Cosine Function: $y = \cos(x)$

• Domain: All real numbers $(-\infty, \infty)$
• Range: $[-1, 1]$
• Period: $2\pi$
• Amplitude: 1
• Key points: $(0, 1)$, $(\frac{\pi}{2}, 0)$, $(\pi, -1)$, $(\frac{3\pi}{2}, 0)$, $(2\pi, 1)$
• Zeros: $x = \frac{\pi}{2} + n\pi$ where $n$ is an integer

Tangent Function: $y = \tan(x)$

• Domain: All real numbers except $x = \frac{\pi}{2} + n\pi$ where $n$ is an integer
• Range: All real numbers $(-\infty, \infty)$
• Period: $\pi$
• Vertical asymptotes: $x = \frac{\pi}{2} + n\pi$
• Key points: $(0, 0)$, $(\frac{\pi}{4}, 1)$, $(-\frac{\pi }{4},$
• Zeros: $x = n\pi$ where $n$ is an integer

General Form and Transformations

The general form of a sinusoidal function is:

$y = A\sin(B(x - C)) + D \quad \text{or} \quad y = A\cos(B(x - C)) + D$

Where:

• $|A|$ = Amplitude (vertical stretch/compression)
  • If $A < 0$, the graph is reflected over the x-axis
• $B$ affects the Period: Period = $\frac{2\pi}{|B|}$
  • If $B > 1$, horizontal compression (shorter period)
  • If $0 < B < 1$, horizontal stretch (longer period)
• $C$ = Phase shift (horizontal translation)
  • If $C > 0$, shift right
  • If $C < 0$, shift left
• $D$ = Vertical shift (midline)
  • The midline is $y = D$

For Tangent Functions

$y = A\tan(B(x - C)) + D$

• Period = $\frac{\pi}{|B|}$
• Vertical asymptotes at $x = C + \frac{\pi}{2B} + \frac{n\pi}{B}$ where $n$ is an integer

Finding Key Features from Equations

Given $y = A\sin(B(x - C)) + D$:

1. Amplitude: $|A|$
2. Period: $\frac{2\pi}{|B|}$
3. Phase Shift: $C$
4. Vertical Shift/Midline: $y = D$
5. Maximum value: $D + |A|$
6. Minimum value: $D - |A|$

Graphing Strategy

1. Identify $A$, $B$, $C$, and $D$
2. Draw the midline at $y = D$
3. Mark the amplitude (max at $D + |A|$, min at $D - |A|$)
4. Find the period and mark one complete cycle
5. Apply phase shift (start at $x = C$)
6. Plot key points within one period
7. Sketch the curve and extend if needed

Common Patterns

Cosecant and Secant

• $\csc(x) = \frac{1}{\sin(x)}$: Has vertical asymptotes where $\sin(x) = 0$
• $\sec(x) = \frac{1}{\cos(x)}$: Has vertical asymptotes where $\cos(x) = 0$

Cotangent

• $\cot(x) = \frac{1}{\tan(x)}$: Period is $\pi$, asymptotes where $\tan(x) = 0$

Solution:

Given: $y = 2\sin(x) + 1$

Compare to general form: $y = A\sin(B(x - C)) + D$

• $A = 2$
• $B = 1$
• $C = 0$
• $D = 1$

Features:

1. Amplitude: $|A| = |2| = 2$
2. Period: $\frac{2\pi}{|B|} = \frac{2\pi}{1} = 2\pi$

Key Points (one period from $x = 0$ to $x = 2\pi$):

• Start: $(0, 1)$ (midline)
• Max: $(\frac{\pi}{2}, 3)$
• Midline: $$

The graph oscillates between $y = -1$ and $y = 3$, centered on the midline $y = 1$.

Our function: $f(x) = 3\sin\left(2x - \frac{\pi}{3}\right) + 1$

Rewrite in standard form: $f(x) = 3\sin\left(2\left(x - \frac{\pi}{6}\right)\right) + 1$

Part (a): Amplitude $= |A| = |3| = 3$

Part (b): Period $= \frac{2\pi}{|B|} = \frac{2\pi}{2} = \pi$

Part (c): Phase shift $= C = \frac{\pi}{6}$ (to the right)

Alternatively, from $2x - \frac{\pi}{3} = 0$, we get $x = \frac{\pi}{6}$

Part (d): Vertical shift $= D = 1$

The midline is $y = 1$.

Find $B$: $\frac{2\pi}{B} = \pi$ $2\pi = \pi B$ $B = 2$

Equation: $y = 3\cos(2(x - \frac{\pi}{4})) - 2$

Or simplified: $y = 3\cos(2x - \frac{\pi}{2}) - 2$

Verification:

• Amplitude: $|3| = 3$ ✓
• Period: $\frac{2\pi}{2} = \pi$ ✓
• Phase shift: $\frac{\pi}{4}$ right ✓
• Vertical shift: $-2$ ✓
• Range: $[-2-3, -2+3] = [-5, 1]$

Here: $A = -2$, $B = \frac{\pi}{4}$, $C = -4$, $D = -3$

• Amplitude $= |A| = |-2| = 2$
• Period $= \frac{2\pi}{|B|} = \frac{2\pi}{\pi/4} = 8$
• Phase shift $= C = -4$ (4 units to the left)
• Midline: $y = D = -3$

Part (b): For cosine, the range is $[-1, 1]$.

With amplitude 2 and reflection: $-2\cos(...) \in [-2, 2]$

Adding vertical shift of $-3$:

Maximum: $2 + (-3) = -1$ Minimum: $-2 + (-3) = -5$

Part (c): Set $g(x) = 0$:

$-2\cos\left(\frac{\pi x}{4} + \pi\right) - 3 = 0$

$-2\cos\left(\frac{\pi x}{4} + \pi\right) = 3$

$\cos\left(\frac{\pi x}{4} + \pi\right) = -\frac{3}{2}$

But wait! The range of cosine is $[-1, 1]$, and $-\frac{3}{2} < -1$.

Therefore, there are no $x$-intercepts (the graph never crosses the $x$-axis).

This makes sense because the maximum value is $-1$, which is still below the $x$-axis.

Compare to general form: $y = A\tan(B(x - C)) + D$

• $A = -\frac{1}{2}$ (negative means reflection over x-axis)
• $B = \frac{1}{2}$
• $C = 0$
• $D = 3$

Features:

1. Period: $\frac{\pi}{|B|} = \frac{\pi}{1/2} = 2\pi$
2. Vertical shift: $y = 3$ (midline)
3. Reflection: Negative $A$ means graph is reflected over the midline
4. Vertical asymptotes occur at $x = \frac{\pi}{2B} + \frac{n\pi}{B}$

Find asymptotes: $x = \frac{\pi}{2(1/2)} + \frac{n\pi}{1/2} = \pi + 2n\pi$

For $n = 0, 1, 2$:

• $n = 0$: $x = \pi$
• $n = 1$: $x = 3\pi$
• $n = 2$: $x = 5\pi$ (outside interval)

In $[0, 4\pi]$, asymptotes are at $x = \pi$ and $x = 3\pi$.

Key points between asymptotes:

Between $x = -\pi$ and $x = \pi$:

• $(0, 3)$ (midline, where tangent crosses zero)
• $(-\frac{\pi}{2}, 3.5)$ (quarter period from center)
• $(\frac{\pi}{2}, 2.5)$ (quarter period from center)

The graph decreases from top to bottom (due to negative $A$) within each period, with the center line at $y = 3$.