`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.`

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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` .**