Answer:
Step-by-step explanation:
Use the formulas supplied in the picture attached. There's nothing to really learn here, it is just a plug-and-chug formula:
[tex]sin(3x)sin(2x)=\frac{1}{2} [cos(3x-2x)-cos(3x+2x)][/tex]
[tex]sin(3x)sin(2x)=\frac{1}{2} [cos(x)-cos(5x)][/tex]
Moving on to the second problem:
[tex]cos(a)sin(b)=\frac{1}{2}[ sin(a+b)-sin(a-b)][/tex]
Therefore:
[tex]2cos(a)sin(b)=sin(a+b)-sin(a-b)[/tex]
Where:
[tex]a+b=5x[/tex]
[tex]a-b=3x[/tex]
Solve for 'a' using the second equation to get:
[tex]a=3x+b[/tex]
Plug this into the first equation to get:
[tex]3x+b+b=5x[/tex]
[tex]3x+2b=5x[/tex]
[tex]2b=2x[/tex]
[tex]b=x[/tex]
Therefore:
[tex]a+x=5x[/tex]
[tex]a=4x[/tex]
Finally:
[tex]sin(a+b)-sin(a-b)=2cos(4x)sin(x)[/tex]
