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

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.