You may also use this althernative method, hence, you should find the derivatives of the functions `f(x) = cos2x` and `g(x) = 2cos^2x-1` and then you need to compare these derivatives.
If `f'(x) = g'(x), ` then the diference `f(x) - g(x)` is a constant.
Differentiating `f(x)` with respect to x yields:
`f'(x) = -2sin2x`
Differentiating `g(x)` with respect to x yields:
`g'(x) = 4cos x*(-sin x) => g'(x) = -2*(2 sin x*cos x)`
Notice that expanding `sin 2x` yields `2in x*cos x` , hence, `g'(x) = -2sin 2x` .
Notice that `f'(x) = g'(x),` hence, f(x) - g(x) must be a constant.
Checking if the difference `f(x) - g(x)` yields a constant gives:
`cos2x - 2cos^2x + 1 = 0`
Since the derivatives `f'(x) = g'(x)` and the difference f(x) - g(x) is a constant, then `cos2x= 2cos^2x- 1` .
`cos(2x)=2cos^2x - 1`
To prove the given equation without using trigonometric identities, we may refer to the graph of each side of the equation separately.
The graph of `y=cos (2x)` is:
And the graph of `y=2cos^2x-1` is:
Since, the graph of `y=cos(2x) ` and `y=2cos^2x-1` are the same, then this proves that `cos(2x)=2cos^2x-1.`