Express the product as a sum or difference.
-$\sin 6 x \sin 2 x$
A) $\frac { 1 } { 2 } ( \cos 4 x - \cos 8 x )$
B) $\sin ^ { 2 } 12 x ^ { 2 }$
C) $\frac { 1 } { 2 } ( \cos 8 x - \cos 4 x )$
D) $\frac { 1 } { 2 } ( \sin 8 x + \cos 4 x )$

Question

Quizplus · Accepted Answer

The product-to-sum formula for sine functions is $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$. Applying this to $\sin 6x \sin 2x$ gives $\frac{1}{2}[\cos(6x-2x) - \cos(6x+2x)] = \frac{1}{2}(\cos 4x - \cos 8x)$.

Express the Product as a Sum or Difference sin⁡6xsin⁡2x\sin 6 x \sin 2 xsin6xsin2x

Express the Product as a Sum or Difference $\sin 6 x \sin 2 x$