Question

write the product as a sum.
19 sin(\\(\frac{x}{2}\\)) cos(\\(\frac{x}{4}\\))

Explanation:

Step1: Use the product - to - sum formula

The formula for \( \sin A\cos B=\frac{1}{2}[\sin(A + B)+\sin(A - B)]\). Here \(A=\frac{x}{2}\) and \(B = \frac{x}{4}\).
So, \(19\sin(\frac{x}{2})\cos(\frac{x}{4})=19\times\frac{1}{2}[\sin(\frac{x}{2}+\frac{x}{4})+\sin(\frac{x}{2}-\frac{x}{4})]\)

Step2: Simplify the arguments of the sine functions

Simplify \(\frac{x}{2}+\frac{x}{4}=\frac{2x + x}{4}=\frac{3x}{4}\) and \(\frac{x}{2}-\frac{x}{4}=\frac{2x-x}{4}=\frac{x}{4}\)
Then \(19\times\frac{1}{2}[\sin(\frac{3x}{4})+\sin(\frac{x}{4})]=\frac{19}{2}\sin(\frac{3x}{4})+\frac{19}{2}\sin(\frac{x}{4})\)

Answer:

\(\frac{19}{2}\sin(\frac{3x}{4})+\frac{19}{2}\sin(\frac{x}{4})\)