Answer:
-2sin(x) * sin(2x)
or any equivalent form, such as 4cos(x)[cos²(x)-1]
Explanation:
simplify cos(3x) first:
cos(3x) = cos(2x+x) = cos(2x)cos(x) -sin(2x)sin(x)
using trig identities
= [2cos²(x)-1]cos(x) - [2sin(x)cos(x)]sin(x)
= 2cos³(x) - cos(x) - 2sin²(x)cos(x)
substituting using trig identity sin²(x) + cos²(x) = 1
2cos³(x) - cos(x) - 2[1-cos²(x)]cos(x)
2cos³(x) - cos(x) - 2cos(x)+2cos³(x)
4cos³(x) - 3cos(x)
remember this cos(3x), we still have to subtract cos(x)
4cos³(x) - 3cos(x) - cos(x) = 4cos³(x) - 4cos(x)
we can factor 4cos(x) to write this as a product of:
4cos(x)[cos²(x)-1]
further simplification if you want
trig identity sin²(x) + cos²(x) = 1
simplifying: sin²(x) = 1-cos²(x)
simplifying: -sin²(x) = cos²(x)-1
4cos(x)[cos²(x)-1]
4cos(x)[-sin²(x)]
-4cos(x)sin²(x)
trig identity: sin(2a) = 2cos(a)sin(a)
-2sin(x) * 2cos(x)sin(x)
-2sin(x)*sin(2x)