NOTE: You may view the original PDFs of the notes written in class by clicking on the link for the 1:30 Section or 2:30 Section, respectively.

Lecture 1: Review of Precalculus

GOAL: Recall the basic properties of trig, exponential, and logarithmic functions.

Unit circle on x y axes with radius 1 to point (x, y), angle theta, and legs x and y
Figure 1
A hand-drawn unit circle centered at the origin of a set of x and y axes, with arrows on the positive axes. A point on the circle in the first quadrant is marked with a filled dot and labeled (x, y). A radius drawn from the origin to that point is labeled 1. The angle between the positive x-axis and the radius is labeled theta. A dashed vertical segment drops from the point to the x-axis; it is labeled y, and the horizontal segment from the origin to the foot of that vertical is labeled x, forming a right triangle inside the circle.

Trig Functions

\[ \begin{gathered} \sin \theta = \frac{\text{opp}}{\text{hyp}} = \frac{y}{1} = y \\[0.5em] \cos \theta = \frac{\text{adj}}{\text{hyp}} = \frac{x}{1} = x \\[0.5em] \tan \theta = \frac{\text{opp}}{\text{adj}} = \frac{y}{x} = \frac{\sin \theta}{\cos \theta} \end{gathered} \]

Reciprocal Functions

\[ \frac{1}{\sin \theta} = \csc \theta \qquad \frac{1}{\cos \theta} = \sec \theta \qquad \frac{1}{\tan \theta} = \cot \theta \]

Pythagorean Identity

\[ \begin{gathered} x^2 + y^2 = 1 \\[0.5em] \cos^2 \theta + \sin^2 \theta = 1 \end{gathered} \]

We note that \(\sin \theta^2\) is ambigous since it could either mean \([\sin\theta]^2\) or \(\sin(\theta^2)\). To avoid this, we write \(\sin^2 \theta \) to mean \([\sin \theta]^2\)

Special Right Triangles:

Right triangle with hypotenuse 2, legs 1 and square root of 3, angles pi over 6 and pi over 3
Figure 2
A hand-drawn right triangle drawn wide and flat. The long slanted upper side (hypotenuse) is labeled 2. The horizontal bottom side is labeled square root of 3. The short vertical right side is labeled 1, and the right angle between the bottom and vertical sides is marked with a small square. The acute angle at the lower-left vertex is labeled pi over 6, and the acute angle at the top-right, between the hypotenuse and the vertical side, is labeled pi over 3.
Right triangle with hypotenuse square root of 2, legs 1 and 1, both acute angles pi over 4
Figure 3
A hand-drawn right triangle. The slanted side rising from the lower-left vertex to the top-right is labeled square root of 2. The horizontal bottom side is labeled 1, and the vertical right side is labeled 1, with a small square marking the right angle between them. The acute angle at the lower-left vertex is labeled pi over 4, and the acute angle at the top vertex is also labeled pi over 4.

Example 1

If \( \sin \theta = \frac{5}{13} \) and \( \theta \) is in the \( 2^{\text{nd}} \) quadrant. What 3 the value of the other trig functions?

SOLUTION: We know \( \sin \theta = \frac{5}{13} \). We need to know \( \cos \theta \)

Second-quadrant right triangle sketch with legs 5 and negative 12, hypotenuse 13, angle theta
Figure 4
A hand-drawn sketch on a pair of coordinate axes. In the second quadrant, a right triangle is drawn with a vertical red leg of length 5 rising from the negative x-axis, a horizontal leg along the x-axis labeled -12, and a hypotenuse of length 13 running from the origin up and to the left. The angle theta is marked at the origin, opening from the positive x-axis. The vertical leg is labeled Sine in red at the upper left, the horizontal direction is labeled Cosine in red to the right along the x-axis, and Tangent is labeled in red below the negative x-axis.
\[ \begin{gathered} \sin^2 \theta + \cos^2 \theta = 1 \\[0.5em] \left(\frac{5}{13}\right)^2 + \cos^2 \theta = 1 \\[0.5em] \frac{25}{169} + \cos^2 \theta = 1 \\[0.5em] \cos^2 \theta = 1 - \frac{25}{169} = \frac{144}{169} \\[0.5em] \cos \theta = \pm\sqrt{\frac{144}{169}} = \pm\frac{12}{13} \end{gathered} \]

Since \( \theta \) is in the second quadrant, \( \cos \theta = \frac{-12}{13}\). We use \(\sin\theta\) and \(\cos \theta\) to find the values for the remaining trig functions.

\[ \begin{gathered} \sin \theta = \frac{5}{13} \\[0.5em] \cos \theta = \frac{-12}{13} \\[0.5em] \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{5/13}{-12/13} = -5/12 \\[0.5em] \csc \theta = \frac{1}{\sin \theta} = \frac{1}{5/13} = 13/5 \\[0.5em] \sec \theta = \frac{1}{\cos \theta} = \frac{1}{-12/13} = -13/12 \\[0.5em] \cot \theta = \frac{1}{\tan \theta} = \frac{1}{-5/12} = -12/5 \end{gathered} \]

The Exponential Function \( y = e^x \); \( e \approx 2.718281\ldots \)

Graph of y equals e to the x, increasing curve passing through the point (0, 1)
Figure 5
A hand-drawn graph on horizontal and vertical axes showing the exponential curve labeled y equals e to the x. The curve approaches the horizontal axis asymptotically on the left, crosses the vertical axis at a marked dot labeled (0, 1), and rises steeply upward to the right.

Properties

  1. \( e^0 = 1 \)
  2. \( e^x > 0 \)
  3. \( e^a e^b = e^{a+b} \)
  4. \( e^a / e^b = e^{a-b} \)
  5. \( [e^a]^b = e^{ab} \)

Ex 2

  1. \[ e^x e^{-5} = e^{x-5} \]
  2. \[ \frac{e^{3x}}{e^{5x}} = e^{3x-5x} = e^{-2x} = \frac{1}{e^{2x}} \]

    We note that in the last example, we used the property \( e^{-a} = \frac{1}{e^a} \).

  3. \[ \left(e^x\right)^2 = e^{2 \cdot x} = e^{2x} \]

    A very common mistake is to assume \( [e^x]^2 = e^{x^2}\). THIS IS NOT TRUE.

  4. \[ \frac{e^{2x} e^3}{e^x} = \frac{e^{2x+3}}{e^x} = e^{(2x+3)-x} = e^{x+3} \]

The Natural Log Function

\[ y = \ln(x) = \log_e(x) = \text{The exponent such that } e^y = x \]
Graph of the natural logarithm curve passing through the point (1, 0)
Figure 6
A hand-drawn sketch of the graph of \( y = \ln(x) \) on a set of horizontal and vertical axes. The curve rises from far down along the right side of the vertical axis, crosses the horizontal axis at the marked point labeled \( (1,0) \), and then increases slowly to the upper right, ending with an arrowhead. A downward arrow at the bottom indicates the curve falls steeply toward negative infinity as \( x \) approaches zero from the right.

Properties

  1. \( \ln(1) = 0 \)
  2. \( \ln(ab) = \ln(a) + \ln(b) \)
  3. \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
  4. \( \ln\left(a^b\right) = b \ln a \)

Ex 3

  1. \( \ln(e) = 1 \)

    2. To determine the value of \(\ln e\), we need to find what number goes into the box in the equation \( e^{\square} = e \). Which, in this case, is 1.

  2. \[ \ln\left(e^{2x}\right) = 2x \ln e = 2x \]
  3. \[ e^{18 + \ln(20x)} = e^{18} e^{\ln(20x)} = e^{18}(20x) \]
  4. \[ \begin{gathered} e^{6 \ln\left(\sqrt{5x}\right)} = e^{6 \ln\left([5x]^{\frac{1}{2}}\right)} \\[0.5em] = e^{6\left(\frac{1}{2}\right) \ln(5x)} = e^{3 \ln(5x)} \\[0.5em] = \left[e^{\ln(5x)}\right]^3 = (5x)^3 = 125x^3 \end{gathered} \]
  5. \[ \ln\left(\frac{xy}{z}\right) = \ln(xy) - \ln(z) = \ln(x) + \ln(y) - \ln(z) \]

Ex 4

Solve for \( x \) in the equation \(\ln\left(x^2\right) = 5\).

SOLUTION: Here we need to use the fact \(e^x\) and \(\ln x\) are inverses. I.e., \[ e^{\ln(x)} = x = \ln(e^x) \]

With this in mind,

\[ \begin{gathered} \ln\left(x^2\right) = 5 \\[0.5em] e^{\ln\left(x^2\right)} = e^5 \\[0.5em] x^2 = e^5 \\[0.5em] x = \pm\sqrt{e^5} \\[0.5em] = \pm\left(e^5\right)^{\frac{1}{2}} = \pm e^{\frac{5}{2}} \end{gathered} \]