Use the product-to-sum formulas to rewrite the product as a sum or difference. $$8 \sin 65 ^ { \circ } \cos 25 ^ { \circ }$$
A) $8 \left( \sin 90 ^ { \circ } + \sin 40 ^ { \circ } \right)$
B) $4 \left( \cos 90 ^ { \circ } + \cos 40 ^ { \circ } \right)$
C) $4 \left( \cos 90 ^ { \circ } - \cos 40 ^ { \circ } \right)$
D) $\left( \sin 90 ^ { \circ } + \sin 40 ^ { \circ } \right)$
E) $4 \left( \sin 90 ^ { \circ } + \sin 40 ^ { \circ } \right)$

Question

Use the product-to-sum formulas to rewrite the product as a sum or difference.​ $$8 \sin 65 ^ { \circ } \cos 25 ^ { \circ }$$ ​
A)​ $8 \left( \sin 90 ^ { \circ } + \sin 40 ^ { \circ } \right)$
B)​ $4 \left( \cos 90 ^ { \circ } + \cos 40 ^ { \circ } \right)$
C)​ $4 \left( \cos 90 ^ { \circ } - \cos 40 ^ { \circ } \right)$
D)​ $\left( \sin 90 ^ { \circ } + \sin 40 ^ { \circ } \right)$
E)​ $4 \left( \sin 90 ^ { \circ } + \sin 40 ^ { \circ } \right)$

Quizplus · Accepted Answer

The product-to-sum formula for $ \sin A \cos B $ is $\frac{1}{2} [\sin(A+B) + \sin(A-B)]$. Applying this to $8 \sin 65^\circ \cos 25^\circ$, we get $8 \times \frac{1}{2} [\sin(90^\circ) + \sin(40^\circ)] = 4 [\sin(90^\circ) + \sin(40^\circ)]$.

Use the Product-To-Sum Formulas to Rewrite the Product as a Sum