Induction Problem help

Hello Dear Experts,

I have the following to prove by induction:

cos(n*theta) = sigma from i = 0 to n C_i * cos^i (theta).

Please help me to prove it.
Thanks a lot in advance.

The only approach I see is to tackle a similar theorem for sines at the same time. This is basically proving that something like De Moivre's formula works, only stated entirely in terms of real variables.

Suppose BOTH of the following are true:

cos nθ = (cos θ)^n - C(n,2)(cos θ)^(n-2) (sin θ)^2 + C(n,4)(cos θ)^(n-4) (sin θ)^4 - ...
sin nθ = S(n,1)(sin θ)(cos θ)^(n-1) - S(n,3)(sin θ)^3(cos θ) + S(n,5)(sin θ)^5(cos θ)^(n-5) - ...

The C(n,k) and S(n,k) are constants. This proof won't worry about it, but they are both equal to the "nCk" binomial coefficient. Note that the cosine formula only has even powers of the sines, and the sine formula has only odd powers of the sines. The sum of the sine and cosine exponents in each term of each sum is n.

This is trivially true for n=1:

cos θ = C(1,0) cos θ
sin θ = S(1,1) sin θ

...with C(1,0) = S(1,1) = 1.
For n=2 you have the double-angle identities:

cos 2θ = C(2,0)(cos θ)^2 - C(2,2)(sin θ)^2
sin 2θ = S(2,1) (cos θ)(sin θ)

...with C(2,0) = C(2,2) = 1 and S(2,1) = 2.

Suppose that both are true for some arbitrary integer n>0. Then write the functions of (n+1)θ using the angle sum identities:

cos (n+1)θ = cos (nθ + θ) = (cos θ)(cos nθ) - (sin θ)(sin nθ)
sin (n+1)θ = sin (nθ + θ) = (sin θ)(cos nθ) + (cos θ)(sin nθ)

All that's important here is the even/oddness of sine powers, and the total degree of each term. The (cos θ)(cos nθ) terms all have even powers of sin θ, because all of the sine factors come from cos nθ, which only has even powers of sin θ. Similarly -(sin θ)(sin nθ) has only even powers because each term of sin nθ has only odd powers of the sine, and the -sin θ factor adds another factor to make an even exponent. All terms have "total degree" (n+1) in sines and cosines.