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Chapter 5: Problem 47

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

Short Answer

Expert verified
The identity \( \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=\frac{\tan \alpha+\tan \beta}{\tan \alpha-\tan \beta} \) is verified.

Step by step solution

01

Apply Addition and Subtraction Formulas for Sine

First recall the addition and subtraction formulas for sine:\( \sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta \)\( \sin(\alpha - \beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta \)Substitute these formulas into the original expression, such that the left side of the equation becomes:\( \frac{\sin\alpha \cos\beta + \cos\alpha \sin\beta}{\sin\alpha \cos\beta - \cos\alpha \sin\beta} \)
02

Rewrite Sine in Terms of Tangent

Now, express \( \sin\alpha \) and \( \sin\beta \) as \( \frac{\tan\alpha}{\sqrt{1+\tan^2\alpha}} \) and \( \frac{\tan\beta}{\sqrt{1+\tan^2\beta}} \) respectively, and \( \cos\alpha \) and \( \cos\beta \) as \( \frac{1}{\sqrt{1+\tan^2\alpha}} \) and \( \frac{1}{\sqrt{1+\tan^2\beta}} \), respectively. Substituting these expressions into the equation from Step 1, we have:\( \frac{\frac{\tan\alpha}{\sqrt{1+\tan^2\alpha}}\frac{1}{\sqrt{1+\tan^2\beta}}+\frac{1}{\sqrt{1+\tan^2\alpha}}\frac{\tan\beta}{\sqrt{1+\tan^2\beta}}}{\frac{\tan\alpha}{\sqrt{1+\tan^2\alpha}}\frac{1}{\sqrt{1+\tan^2\beta}}-\frac{1}{\sqrt{1+\tan^2\alpha}}\frac{\tan\beta}{\sqrt{1+\tan^2\beta}}} \)
03

Simplify the Expression

Further simplifying we get \( \frac{\tan\alpha+\tan\beta}{\tan\alpha-\tan\beta} \)The result is identical with the right hand side of the original identity, which confirms it.

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Most popular questions from this chapter

Find the exact value of each expression. Do not use a calculator. $$\sin \left(\cos ^{-1} \frac{1}{2}+\sin ^{-1} \frac{3}{5}\right)$$

Use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$\tan x \sec x=2 \tan x$$

Exercises \(116-118\) will help you prepare for the material covered in the next section. In each exercise, use exact values of trigonometric functions to show that the statement is true. Notice that each statement expresses the product of sines and/or cosines as a sum or a difference. $$\sin \pi \cos \frac{\pi}{2}=\frac{1}{2}\left[\sin \left(\pi+\frac{\pi}{2}\right)+\sin \left(\pi-\frac{\pi}{2}\right)\right]$$

Use a graphing utility to approximate the solutions of each equation in the interval \([0,2 \pi) .\) Round to the nearest hundredth of a radian. $$\sin x+\sin 2 x+\sin 3 x=0$$

Determine whether each -statement makes sense or does not make sense, and explain your reasoning. There are similarities and differences between solving \(4 x+1=3\) and \(4 \sin \theta+1=3:\) In the first equation, I need to isolate \(x\) to get the solution. In the trigonometric equation, I need to first isolate \(\sin \theta,\) but then \(I\) must continue to solve for \(\theta\)
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