h}-\cos x \cdot \frac{1-\cos h}{h} \end{aligned} $$

Solve each equation on the interval \(0 \leq \theta<2 \pi\). $$ \tan \theta+\sqrt{3}=\sec \theta $$

Challenge Problem Show that \(\tan ^{-1} v+\cot ^{-1} v=\frac{\pi}{2}\)

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Write \(f(x)=\frac{1}{4} x^{2}+x-2\) in vertex form.
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