Answer: cos(x)
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