Question

How do you prove `cos2x=cos^2x-sin^2` using other trigonometric identities?

Answer 1

Apply the angle-sum identity for cosine to `cos(x+x)`.

Explanation:

The identity needed is the angle-sum identity for cosine.

`cos(alpha + beta) = cos(alpha)cos(beta) - sin(alpha)sin(beta)`

With that, we have

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

`= cos(x)cos(x) - sin(x)sin(x)`

`= cos^2(x) - sin^2(x)`

Answer 2

Alternatively, you can use De Moivre's Theorem of complex numbers to prove the identity.

Explanation:

This maybe is not a very nice proof for the identities themselves for a trigonometry student, but I find it a very useful way to derive the formula if you can't remember it.

De Moivre's Theorem says that if you have a complex number
`z=r(cos(theta)+isin(theta))`

Exponent of that complex number can be expressed as:
`z^n=r^n(cos(ntheta)+isin(ntheta))`

If we let
`omega=cos(theta)+isin(theta)`

We can than use De Moivre's theorem to say:

`omega^2=cos(2theta)+isin(2theta))`

We can also express `omega^2` in the following way:
`omega^2=(cos(theta)+isin(theta))^2`

These two expressions both equal `omega^2`, so we can set them equal to each other:
`cos(2theta)+isin(2theta)=(cos(theta)+isin(theta))^2`

Expanding the right hand side, we get:
`cos(2theta)+isin(2theta)=cos^2(theta)+2icos(theta)sin(theta)-sin^2(theta)`

Since the imaginary parts on the left must equal the imaginary parts on the right and the same for the real, we can deduce the following relationships:

`cos(2theta)=cos^2(theta)-sin^2(theta)`

`sin(2theta)=2sin(theta)cos(theta)`

And with that, we've proved both the double angle identities for `sin` and `cos` at the same time.

In fact, using complex number results to derive trigonometric identities is a quite powerful technique. You can for example prove the angle sum and difference formulas with just a few lines using Euler's identity.