13.1 Graphs of the Six Trigonometric Functions & Periodic Modeling

13.1 Graphs of the Six Trigonometric Functions & Periodic Modeling

Geometric and Analytic Properties of the Six Trigonometric Functions

The Cartesian graphs of the six trigonometric functions represent the unwrapping of circular arc lengths onto the linear real number line. By convention, the independent variable $x$ represents an angle measured in dimensionless radians, while the dependent variable $y$ represents the corresponding trigonometric ratio.

The Fundamental Sinusoids: Sine and Cosine

The primary trigonometric functions, $f(x) = \sin x$ and $g(x) = \cos x$, are continuous, infinitely differentiable wave functions defined across the entire real line ($x \in \mathbb{R}$). Both functions oscillate smoothly between $-1$ and $1$, establishing their common range as $[-1, 1]$. The distance along the horizontal axis required for either wave to complete one full cycle of oscillation is its fundamental period, which is exactly $2\pi$ radians ($360^\circ$):

$$\sin(x + 2\pi) = \sin x, \quad \cos(x + 2\pi) = \cos x$$

The sine function is an odd function, exhibiting rotational point symmetry about the origin:

$$\sin(-x) = -\sin x$$

Its graph passes directly through the origin $(0, 0)$, reaches a local maximum of $1$ at $x = \frac{\pi}{2}$, crosses the x-axis at $x = \pi$, reaches a local minimum of $-1$ at $x = \frac{3\pi}{2}$, and completes its fundamental cycle at $x = 2\pi$. The zeros of $y = \sin x$ occur at all integer multiples of $\pi$: $x = k\pi$ for $k \in \mathbb{Z}$.

The cosine function is an even function, exhibiting bilateral line symmetry across the vertical y-axis:

$$\cos(-x) = \cos x$$

Its graph starts at its peak value $(0, 1)$, crosses the x-axis at $x = \frac{\pi}{2}$, reaches a local trough of $-1$ at $x = \pi$, crosses the x-axis at $x = \frac{3\pi}{2}$, and returns to its peak of $1$ at $x = 2\pi$. The zeros of $y = \cos x$ occur at odd half-integer multiples of $\pi$: $x = \frac{\pi}{2} + k\pi$ for $k \in \mathbb{Z}$.

Because cosine is simply a horizontal phase translation of sine by a quarter-cycle, they are linked by the cofunction shift identity:

$$\cos x = \sin\left(x + \frac{\pi}{2}\right)$$

The Tangent Function

The tangent function is defined analytically as the quotient $f(x) = \tan x = \frac{\sin x}{\cos x}$. Because division by zero is undefined, $\tan x$ ceases to exist whenever its denominator vanishes, which occurs at the zeros of $\cos x$. Consequently, the domain of tangent excludes all points $x = \frac{\pi}{2} + k\pi$ ($k \in \mathbb{Z}$), where the graph exhibits vertical asymptotes:

$$\lim_{x \to (\frac{\pi}{2})^-} \tan x = +\infty, \quad \lim_{x \to (\frac{\pi}{2})^+} \tan x = -\infty$$

Unlike sine and cosine, the fundamental period of tangent is $\pi$ radians:

$$\tan(x + \pi) = \tan x$$

Between consecutive vertical asymptotes, tangent is strictly increasing and surjectively maps onto the entire set of real numbers, giving a range of $(-\infty, \infty)$. Tangent is an odd function ($\tan(-x) = -\tan x$), passing through points of inflection with slope $1$ at $x = k\pi$.

The Reciprocal Trigonometric Functions

1. Cosecant ($y = \csc x = \frac{1}{\sin x}$): Because $\sin x = 0$ at $x = k\pi$, cosecant has vertical asymptotes at $x = k\pi$ ($k \in \mathbb{Z}$). Its range is $(-\infty, -1] \cup [1, \infty)$. The graph consists of alternating U-shaped branches that share local extrema with $y = \sin x$: local minima of $1$ at $x = \frac{\pi}{2} + 2k\pi$ and local maxima of $-1$ at $x = \frac{3\pi}{2} + 2k\pi$. The fundamental period is $2\pi$, and cosecant is an odd function.
2. Secant ($y = \sec x = \frac{1}{\cos x}$): Because $\cos x = 0$ at $x = \frac{\pi}{2} + k\pi$, secant has vertical asymptotes at $x = \frac{\pi}{2} + k\pi$ ($k \in \mathbb{Z}$). Its range is $(-\infty, -1] \cup [1, \infty)$. Its U-shaped branches touch the peaks and troughs of $y = \cos x$. The fundamental period is $2\pi$, and secant is an even function ($\sec(-x) = \sec x$).
3. Cotangent ($y = \cot x = \frac{\cos x}{\sin x} = \frac{1}{\tan x}$): Cotangent has vertical asymptotes at the zeros of sine: $x = k\pi$ ($k \in \mathbb{Z}$). Its range is $(-\infty, \infty)$, its fundamental period is $\pi$, and it is strictly decreasing on each open interval $(k\pi, (k+1)\pi)$. Its zeros occur at $x = \frac{\pi}{2} + k\pi$, and cotangent is an odd function.

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Comparison of the Six Trigonometric Functions

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The General Sinusoidal Transformation Model

Real-world physical oscillations rarely have an amplitude of $1$ or a period of $2\pi$. To model arbitrary harmonic motion, we apply geometric transformations to the parent sine or cosine curves. The standard transformed sinusoidal model is formulated as:

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

Often, the interior argument is expanded as $(Bx - \phi)$. In this alternate format, factoring out the coefficient $B$ is necessary to identify the true horizontal phase shift: $Bx - \phi = B\left(x - \frac{\phi}{B}\right)$, meaning the true horizontal shift is $C = \frac{\phi}{B}$.

The four transformation parameters govern distinct physical and geometric attributes:

1. Amplitude ($|A|$): The amplitude measures the vertical dilation from the central equilibrium axis to the maximum crest (or minimum trough). It is strictly non-negative: $$\text{Amplitude} = |A| = \frac{y_{\max} - y_{\min}}{2}$$ If $A < 0$, the wave undergoes a vertical reflection across its midline.
2. Frequency Coefficient ($B$) and Period ($P$): The coefficient $B$ represents the angular frequency (in radians per unit horizontal change). It compresses the parent period $2\pi$ horizontally by a factor of $|B|$: $$\text{Period } P = \frac{2\pi}{|B|} \iff |B| = \frac{2\pi}{P}$$ For tangent and cotangent, because their parent period is $\pi$, the transformed period is $P = \frac{\pi}{|B|}$.
3. Phase Shift ($C$): The parameter $C$ dictates the horizontal translation. If $C > 0$, the waveform shifts to the right by $C$ units; if $C < 0$, it shifts to the left by $|C|$ units.
4. Midline / Vertical Displacement ($D$): The midline is the horizontal average line around which the sinusoid oscillates: $$y = D = \frac{y_{\max} + y_{\min}}{2}$$

The Five Key Points Partition

To accurately sketch or reconstruct any sinusoidal wave over one period $P$, divide the period into four equal quarter-period subintervals of width $\Delta x = \frac{P}{4} = \frac{\pi}{2|B|}$. Evaluating the function at $x_0 = C$, $x_1 = C + \Delta x$, $x_2 = C + 2\Delta x$, $x_3 = C + 3\Delta x$, and $x_4 = C + 4\Delta x$ generates the signature five-point sequence:

• Standard Sine ($A > 0$): Midline $\to$ Crest $\to$ Midline $\to$ Trough $\to$ Midline.
• Standard Cosine ($A > 0$): Crest $\to$ Midline $\to$ Trough $\to$ Midline $\to$ Crest.
• Reflected Cosine ($A < 0$): Trough $\to$ Midline $\to$ Crest $\to$ Midline $\to$ Trough.

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Modeling Real-World Periodic Phenomena

Many natural phenomena repeat at regular temporal intervals: ocean tides governed by lunar gravitational pull, diurnal and seasonal ambient temperatures, the vertical height of a rider on a rotating Ferris wheel, and simple harmonic mechanical systems (such as a mass suspended on an ideal spring).

General Modeling Protocol from Empirical Extrema

When given observed maximum and minimum values and their corresponding times:

1. Compute the vertical midline: $D = \frac{y_{\max} + y_{\min}}{2}$.
2. Compute the amplitude: $A = \frac{y_{\max} - y_{\min}}{2}$.
3. Determine the period $P$: the time interval between consecutive identical peaks equals $P$, while the time between an adjacent peak and trough equals a half-period ($\frac{P}{2}$). Then compute $B = \frac{2\pi}{P}$.
4. Select the optimal parent model to minimize phase shift algebra:
   • If the curve starts at a crest at $t = t_{\text{peak}}$, use a positive cosine model with $C = t_{\text{peak}}$: $y = A\cos(B(t - t_{\text{peak}})) + D$.
   • If the curve starts at a trough at $t = t_{\text{trough}}$, use a negative cosine model with $C = t_{\text{trough}}$: $y = -A\cos(B(t - t_{\text{trough}})) + D$.
   • If the curve crosses the midline ascending at $t = t_{\text{mid}}$, use a positive sine model with $C = t_{\text{mid}}$.

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Worked Exemplar: Hydrodynamic Tidal Height Modeling

Problem: Oceanographic monitors at a coastal estuary record water depths throughout a lunar cycle. On a particular day, high tide occurs at 4:00 AM ($t = 4.0$ hours after midnight) with an observed water depth of $14.6$ feet. The subsequent low tide occurs at 10:15 AM ($t = 10.25$ hours after midnight) with an observed water depth of $6.2$ feet. Assuming water depth oscillates sinusoidally over time:

1. Construct a sinusoidal function $h(t) = A\cos(B(t - C)) + D$ modeling the water depth in feet as a function of hours $t$ past midnight.
2. Calculate the exact predicted water depth at 1:00 PM ($t = 13.0$ hours).
3. Determine the first time after midnight when the water depth reaches exactly $12.5$ feet.

Step 1: Calculate Midline and Amplitude

From the given extrema, $y_{\max} = 14.6\text{ ft}$ and $y_{\min} = 6.2\text{ ft}$:

$$D = \frac{y_{\max} + y_{\min}}{2} = \frac{14.6 + 6.2}{2} = \frac{20.8}{2} = 10.4\text{ feet}$$ $$A = \frac{y_{\max} - y_{\min}}{2} = \frac{14.6 - 6.2}{2} = \frac{8.4}{2} = 4.2\text{ feet}$$

Step 2: Determine Period and Frequency Factor $B$

The elapsed time between high tide ($t = 4.0$) and the consecutive low tide ($t = 10.25$) is a half-period:

$$\frac{P}{2} = 10.25 - 4.0 = 6.25\text{ hours} \implies P = 2 \times 6.25 = 12.5\text{ hours}$$

Now compute the angular frequency coefficient $B$:

$$B = \frac{2\pi}{P} = \frac{2\pi}{12.5} = \frac{4\pi}{25} = 0.16\pi\text{ rad/hr}$$

Step 3: Identify Phase Shift and Formulate Equation

Because high tide (a maximum crest) occurs at $t = 4.0$, a standard positive cosine function aligns with zero initial phase offset at $C = 4.0$:

$$h(t) = 4.2\cos\left(\frac{4\pi}{25}(t - 4)\right) + 10.4$$

Step 4: Evaluate Depth at 1:00 PM ($t = 13.0$)

Substitute $t = 13.0$ into the model:

$$h(13) = 4.2\cos\left(\frac{4\pi}{25}(13 - 4)\right) + 10.4 = 4.2\cos\left(\frac{4\pi}{25}(9)\right) + 10.4 = 4.2\cos\left(\frac{36\pi}{25}\right) + 10.4$$

Evaluating $\frac{36\pi}{25} = 1.44\pi$ radians (which is in Quadrant III, where cosine is negative):

$$\cos(1.44\pi) = \cos(259.2^\circ) \approx -0.18738$$ $$h(13) \approx 4.2(-0.18738) + 10.4 = -0.787 + 10.4 = 9.613 \approx 9.61\text{ feet}$$

Step 5: Solve for Time When Depth Reaches $12.5$ Feet

Set $h(t) = 12.5$ and isolate the cosine term:

$$4.2\cos\left(\frac{4\pi}{25}(t - 4)\right) + 10.4 = 12.5$$ $$4.2\cos\left(\frac{4\pi}{25}(t - 4)\right) = 12.5 - 10.4 = 2.1$$ $$\cos\left(\frac{4\pi}{25}(t - 4)\right) = \frac{2.1}{4.2} = 0.5$$

Recall that $\cos\theta = 0.5 = \frac{1}{2}$ at principal angles $\theta = \pm\frac{\pi}{3} + 2k\pi$:

$$\frac{4\pi}{25}(t - 4) = \pm\frac{\pi}{3} + 2k\pi$$

Multiply both sides by $\frac{25}{4\pi}$:

$$t - 4 = \pm\frac{\pi}{3} \times \frac{25}{4\pi} + 2k\pi \times \frac{25}{4\pi} = \pm\frac{25}{12} + 12.5k$$ $$t = 4 \pm \frac{25}{12} + 12.5k$$

Converting $\frac{25}{12}$ hours gives $2\text{ hours } 5\text{ minutes}$ ($2.0833\text{ hrs}$):

• Taking the negative branch with $k = 0$: $t = 4 - \frac{25}{12} = \frac{23}{12} \approx 1.9167\text{ hours}$ past midnight ($1:55\text{ AM}$).
• Taking the positive branch with $k = 0$: $t = 4 + \frac{25}{12} = \frac{73}{12} \approx 6.0833\text{ hours}$ past midnight ($6:05\text{ AM}$).

Thus, the very first time after midnight when the water reaches $12.5$ feet is $t = \frac{23}{12}\text{ hours}$, or approximately $1:55\text{ AM}$.