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

2 Answers
Dec 17, 2015

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)

Jan 1, 2018

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.