Question

Sin(5x)sin(3x)

Answer 1

$\bf \textit{Product to Sum Identities} \\\\ sin(\alpha)sin(\beta)=\cfrac{1}{2}[cos(\alpha-\beta)\quad -\quad cos(\alpha+\beta)]\qquad \leftarrow \textit{we'll use this one} \\\\\\ cos(\alpha)cos(\beta)=\cfrac{1}{2}[cos(\alpha-\beta)\quad +\quad cos(\alpha+\beta)] \\\\\\ sin(\alpha)cos(\beta)=\cfrac{1}{2}[sin(\alpha+\beta)\quad +\quad sin(\alpha-\beta)]$

$\bf cos(\alpha)sin(\beta)=\cfrac{1}{2}[sin(\alpha+\beta)\quad -\quad sin(\alpha-\beta)] \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ sin(5x)sin(3x)\implies \cfrac{cos(5x-3x)-cos(5x+3x)}{2}\implies \cfrac{cos(2x)-cos(8x)}{2}$

Answer 2

Answer:$\frac{1}{2}\left ( cos\left ( 2x\right )-cos\left ( 8x\right )\right )$

Step-by-step explanation:

Solution

$Sin\left ( 5x\right )Sin\left ( 3x\right )$

We know

$2sin\left ( a\right )sin\left ( b\right )=cos\left ( a-b\right )-cos\left ( a+b\right )$

Applying formula

$Sin\left ( 5x\right )Sin\left ( 3x\right )=\frac{1}{2}\left ( cos\left ( 5x-3x\right )-cos\left ( 5x+3x\right )\right )$

=$\frac{1}{2}\left ( cos\left ( 2x\right )-cos\left ( 8x\right )\right )$