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

Then teach the underlying concepts
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#### Explanation

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Dec 17, 2015

Apply the angle-sum identity for cosine to $\cos \left(x + x\right)$.

#### Explanation:

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

$\cos \left(\alpha + \beta\right) = \cos \left(\alpha\right) \cos \left(\beta\right) - \sin \left(\alpha\right) \sin \left(\beta\right)$

With that, we have

$\cos \left(2 x\right) = \cos \left(x + x\right)$

$= \cos \left(x\right) \cos \left(x\right) - \sin \left(x\right) \sin \left(x\right)$

$= {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$

Then teach the underlying concepts
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#### Explanation

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#### Explanation:

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4
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
$z = r \left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)$

Exponent of that complex number can be expressed as:
${z}^{n} = {r}^{n} \left(\cos \left(n \theta\right) + i \sin \left(n \theta\right)\right)$

If we let
$\omega = \cos \left(\theta\right) + i \sin \left(\theta\right)$

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:
${\omega}^{2} = {\left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)}^{2}$

These two expressions both equal ${\omega}^{2}$, so we can set them equal to each other:
$\cos \left(2 \theta\right) + i \sin \left(2 \theta\right) = {\left(\cos \left(\theta\right) + i \sin \left(\theta\right)\right)}^{2}$

Expanding the right hand side, we get:
$\cos \left(2 \theta\right) + i \sin \left(2 \theta\right) = {\cos}^{2} \left(\theta\right) + 2 i \cos \left(\theta\right) \sin \left(\theta\right) - {\sin}^{2} \left(\theta\right)$

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:

$\cos \left(2 \theta\right) = {\cos}^{2} \left(\theta\right) - {\sin}^{2} \left(\theta\right)$

$\sin \left(2 \theta\right) = 2 \sin \left(\theta\right) \cos \left(\theta\right)$

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.

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