**Problem**: Derive the double angle identities

$ \sin(2\theta) = 2\sin(\theta)\cos(\theta)\\ \cos(2\theta) = \cos^2(\theta) – \sin^2(\theta)$

**Solution**: Recall from linear algebra how one rotates a point in the plane. The matrix of rotation (derived by seeing where $ (1,0)$ and $ (0,1)$ go under a rotation by $ \theta$, and writing those coordinates in the columns) is

$ A = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$

Next, note that to rotate a point twice by $ \theta$, we simply multiply the point (as a vector) by $ A$ twice. That is, multiply by $ A^2$:

$ AAv = A^2v$

Computing $ A^2$ gives the following matrix:

$ A^2 = \begin{pmatrix} \cos^2(\theta) – \sin^2(\theta) & -2\sin(\theta)\cos(\theta) \\ 2\sin(\theta)\cos(\theta) & \cos^2(\theta) – \sin^2(\theta) \end{pmatrix}$

But rotating twice by $ \theta$ is the same as rotating once by $ 2\theta$, so we have the equality:

$ \begin{pmatrix} \cos(2\theta) & -\sin(2\theta) \\ \sin(2\theta) & \cos(2\theta) \end{pmatrix} = \begin{pmatrix} \cos^2(\theta) – \sin^2(\theta) & -2\sin(\theta)\cos(\theta) \\ 2\sin(\theta)\cos(\theta) & \cos^2(\theta) – \sin^2(\theta) \end{pmatrix}$

The matrices are equal, so they must be equal entrywise, giving the identities we desire. $ \square$

**Discussion**: There are (painful, messy) ways to derive these identities by drawing triangles on the unit circle and cultishly chanting “soh-cah-toa.” The key idea in this proof that one might study geometric *transformations*, and it is a truly mature viewpoint of mathematics. Specifically, over the last two hundred years the field of mathematics has changed focus from the study of mathematical “things” to the study of transformations of mathematical things. This proof is an elementary example of the power such perspective can provide. If you want to be really high-brow, start asking about transformations of transformations of things, and transformations of *those *transformations, and recurse until you’re doing something original.

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I usually just derive all the trig identities from Euler’s Identity: e^ix = cos(x) + i*sin(x)

In a sense that is the same thing: multiplication by such a complex exponential is a rotation. But adding complex numbers into the mix makes it slightly less elementary. (And I want to make *some* things elementary on this blog…)

Great proof, Jeremy! I’m going to introduce this to the linear class I’m T.A.ing.

I never thought about using rotation matrices to prove this! This is SO MUCH easier than what I used to do: geometrically prove the sum identity, set the sum to be equal, then simplify with the Pythagorean identity.

This generalizes to prove the sum/difference identities. Write down the matrix for rotation by a and +/- b separately, multiply them together, and then equate that with the matrix for rotation by a +/- b. And, as Allan said, the same thing applies to doing multiplication of complex exponentials (but I think the matrix way gives more geometric intuition as to what’s going on).

Thank you so much Jeremy. Your posts always explain complex topics in the clearest of ways.