Problem 39
Question
Verify each identity. $$\sin (\alpha+\beta)+\sin (\alpha-\beta)=2 \sin \alpha \cos \beta$$
Step-by-Step Solution
Verified Answer
The given identity is true as the simplified left side of the equation is exactly the same as the right side of the equation.
1Step 1: Apply the sum and difference identities
Begin by replacing \( \sin (\alpha + \beta) \) and \( \sin (\alpha - \beta) \) using the respective sum and difference identities for sine. These identities are: \( \sin (A + B) = \sin A \cos B + \cos A \sin B \) and \( \sin (A - B) = \sin A \cos B - \cos A \sin B \). So the left side of the equation becomes \( \sin A \cos B + \cos A \sin B + \sin A \cos B - \cos A \sin B \)
2Step 2: Simplify
The terms \(\cos A \sin B \) in the expression cancel each other. This leaves us with the equation \(2 \sin A \cos B \)
3Step 3: Compare with the right side of the equation
The simplified left side of the equation is exactly the same as the right side of the equation. Therefore, the given identity has been verified.
Other exercises in this chapter
Problem 39
Involve trigonometric equations quadratic in form. Solve each equation on the interval \([0,2 \pi)\) $$2 \sin ^{2} x-\sin x-1=0$$
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In Exercises \(39-46,\) use a half-angle formula to find the exact value of each expression. $$\sin 15^{\circ}$$
View solution Problem 40
Use words to describe the given formula. $$\cos \alpha \cos \beta=\frac{1}{2}[\cos (\alpha-\beta)+\cos (\alpha+\beta)]$$
View solution Problem 40
Verify each identity. $$\cot ^{2} 2 x+\cos ^{2} 2 x+\sin ^{2} 2 x=\csc ^{2} 2 x$$
View solution