Use the sum-to-product formula to simplify the expression:

If $\sin 39^\circ + \sin 21^\circ = \sin A^\circ$, where $0^\circ < A < 90^\circ$, then:

$A = \_\_$ degrees.

Neetesh Kumar | October 16, 2024

Pre-Calculus Homework Help

Sum-to-Product and Product-to-Sum Formulas Assignment 9.4 Question 05
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Step-by-step solution:

We will use the sum-to-product formula for $\sin(A) + \sin(B)$:

$\sin(A) + \sin(B) = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)$

Trignometric-ratio Formula Sheet

Given expression: $\sin 39^\circ + \sin 21^\circ$

Using the sum-to-product formula:

$\sin 39^\circ + \sin 21^\circ = 2 \sin\left(\frac{39^\circ + 21^\circ}{2}\right) \cos\left(\frac{39^\circ - 21^\circ}{2}\right)$

Simplify:

$= 2 \sin\left(\frac{60^\circ}{2}\right) \cos\left(\frac{18^\circ}{2}\right)$

$= 2 \sin(30^\circ) \cos(9^\circ)$

We know that:

$\sin(30^\circ) = \frac{1}{2}$

Thus:

$= 2 \times \frac{1}{2} \times \cos(9^\circ)$

$= \cos(9^\circ)$

This matches the given equation $\sin A^\circ$. Therefore, we compare:

$\cos(9^\circ) = \sin A^\circ$

Since $\sin(A^\circ) = \cos(90^\circ - A^\circ)$, we get:

$A^\circ = 90^\circ - 9^\circ = 81^\circ$

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Final answer:

$A = \boxed{81^\circ}$

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