Question

PLEASE HELP W BOTH PRECALC QUESTIONS!!!

Answer 1

Explanation:

Use the formulas supplied in the picture attached. There's nothing to really learn here, it is just a plug-and-chug formula:

`sin(3x)sin(2x)=(1)/(2) [cos(3x-2x)-cos(3x+2x)]`

`sin(3x)sin(2x)=(1)/(2) [cos(x)-cos(5x)]`

Moving on to the second problem:

`cos(a)sin(b)=(1)/(2)[ sin(a+b)-sin(a-b)]`

Therefore:

`2cos(a)sin(b)=sin(a+b)-sin(a-b)`

Where:

`a+b=5x`

`a-b=3x`

Solve for 'a' using the second equation to get:

`a=3x+b`

Plug this into the first equation to get:

`3x+b+b=5x`

`3x+2b=5x`

`2b=2x`

`b=x`

Therefore:

`a+x=5x`

`a=4x`

Finally:

`sin(a+b)-sin(a-b)=2cos(4x)sin(x)`