You can
put this solution on YOUR website! Use sum/difference formulas to derive the formula for cos(3x) in terms of cosx. PLEASE HELP!
cos(3x)
= cos(2x+x)
= cos(2x)cos(x) - sin(2x)sin(x)
= [2cos˛(x)-1]cos(x) - [2sin(x)cos(x)]sin(x)
= 2cosł(x) - cos(x) - 2sin˛(x)cos(x)
= 2cosł(x) - cos(x) - 2[1 - cos˛(x)]cos(x)
= 2cosł(x) - cos(x) - 2cos(x)[1 - cos˛(x)]
= 2cosł(x) - cos(x) - 2cos(x) + 2cosł(x)
= 4cosł(x) - 3cos(x)
Edwin
