# 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

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### 1 Answer

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.**