Lemma 1: In a right triangle, given angle $\theta$, $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$#
But it’s definitely best to visualize:
Basic geometric definition of sine and cosine
From the definitions, we get that the green side is $h\cos \theta$ because $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \implies \text{adjacent} = h\cos \theta$. Similarly, the blue side is $h\sin \theta$. Letting $h=1$, the adjacent side is just $\cos \theta$, and the opposite side is $\sin \theta$. With this setup, we can show Lemma 2.
Lemma 2: $\cos \theta = \sin(\frac{\pi}{2} - \theta)$ and $\sin \theta = \cos(\frac{\pi}{2} - \theta)$ for arbitrary $\theta$#
Proof. Consider $\alpha=\frac{\pi}{2}-\theta$. In the triangle with hypotenuse 1, the green side is $\cos \theta = \frac{\text{adjacent}}{1}$. It is also $\sin \alpha = \frac{\text{opposite}}{1}$. Hence, $\cos \theta = \sin(\frac{\pi}{2} - \theta)$. Similar reasoning holds for the latter identity and for arbitrary $h$.
Geometric proof of separation by 90 degrees
Here’s a Desmos graph where you can see the identity for yourself: 1
Back to Lemma 1, why should the basic definition be limited to a static right triangle? What about a dynamic right triangle, perhaps one like this: 2
In this graph, the x coordinates always involve $\cos \theta$, and the y coordinates always involve $\sin \theta$. Thinking of the black line as a vector 3, we see that to get the horizontal component of the vector, we multiply its length by $\cos \theta$. To get its vertical component, we multiply its length by $\sin \theta$. In other words, $\cos \theta$ projects to $x$, and $\sin \theta$ projects to $y$.
This applies to the graph of a circle as well. Note that, again, $x$ always goes with $\cos \theta$, and $y$ always goes with $\sin \theta$.
There is an algebraic symmetry that is, in my opinion, as beautiful as the symmetries above. This symmetry also allows us to prove $e^{ix} = \cos x + i\sin x$ for any $x$.
Lemma 3: $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$, and $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$ 5#
Proof. Consider that the Taylor series6 expansion of $f(x)$ about the point $a$ is $$f(a)+f’(a)(x-a) + \frac{f’’(a)}{2!}(x-a)^2 + \frac{f’’’(a)}{3!}(x-a)^3 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \cdots$$
The Maclaurin expansion is the Taylor series centered at $a=0$, so the Maclaurin expansion of $f(x)$ is $$f(0) + f’(0)x + \frac{f’’(0)}{2!}x^2 + \frac{f’’’(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + \cdots$$
Using that $\cos(0) = 1$ and $\sin(0) = 0$, we have that the first few terms of the Maclaurin series are $1 + 0x -\frac{x^2}{2} + 0x^3 + \frac{x^4}{4!}$. Now, we write the pattern: $$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n!} x^{2n} $$
Any odd power will have $\sin(0)=0$ and so can be ignored. Hence, we use only the even numbers $2n$ in our summation. Using that $f(x)=\cos x$ and $f’’(x) = -\cos x$, we note that the signs alternate each (even) term, so we also have $(-1)^n$ in the summation.
Similar reasoning holds for $\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}$.
Lastly, there are also two trigonometric identities that I have always found interesting, in terms of symmetry.
I won’t prove these identities here, but I will write about the symmetries. Note that the $\sin$ identities have matching signs; that is, if there is a $+$ on the left side, there’s a $+$ on the right side. However, the multiplications are mismatched in that $\sin$ is multiplied by $\cos$ and vice versa.
On the other hand, the $\cos$ identities have matching multiplication in that $\cos$ multiplies $\cos$, and $\sin$ multiplies $\sin$. The signs, however, are mismatched.
You can press the ▶️ on the left to view the animation. ↩︎
Note that there is no magic going on here. There’s nothing in Desmos that specifically generates an arbitrary right triangle - this graph uses only the sine and cosine knowledge shown above! ↩︎
Since we can vary its length $r$ and its angle $t$ ↩︎
A still screenshot of unit circle animation on Khan Academy
Note that the cosine expansion contains only even powers of $x$ (including $x^0 = 1$) divided by even factorials, and the sine expansion contains only odd powers of $x$ divided by odd factorials. ↩︎
I might write an article about the Taylor series! For now, the tl;dr is that the Taylor expansion of a complicated function is a highly accurate approximation that is made up of less complicated polynomial terms, which are much easier to work with. ↩︎
Note here that the cycle length is 4, which is the same as that of the powers of $i$. ↩︎
