Alternative Methods for Trigonometry Proofs (Varying Angles)

Question

Consider all the ways to prove $\sin(2x)=2\sin(x)\cos(x)$. There are many!

I personally would use Euler's formula involving complex numbers due to it being extremely simple and straightforward. We know $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$. The LH and RH sides become $\frac{e^{i2x}-e^{-i2x}}{2i}=2(\frac{e^{ix}-e^{-ix}}{2i})(\frac{e^{ix}+e^{-ix}}{2})$. And now this proof can be done with some basic algebra!

At the same time, I have always been bothered by rewriting $\sin(x)$ and $\cos(x)$ this way. To rewrite everything in terms of $e$ feels like an utter avoidance to use the words "sine" and "cosine." It also seems like an avoidance to represent such formulas along the unit circle to prove them. The proofs involving the sum and difference formulas that use the unit circle have always appeared so clean to me.

This got me thinking... is there a dependable way to prove results like these using the unit circle? Or is there a way to rewrite everything in terms of sine and cosine another way?

I have been curious about this for a long time, and I would be curious as to what other dependable ways are used to prove trigonometric results such as these without $i$ and infinite sequences. What do other people do?

Answer 1

A delightful application of linear algebra:

Let $R(\theta)$ be the linear map $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ that rotates points counterclockwise an angle $\theta$ about the origin. It is well-known that $R(\theta)$ has the matrix representation:

$$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$

It is obvious from inspection that $R(\theta + \psi) = R(\theta) \circ R(\psi)$. That is to say:

$$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix} \cos \psi & -\sin \psi \\ \sin \psi & \cos \psi \end{bmatrix} = \begin{bmatrix} \cos (\theta+\psi) & -\sin (\theta+\psi) \\ \sin (\theta+\psi) & \cos (\theta +\psi)\end{bmatrix}$$

Hence matrix multiplication gives you the angle addition identities:

$$\sin (\theta + \psi) = \sin\theta\cos\psi + \cos\theta\sin\psi$$

for example. Of course, identifying $a+bi \equiv \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ takes a little bit of the shock out of this, but I would argue that this is an elegant way of deriving trigonometric identities without relying on Euler's formula.

Answer 2

Partial Answer

Firstly,

The proofs involving the sum and difference formulas that use the unit circle have always appeared so clean to me.

$$\sin(2x)=\sin(x+x)=\sin(x)\cos(x)+\cos(x)\sin(x)=2\sin(x)\cos(x)$$

Most of the trigonometry formulas can be derived after establishing the basic ones (especially the sum and difference identities). In the unit circle, by rotating an angle (in the standard position) by the multiples of 90 degrees counterclockwise around the origin (or by using congruent triangles), you can prove the phase shift identities and much more. Sketching graphs often helps too. For example, $\sin(x-180°)=-\sin(x)$ can be seen by sketching $\sin(x-180°)$, noting that it is simply a sine curve with a horizontal shift of $180°$. Complex numbers are especially useful for double/triple angle formulae and power reduction etc.

Answer 3

$\def\a{\alpha} \def\b{\beta} \def\t{\theta} \def\ar{\text{area}}$

Here's an unusual way to derive the addition formula for $\sin(\a+\b)$, based on areas of triangles. First, recall that the area of a triangle with two sides $a$ and $b$ and included angle $\t$ is $A=\frac12ab\sin\t$. The steps are outlined.

Consider the diagram below:

The points $D$ and $E$ are obtained by dropping perpendiculars from $B$ and $A$, respectively, onto $\overline{OC}$.

$\bullet$ Show that $\triangle ACE \sim \triangle BCD$.

$\bullet$ Show that $(AE)(CD) = (BD)(CE)$ and conclude that $\ar\triangle ACD = \ar\triangle BCE$.

$\bullet$ Explain why \begin{align*} \sin(\a+\b) &= 2\,\ar\triangle OAB = 2\big(\ar\triangle OAC+\ar\triangle OCB\big) \\ &= 2\big(\ar\triangle OAD + \ar\triangle OEB\big). \end{align*}

$\bullet$ Find $OD$ (in terms of $\b$) and $OE$ (in terms of $\a$), and deduce the formula for $\sin(\a+\b)$.

Answer 4

Let $R_{\theta}$ be a rotation matrix that rotates an arbitrary vector by an angle of $\theta$ radians anticlockwise about the origin: $$ R_{\theta} = \begin{bmatrix} \cos \theta & -\sin\theta \\ \sin \theta & \cos\theta \end{bmatrix} $$ Clearly, $R_{\theta}R_{\theta}=R_{2\theta}$, since two rotations of size $\theta$ has the same result as a rotation of size $2\theta$: $$ R_{2\theta}= \begin{bmatrix} \cos 2\theta & -\sin2\theta \\ \sin 2\theta & \cos2\theta \end{bmatrix} \, . $$ On the other hand, if we multiply out the matrices the long way, we get $$ R_{\theta}R_{\theta}= \begin{bmatrix} \cos \theta & -\sin\theta \\ \sin \theta & \cos\theta \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin\theta \\ \sin \theta & \cos\theta \end{bmatrix} = \begin{bmatrix} \cos^2\theta-\sin^2\theta & -2\sin\theta\cos\theta \\ 2\sin\theta\cos\theta & \cos^2\theta - \sin^2\theta \end{bmatrix} \, , $$ and comparing the two gives us $\cos 2\theta \equiv \cos^2\theta - \sin^2\theta$ and $\sin 2\theta \equiv 2\sin\theta\cos\theta$.

Answer 5

A rather twisted idea: $y = \sin 2x$ and $y = 2\sin x\cos x$ are solutions of the ODE $y'' + 4y = 0$ with initial conditions $y(0) = 0$, $y'(0) = 2$. Now, apply unicity...