Respuesta :
[tex]\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)][/tex]
[tex]\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}[/tex]
Answer:[tex] \frac{1}{2}\left ( cos\left ( 2x\right )-cos\left ( 8x\right )\right )[/tex]
Step-by-step explanation:
Solution
[tex]Sin\left ( 5x\right )Sin\left ( 3x\right )[/tex]
We know
[tex] 2sin\left ( a\right )sin\left ( b\right )=cos\left ( a-b\right )-cos\left ( a+b\right ) [/tex]
Applying formula
[tex]Sin\left ( 5x\right )Sin\left ( 3x\right )=\frac{1}{2}\left ( cos\left ( 5x-3x\right )-cos\left ( 5x+3x\right )\right )[/tex]
=[tex] \frac{1}{2}\left ( cos\left ( 2x\right )-cos\left ( 8x\right )\right ) [/tex]
