Week 0 Discussion Notes

Table of Contents

• Trig

  • Unit Circle
  • Identities

• Calculus

  • Differentiation

Trig

Unit Circle

The unit circle is a circle with radius $1$. If we have an angle $\theta$, we can draw this picture:

In the image above, the coordinate of the black dot is $\p{\cos\theta, \sin\theta}$ by definition. Once you have $\cos\theta$ and $\sin\theta$, we can define every other trig function that we care about:

So the moral of the story is that if you know what to do with $\cos\theta$ and $\sin\theta$, you essentially know what to do with every trig function.

Example 1.

Calculate $\tan\frac{2\pi}{3}$.

Solution.

If you're not comfortable with radians, it'll be helpful to convert things to degrees (though I recommend that you try to get comfortable with radians eventually). $2\pi~\mathrm{rad} = 360^\circ$ (both represent one full rotation), so

$$\frac{2\pi}{3}~\mathrm{rad} \cdot \frac{360^\circ}{2\pi~\mathrm{rad}} = 120^\circ.$$

This gives us the following triangle:

This is a $30, 60, 90$ triangle, which has ratios $1, \sqrt{3}, 2$. Since the hypotenuse is $1$, we need to divide everything by $2$ to get the lengths of the sides of the triangle: $\frac{1}{2}, \frac{\sqrt{3}}{2}, 1$. So, the coordinate of the point is $\p{-\frac{1}{2}, \frac{\sqrt{3}}{2}}$, which means

$$\begin{aligned} \cos\frac{2\pi}{3} &= -\frac{1}{2} \\ \sin\frac{2\pi}{3} &= \frac{\sqrt{3}}{2}. \end{aligned}$$

Looking at the definition above, we can calculate $\tan\frac{2\pi}{3}$ from these:

$$\tan\frac{2\pi}{3} = \frac{\sin\frac{2\pi}{3}}{\cos\frac{2\pi}{3}} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \boxed{-\sqrt{3}}.$$

Identities

From the first picture above, we can use the Pythagorean theorem to get

If we divide everything by $\cos^2\theta$, we then get

Similarly, if we divide everything by $\sin^2\theta$, we get the identity

These are called the Pythagorean identities because they all come from the Pythagorean theorem. In addition to these, we have the double-angle identities:

The first one isn't super useful, but the second one is. If we solve for $\cos^2\theta$ and $\sin^2\theta$, we get

These are very useful identities. When my previous students were stuck on a trig-related integral, these identities were usually what they were missing.

Calculus

Differentiation

For this section, I basically followed my notes from a previous class up to the Integration section. I added a few examples, though:

Example 2.

Let $f\p{x} = x^2$. Calculate $f'\p{x}$ using the definition.

Solution.

We can essentially plug everything into the formula for the definition and simplify:

$$\begin{aligned} f'\p{x} &= \lim_{h\to0} \frac{f\p{x + h} - f\p{x}}{h} \\ &= \lim_{h\to0} \frac{\p{x + h}^2 - x^2}{h} \\ &= \lim_{h\to0} \frac{x^2 + 2xh + h^2 - x^2}{h} \\ &= \lim_{h\to0} \frac{2xh + h^2}{h} \\ &= \lim_{h\to0} \frac{h\p{2x + h}}{h} \\ &= \lim_{h\to0} \p{2x + h} \\ &= 2x. \end{aligned}$$

For the last "$=$" sign, I used the fact that $2x + h$ is continuous when $h$ is the variable.

Example 3.

Prove the quotient rule from the product and chain rules.

Solution.

First, I'm going to let $h\p{x} = \frac{1}{x}$ so that

$$h\p{g\p{x}} = \frac{1}{g\p{x}} \quad\text{and}\quad h'\p{x} = -\frac{1}{x^2}.$$

Then

$$\begin{aligned} \deriv{}{x} \frac{f\p{x}}{g\p{x}} &= \deriv{}{x} \p{f\p{x} \cdot \frac{1}{g\p{x}}} \\ &= \deriv{}{x} \p{f\p{x} \cdot h\p{g\p{x}}} \\ &= f'\p{x}h\p{g\p{x}} + f\p{x}\p{\deriv{}{x} h\p{g\p{x}}} && \p{\text{product rule}} \\ &= f'\p{x}h\p{g\p{x}} + f\p{x}h'\p{g\p{x}}g'\p{x} && \p{\text{chain rule}} \\ &= f'\p{x} \cdot \frac{1}{g\p{x}} + f\p{x}\p{-\frac{1}{\br{g\p{x}}^2}}g'\p{x} \\ &= \frac{f'\p{x}}{g\p{x}} \cdot \frac{g\p{x}}{g\p{x}} - \frac{f\p{x}g'\p{x}}{\br{g\p{x}}^2} \\ &= \frac{f'\p{x}g\p{x} - f\p{x}g'\p{x}}{\br{g\p{x}}^2}. \end{aligned}$$