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

2 Answers
Dec 17, 2015

Answer:

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

Answer:

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.