Show that `cos(2x)=2cos^2x-1` :

The typical proof relies on the sum formulas. Another proof, more geometric in nature, requires the law of cosines. Here is a proof that only requires the Pythagorean identity:

Form Euler's identity we have:

`cos(2x)+isin(2x)=e^((2x)i)` Using laws of exponents we get on the RHS

`=(e^(ix))^2` Then applying Euler's identity once more we get

`=(cos(x)+isin(x))^2`

`=cos^2(x)+2icos(x)sin(x)+i^2sin^2(x)`

`=cos^2x-sin^2x+i2sinxcosx`

Since the real parts must be equal we get

`cos(2x)=cos^2x-sin^2x` Using the Pythagorean identity we get

`cos(2x)=cos^2x-(1-cos^2x)` or

`cos(2x)=2cos^2x-1` as required.

** Note that we also have, since the imaginary parts are equal, `sin(2x)=2sinxcosx` **

You need to move all terms to the left side, such that:

`cos2x - 2cos^2 x + 1 = 0`

You may consider the equation `cos2x - 2cos^2 x + 1` as equation of the function `f(x) = cos2x - 2cos^2 x + 1` .

Notice that you need to prove that the function `f(x) = cos2x - 2cos^2 x + 1` is constant, hence, you need to find derivative of the function to check if it is equal to zero since the derivative of a constant function is equal to zero.

`f'(x) = (cos2x - 2cos^2 x + 1)'`

You need to differentiate with respect to x using the chain rule, such that:

`f'(x) = -(sin 2x)*(2x)' - 2*2 cos x*(- sin x) + 0`

`f'(x) = -2(sin 2x) +2*2sin x*cos x`

You need to use the following trigonometric identity such that:

`sin 2x = 2 sin x*cos x`

`f'(x) = -2*2 sin x*cos x + 2*2 sin x*cos x`

`f'(x) = 0`

**Hence, evaluating the derivative of the function yields `f'(x) = 0` , thus, the given function is a constant `f(x) = 0` , such that `cos2x = 2cos^2 x - 1` .**

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