Question

Sin(5x)sin(3x)

Express the given product as a sum or difference containing only sines or cosines.

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:
`(1)/(2)\left ( cos\left ( 2x\right )-cos\left ( 8x\right )\right )`

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 )=(1)/(2)\left ( cos\left ( 5x-3x\right )-cos\left ( 5x+3x\right )\right )`

=
`(1)/(2)\left ( cos\left ( 2x\right )-cos\left ( 8x\right )\right )`