That isn't generally true, but it's close to something that is true. Drop the "cos kθ =" at the front and:
cos kθ + i sin kθ = (cos θ + i sin θ)^k
...is de Moivre's formula, and is true for all θ.
What you typed requires sin kθ to be 0, or θ = Ïn/k for the left equation to be true. (n is any integer, not just 0.) The right equation is still de Moivre, and still true.
So, I guess this shows that θ = Ïn/k are the values that make cos kx = (cos x)^k true, if that is what you were after.
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Verified answer
From Euler's equation
e^(iθ) = cosθ+isinθ
[e^(iθ)]^k = e^(ikθ) = coskθ+isinkθ=[cosθ+isinθ]^k
That isn't generally true, but it's close to something that is true. Drop the "cos kθ =" at the front and:
cos kθ + i sin kθ = (cos θ + i sin θ)^k
...is de Moivre's formula, and is true for all θ.
What you typed requires sin kθ to be 0, or θ = Ïn/k for the left equation to be true. (n is any integer, not just 0.) The right equation is still de Moivre, and still true.
So, I guess this shows that θ = Ïn/k are the values that make cos kx = (cos x)^k true, if that is what you were after.