Demonstrate that the identity is true for x a real number and n a natural number, n>=2.cosnx + 2cos(n-1)x + cos(n-2)*x=4cos^2(x/2)*cos(n-1)x

We'll use the following
identity:

cos a + cos b = 2 cos[(a+b)/2]*cos [(a-b)/2]
(*)

We'll manage the left side and we'll create an even number of
terms, writing the middle term 2cos(n-1)x = cos(n-1)x +
cos(n-1)x.

We'll re-write the left
side:

cosnx + cos(n-1)x + cos(n-1)x +
cos(n-2)*x

We'll group the
terms:

[cosnx + cos(n-1)x] + [cos(n-1)x +
cos(n-2)*x]

We'll apply the identity
(*):

[cosnx + cos(n-1)x] = 2cos
[(n+n-1)x/2]*cos[(n-n+1)x/2]

[cosnx + cos(n-1)x] = 2cos
[(2n-1)x/2]*cos[(n-n+1)x/2]

[cosnx + cos(n-1)x] = 2cos
[(2n-1)x/2]*cos(x/2) (1)

[cos(n-1)x + cos(n-2)*x] = 2cos
[(n-1+n-2)x/2]*cos[(n-1-n+2)x/2]

[cos(n-1)x + cos(n-2)*x] = 2cos
[(2n-3)x/2]*cos(x/2) (2)

We'll add (1) +
(2):

LHS = 2cos [(2n-1)x/2]*cos(x/2) + 2cos
[(2n-3)x/2]*cos(x/2)

LHS = 2cos(x/2){cos [(2n-1)x/2] + cos
[(2n-3)x/2]}

We'll apply the identity
(*):

LHS = 2cos(x/2){2cos [(2n-1+2n-3)x/4]*cos
[(2n-1-2n+3)x/4]}

LHS = 2cos(x/2){2cos [(4n-4)x/4]*cos
[(2x)/4]}

LHS = 2cos(x/2){2cos [(4n-4)x/4]*cos
[(x)/2}

LHS = 4[cos(x/2)]^2*[cos (n-1)x] =
RHS

Since LHS = RHS, then the identity
cosnx + 2cos(n-1)x + cos(n-2)*x=4cos^2(x/2)*cos(n-1)x is verified,
for any real value of x and for any natural number
n>=2.