A Quest for Calculus: Here's Your Sine

I'm going to put off actual calculus for a few more posts. It turns out calculus has some uses for trigonometry. So, just like with the Pythagorean Theorem, there are a few proofs I want to get out of the way so I'll feel good about myself when I use them in later posts. I'm not going to cover all of the identities here, but there are some pieces I want to be clear about.

To get started, we need to dig into the unit circle:

The unit circle is simply a circle with a radius of 1. With just the radius known, we can define a couple of things about the circle. We know that the circumference of any circle is $2{\pi}r$ . This means that for the unit circle, we have a circumference of $2{\pi}$ . We know that there are $360^\circ$ s in the circle. This means that $90^\circ$ s is $\frac{1}{4}$ th the circumference or $\frac{2{\pi}}{4}$ . This means that with regards to the unit circle, ${90^\circ}=\frac{\pi}{2}$ . You can calculate other angles in a similar fashion.

Moving on, we need to introduce the 3 basic trig functions: sine, cosine, and tangent. These three functions are defined on the unit circle as follows:

Notice that the line created between points A and B (represented as $\overline{AB}$ ) is the radius of the circle which is 1. This is true for $\overline{AE}$ as well. The angle $\angle$BAE$ is represented as $x$ . Now we need to define the 3 functions.

Sine of x: We can see that $sin{x}=\overline{BC}$ which is perpendicular to $\overline{AE}$

Cosine of x: It's also clear that $\cos{x}=\overline{AC}$

Tangent of x: Finally $\tan{x}=\overline{DE}$ which is also perpendicular to $\overline{AE}$

So, that's what they are. The why they are is something else entirely and I currently have no interest in looking into it. My sincere apologies if I've let you down. The next piece we're going to look at is the ever popular SOHCAHTOA.

SOHCAHTOA refers to the following triangle:

It's simply a mnemonic to help remember how to calculate sin, cos, and tan of an angle. Breaking down the mnemonic we get:

SOH: $\sin{x}=\frac{Opposite}{Hypotenuse}$

CAH: $\cos{x}=\frac{Adjacent}{Hypotenuse}$

TOA: $\tan{x}=\frac{Opposite}{Adjacent}$

That's all well and good, but if I could take the mnemonic at it's word, I wouldn't have much of a blog. So, going back to our unit circle, we'll make some logical assertions and work things out starting with sine.

SOH:

We know that triangle ABC (represented as $\bigtriangleup{ABC}$ ) is similar in proportion to $\bigtriangleup{ADE}$

This tells us that the ratio between $\overline{BC}$ and $\overline{AB}$ is the same as the ratio between $\overline{DE}$ and $\overline{AD}$ .

This can also be represented as $\frac{BC}{AB}=\frac{DE}{AD}$

We know that $\overline{AB}$ = 1 so, we can reduce this down to $\overline{BC}=\frac{DE}{AD}$

We also know that $\overline{BC}=\sin{x}$ which means: $\sin{x}=\frac{DE}{AD}$

We can note that for $\bigtriangleup{ADE}$ , with regard to $\angle{x}$ , $\overline{DE}$ is the opposite side and $\overline{AD}$ is the hypotenuse.
Finally, we can deduce that for all triangles similar to $\bigtriangleup{ADE}$ (meaning any right triangle built with $\angle{x}$ ), $\sin{x}=\frac{Opposite}{Hypotenuse}$

CAH:

The proof here is the same as the one for sine however we substitute $\overline{AC}$ for $\overline{BC}$ and $\overline{AE}$ for $\overline{DE}$ .

We know that $\frac{AC}{AB}=\frac{AE}{AD}$
This means (skipping a couple steps) $\cos{x}={AE}/{AD}$
Finally, using the same logic as with sine, $\cos{x}=\frac{Adjacent}{Hypotenuse}$