Write as a single term: cos(2x) cos(x) + sin(2x) sin(x)

cos(x)

Step-by-step explanation:

We have  

sin ( x + y ) = sin(x)*cos(y) + cos(x)*sin(y)   (1)   and

cos ( x + y ) = cos(x)*cos(y) - sin(x)*sin(y)  (2)

From eq. (1)

if x = y

sin ( x + x ) = sin(x)*cos(x) + cos(x)*sin(x) ⇒ sin(2x) = 2sin(x)cos(x)

From eq. 2

If x = y

cos ( x + x ) = cos(x)*cos(x) - sin(x)*sin(x)  ⇒ cos²(x) - sin²(x)

cos (2x) = cos²(x) - sin²(x)

Hence:The expression:

cos(2x) cos(x) + sin(2x) sin(x)  (3)

Subtition of sin(2x) and cos(2x)  in eq. 3

[cos²(x)-sin²(x)]*cos(x) + [(2sen(x)cos(x)]*sin(x)

and operating

cos³(x) - sin²(x)cos(x) + 2sin²(x)cos(x) = cos³(x) + sin²(x)cos(x)

cos (x) [ cos²(x) + sin²(x) ]  = cos(x)

since cos²(x) + sin²(x) = 1

4x and 6y

im not sure though