5.4 Graphs and Properties of Sine, Cosine, and Tangent Functions

5.4 Graphs and Properties of Sine, Cosine, and Tangent Functions

The study of trigonometric graphs and their analytical properties (domain, range, periodicity, parity, monotonicity, symmetry, and zero-point distribution) represents one of the most prominent components of the Gaokao Mathematics syllabus. This section presents a systematic investigation of fundamental trigonometric functions and the generalized sinusoidal model $f(x) = A\sin(\omega x + \phi) + B$.

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1. Properties of Fundamental Trigonometric Functions (基本三角函数的性质)

Comprehensive Comparative Table

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2. Graph Transformations of $y = A\sin(\omega x + \phi) + B$ (图象变换规则)

In the model $f(x) = A\sin(\omega x + \phi) + B$ (assuming $A > 0, \omega > 0$):

• $A$: Amplitude (振幅)
• $\omega$: Angular Frequency (圆频率), where Period $T = \frac{2\pi}{\omega}$
• $\omega x + \phi$: Phase (相位)
• $\phi$: Initial Phase (初相)
• $B$: Vertical Shift (中心轴 $y = B$)

To transform the parent graph $y = \sin x$ into $y = A\sin(\omega x + \phi)$, two standard rigorous transformation pathways exist:

Pathway 1: Phase Shift First, Frequency Scaling Second (先平移后缩放)

1. Phase Shift: Shift $y = \sin x$ horizontally by $|\phi|$ units (left if $\phi > 0$, right if $\phi < 0$) to obtain: $$y = \sin(x + \phi)$$
2. Frequency Scaling: Compress/stretch the graph horizontally by a factor of $\frac{1}{\omega}$ (replace $x$ with $\omega x$) to obtain: $$y = \sin(\omega x + \phi)$$
3. Amplitude Scaling: Stretch/compress the graph vertically by a factor of $A$ to obtain: $$y = A\sin(\omega x + \phi)$$

Pathway 2: Frequency Scaling First, Phase Shift Second (先缩放后平移)

1. Frequency Scaling: Compress/stretch $y = \sin x$ horizontally by a factor of $\frac{1}{\omega}$ to obtain: $$y = \sin(\omega x)$$
2. Phase Shift: Shift $y = \sin(\omega x)$ horizontally by $\frac{|\phi|}{\omega}$ units (left if $\phi > 0$, right if $\phi < 0$) to obtain: $$y = \sin\left(\omega\left(x + \frac{\phi}{\omega}\right)\right) = \sin(\omega x + \phi)$$
3. Amplitude Scaling: Stretch/compress vertically by $A$ to obtain: $$y = A\sin(\omega x + \phi)$$

Critical Warning for Gaokao: In Pathway 2, the horizontal shift distance is $\frac{|\phi|}{\omega}$, NOT $|\phi|$! Confusing these two shifts is a frequent source of error in exam questions.

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3. Determining Analytical Expressions from Graphs (由图象求解析式)

When presented with a portion of a sinusoidal curve on the Gaokao exam:

1. Determine Amplitude $A$ and Axis $B$: $$A = \frac{y_{\max} - y_{\min}}{2}, \quad B = \frac{y_{\max} + y_{\min}}{2}$$
2. Determine Period $T$ and Angular Frequency $\omega$:
   • Distance between two adjacent peaks (or troughs) = $T \implies \omega = \frac{2\pi}{T}$.
   • Distance between adjacent peak and trough = $\frac{T}{2} \implies T = 2 \Delta x$.
   • Distance between adjacent zero point and peak = $\frac{T}{4} \implies T = 4 \Delta x$.
3. Determine Initial Phase $\phi$: Substitute a known peak point $(x_0, y_{\max})$ into the phase equation: $$\omega x_0 + \phi = \frac{\pi}{2} + 2k\pi \quad (k \in \mathbb{Z})$$ Select the unique $k$ that satisfies the prescribed range constraint (e.g., $|\phi| < \frac{\pi}{2}$ or $|\phi| < \pi$).

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4. Worked Gaokao Exam Examples (高考典型例题精讲)

Worked Example 1: Full Parameter Extraction and Monotonicity Analysis

Problem: Part of the graph of $f(x) = A\sin(\omega x + \phi)$ ($A > 0, \omega > 0, |\phi| < \frac{\pi}{2}$) is given such that its maximum point is $M\left(\frac{\pi}{6}, 2\right)$ and the adjacent minimum point is $N\left(\frac{2\pi}{3}, -2\right)$.

1. Find the analytical expression of $f(x)$.
2. Find the monotonically increasing intervals of $f(x)$.
3. Describe the transformation steps to obtain $f(x)$ from $y = \sin x$ using Pathway 2 (Frequency scaling first).

Solution:

1. Find $A$, $\omega$, and $\phi$:

   • Amplitude: $A = 2$.
   • Half-period: $\frac{T}{2} = \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \implies T = \pi$.
   • Frequency: $\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi} = 2$.
   • Initial Phase $\phi$: Substitute peak point $M\left(\frac{\pi}{6}, 2\right)$: $$2\left(\frac{\pi}{6}\right) + \phi = \frac{\pi}{2} + 2k\pi \implies \frac{\pi}{3} + \phi = \frac{\pi}{2} \implies \phi = \frac{\pi}{6}$$ Since $|\phi| = \frac{\pi}{6} < \frac{\pi}{2}$, this value is valid.
   • Analytical Expression: $f(x) = 2\sin\left(2x + \frac{\pi}{6}\right)$.

2. Find Monotonically Increasing Intervals: Set $-\frac{\pi}{2} + 2k\pi \le 2x + \frac{\pi}{6} \le \frac{\pi}{2} + 2k\pi$ ($k \in \mathbb{Z}$):

   $$-\frac{\pi}{2} - \frac{\pi}{6} + 2k\pi \le 2x \le \frac{\pi}{2} - \frac{\pi}{6} + 2k\pi$$ $$-\frac{2\pi}{3} + 2k\pi \le 2x \le \frac{\pi}{3} + 2k\pi$$ $$-\frac{\pi}{3} + k\pi \le x \le \frac{\pi}{6} + k\pi \quad (k \in \mathbb{Z})$$

   Thus, the monotonically increasing intervals are $\left[-\frac{\pi}{3} + k\pi, \frac{\pi}{6} + k\pi\right]$ ($k \in \mathbb{Z}$).

3. Graph Transformation Pathway 2:

   • Step 1: Compress $y = \sin x$ horizontally by a factor of $\frac{1}{2}$ to obtain $y = \sin 2x$.
   • Step 2: Shift $y = \sin 2x$ to the left by $\frac{\pi/6}{2} = \frac{\pi}{12}$ units to obtain $y = \sin\left(2\left(x + \frac{\pi}{12}\right)\right) = \sin\left(2x + \frac{\pi}{6}\right)$.
   • Step 3: Stretch the graph vertically by a factor of $2$ to obtain $f(x) = 2\sin\left(2x + \frac{\pi}{6}\right)$.

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Worked Example 2: Zero-Point Count and Parameter Range

Problem: Consider the function $g(x) = \sin\left(\omega x + \frac{\pi}{4}\right)$ with $\omega > 0$. If $g(x)$ has exactly 2 zero points on the closed interval $[0, \pi]$, determine the range of values for $\omega$.

Solution:

1. Determine Phase Range: Since $x \in [0, \pi]$ and $\omega > 0$, the phase $\theta = \omega x + \frac{\pi}{4}$ ranges over: $$\theta \in \left[\frac{\pi}{4}, \omega\pi + \frac{\pi}{4}\right]$$
2. Identify Zero Points of Sine: The zero points of $\sin\theta$ occur at $\theta = k\pi$ ($k \in \mathbb{Z}$), specifically $\pi, 2\pi, 3\pi, \dots$.
3. Apply Zero-Point Count Constraint: For $g(x)$ to have exactly 2 zero points on $[0, \pi]$, the phase interval must include the first two positive zeros ($\pi$ and $2\pi$) but strictly exclude the third positive zero ($3\pi$): $$2\pi \le \omega\pi + \frac{\pi}{4} < 3\pi$$
4. Solve for $\omega$: Subtract $\frac{\pi}{4}$ across the inequality: $$\frac{7\pi}{4} \le \omega\pi < \frac{11\pi}{4}$$ Divide by $\pi$: $$\frac{7}{4} \le \omega < \frac{11}{4}$$ Thus, the range of $\omega$ is $\left[\frac{7}{4}, \frac{11}{4}\right)$.