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Chapter 8: Problem 23

Use Half-angle Formulas to find the exact value of each expression. \(\tan \frac{7 \pi}{8}\)

Short Answer

Expert verified
\(\tan \frac{7 \pi}{8} = \pm (3 - 2\sqrt{2}) \).

Step by step solution

01

Recall the Half-Angle Formula for Tangent

The half-angle formula for tangent is oindent \ \(\tan \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{1 + \cos(\theta)}}\)
02

Identify the Angle

\( \frac{7 \pi}{8} \) can be expressed as \( \frac{\pi - \pi/8}{2} \). Therefore, \( \theta = \pi - \frac{\pi}{8}\) which simplifies \( \theta = \frac{7\pi}{4} \)
03

Calculate \(\cos( \frac{7\pi}{4} )\)

From the unit circle, \ \(\cos(\frac{7\pi}{4}) = \cos(2 \pi - \frac{\pi}{4}) = \cos(\frac{\pi}{4})\). The value is \(\frac{\sqrt{2}}{2}\).
04

Substitute the Cosine Value into the Formula

Using the cosine value from Step 3, substitute into the half-angle formula: \ \(\tan \left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}}}\).
05

Simplify the Expression

Simplify the expression: \(\frac{1 - \frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} = \frac{\frac{2 - \sqrt{2}}{2}}{\frac{2 + \sqrt{2}}{2}} = \frac{2 - \sqrt{2}}{2 + \sqrt{2}}\).
06

Rationalize the Denominator

Rationalize the denominator: \(\frac{2 - \sqrt{2}}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{(2 - \sqrt{2})^2}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{4 - 4\sqrt{2} + 2}{4 - 2} = \frac{6 - 4\sqrt{2}}{2} = 3 - 2 \sqrt{2}.\)

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Key Concepts

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

Tangent Half-Angle Formula
To understand the tangent half-angle formula, remember that it allows us to find the tangent of half of any given angle. The formula is: \ \ \( \tan \bigg(\frac{\theta}{2}\bigg) = \frac{1 - \text{cos}(\theta)}{1 + \text{cos} (\theta)} \). \ \ The plus or minus (\

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Most popular questions from this chapter

Solve each equation on the interval \(0 \leq \theta<2 \pi\). $$ \sqrt{3} \sin \theta+\cos \theta=1 $$

Challenge Problem Show that \(\tan ^{-1}\left(\frac{1}{v}\right)=\frac{\pi}{2}-\tan ^{-1} v\) if \(v>0\).

Find the value of the number \(C\) : $$ \frac{1}{2} \cos ^{2} x+C=\frac{1}{4} \cos (2 x) $$

Show that \(\sin \left(\sin ^{-1} v+\cos ^{-1} v\right)=1\)

Establish each identity. $$ \frac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}=\tan \alpha+\tan \beta $$
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