正弦函数zhèngxián hánshù

Core Concept

The Sine Function is one of the fundamental trigonometric functions, describing the relationship between an angle and the ratio of the opposite side to the hypotenuse in a right triangle. On the unit circle, the sine function represents the y-coordinate of a point.

Mathematical Definition

In a right triangle, for an acute angle $\alpha$: $\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}}$

On the unit circle, if angle $\alpha$ terminates at point $P(x, y)$: $\sin \alpha = y$

Function Form

$y = \sin x, \quad x \in \mathbb{R}$

Graph and Properties

Basic Properties

1. Domain: $\mathbb{R}$ (all real numbers)
2. Range: $[-1, 1]$
3. Period: $T = 2\pi$
4. Parity: Odd function, $\sin(-x) = -\sin x$
5. Symmetry:
   • Symmetric about the origin
   • Symmetric about lines $x = \frac{\pi}{2} + k\pi$ ($k \in \mathbb{Z}$)

Monotonicity

• Increasing intervals: $\left[-\frac{\pi}{2} + 2k\pi, \frac{\pi}{2} + 2k\pi\right]$ ($k \in \mathbb{Z}$)
• Decreasing intervals: $\left[\frac{\pi}{2} + 2k\pi, \frac{3\pi}{2} + 2k\pi\right]$ ($k \in \mathbb{Z}$)

Special Values

Angle	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\pi$
$\sin$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$	$0$

Extreme Values

• Maximum: $1$, at $x = \frac{\pi}{2} + 2k\pi$ ($k \in \mathbb{Z}$)
• Minimum: $-1$, at $x = -\frac{\pi}{2} + 2k\pi$ ($k \in \mathbb{Z}$)

Important Formulas

Fundamental Identity

$\sin^2 x + \cos^2 x = 1$

Reduction Formulas

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

Sum and Difference Formulas

$\sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta$

Double Angle Formula

$\sin 2\alpha = 2\sin \alpha \cos \alpha$

CSCA Practice Problems

💡 Note: The following practice problems are designed according to CSCA exam syllabus and Chinese standardized testing format to help students familiarize with exam question types and problem-solving strategies.

[Example 1] Basic (Difficulty ★★☆☆☆)

Given $\sin \alpha = \frac{3}{5}$ and $\alpha$ is in the second quadrant, find $\cos \alpha$ and $\tan \alpha$.

Solution:

From $\sin^2 \alpha + \cos^2 \alpha = 1$: $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{9}{25} = \frac{16}{25}$

Since $\alpha$ is in the second quadrant, $\cos \alpha < 0$: $\cos \alpha = -\frac{4}{5}$

$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/5}{-4/5} = -\frac{3}{4}$

Answer: $\cos \alpha = -\frac{4}{5}$, $\tan \alpha = -\frac{3}{4}$

---

[Example 2] Intermediate (Difficulty ★★★☆☆)

Find the maximum and minimum values of $y = 2\sin x + 1$ on $[0, 2\pi]$.

Solution:

Since $\sin x \in [-1, 1]$:

Maximum: When $\sin x = 1$ (at $x = \frac{\pi}{2}$), $y_{max} = 2 \times 1 + 1 = 3$

Minimum: When $\sin x = -1$ (at $x = \frac{3\pi}{2}$), $y_{min} = 2 \times (-1) + 1 = -1$

Answer: Maximum is $3$, minimum is $-1$

Common Misconceptions

❌ Misconception 1: Confusing the period

Wrong: Thinking the sine function has period $\pi$

Correct: The sine function has period $2\pi$, $\sin(x + 2\pi) = \sin x$

❌ Misconception 2: Forgetting to consider quadrants

Wrong: From $\sin \alpha = \frac{3}{5}$ directly concluding $\cos \alpha = \frac{4}{5}$

Correct: Must determine the sign of $\cos \alpha$ based on the quadrant

Study Tips

1. ✅ Memorize special values: Sine values for $0°, 30°, 45°, 60°, 90°$
2. ✅ Understand the graph: Draw and understand periodicity and symmetry
3. ✅ Master formulas: Reduction, sum/difference, double angle formulas
4. ✅ Distinguish quadrants: Sine values have different signs in different quadrants

---

💡 Exam Tip: The sine function is core to trigonometry, accounting for about 40% of trigonometric questions in CSCA exams. Must be thoroughly understood and mastered!