Proving the angle sum and difference identities for sine and cosine without involving the functions' geometric meanings

The answer is that it depends how you define the sine and cosine functions; if they have a geometric definition then geometry has to come in somewhere. In fact, they relate closely to the concept of similarity in the Euclidean plane, and are useful in this context because similar triangles have equal angles.

Sine and cosine are also related to the exponential function in the complex plane through the identity $$e^{ia}=\cos a +i\sin a,$$ and we can compute \begin{align} e^{i(a+b)}&=e^{ia}e^{ib}\\ \cos (a+b)+i\sin (a+b)&=(\cos a +i\sin a)(\cos b +i\sin b)\\ &=(\cos a\cos b-\sin a \sin b)+i(\sin a \cos b+\cos a\sin b). \end{align}

Equating real and imaginary parts then gives what we want, and this is applicable generally. Some geometric proofs and constructions apply only to a specific range of values or require considering various cases.

Here's an approach I recently assigned as homework that needs nothing more than simple calculus facts. No complex numbers, in particular.

Suppose you know only the following: $$\sin'=\cos, \quad \cos'=-\sin, \quad \sin 0 = 0, \quad \cos 0 = 1$$

As a warmup, prove the Pythagorean identity $\sin^2 x + \cos^2 x = 1$. (Hint: let $f(x) = \sin^2 x + \cos^2 x$ and compute $f'$.) In particular, $|\sin x|\le 1$ and $|\cos x| \le 1$.

Now fix $a$ and consider the function $$g(x) = \sin(x+a) - \sin x \cos a - \cos x \sin a.$$ Compute $g'$ and $g''$, and note that $g'' = -g$. Verify that $g^{(n)}(0) = 0$ for every $n$, and that $|g^{(n)}(x)| \le 3$ for every $n,x$. Now apply Taylor's theorem with Lagrange remainder (which is really just a consequence of the mean value theorem) to bound the difference $|g(x) - p_n(x)|$, where $p_n$ is the $n$th degree Taylor polynomial of $g$ centered at 0. But $p_n=0$. Letting $n \to \infty$ you can conclude $g \equiv 0$.

For the identity involving $\cos(x+a)$, consider $g'$. The minus versions may be done similarly via the function $h(x) = \sin(a-x) - \sin a\cos x + \cos a \sin x$, or by showing separately that $\sin$ is an odd function and $\cos$ is an even function.

Some variations:

• If you know about real analytic functions, and you know that $\sin x, \cos x,\sin(x+a)$ are all real analytic, then you are done as soon as you show that $g^{(n)}=0$ for every $n$.

• If you know about uniqueness of solutions to ODEs, then just note that $g(0) = g'(0) = 0$ and $g'' = -g$.

Just to share my idea. I will not say that my proof doesn't involve geometry.

Let $A=(\cos\alpha,\sin\alpha)$ and $B=(\cos\beta,\sin\beta)$. Then

$$AB^2=(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2=2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta$$

On the other hand,

$$AB^2=OA^2+OB^2-2(OA)(OB)\cos\angle AOB=1^2+1^2-2(1)(1)\cos(\alpha-\beta)$$

This proves that $\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$. The proof holds for arbitrary $\alpha$ and $\beta$.

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

$\sin(\alpha+\beta)=\cos\left(\frac{\pi}{2}-\alpha-\beta\right)=\cos\left(\frac{\pi}{2}-\alpha\right)\cos\beta+\sin\left(\frac{\pi}{2}-\alpha\right)\sin\beta=\sin\alpha\cos\beta+\cos\alpha\sin\beta$

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