M203 20260214 Trigonometric Functions

1. Graphs of Trigonometric Functions

• $y = A \sin(kx + m) + n$

• $y = A \cos(kx + m) + n$

• $y = A \tan(kx + m) + n$

• $y = A \cot(kx + m) + n$

正弦函数的图像特征 | Image Characteristics of $\sin(x)$

正弦函数 $y = \sin(x)$ 的图像被称为正弦曲线 (Sine Wave),其核心特征如下:

1. 周期性与值域 | Periodicity & Range

2. 对称性 | Symmetry

3. 五点作图法 | The Five-Point Method

在一个周期 $[0, 2\pi]$ 内,图像由以下关键点确定:

1. 原点 (Start): $(0, 0)$

2. 波峰 (Peak): $(\frac{\pi}{2}, 1)$

3. 中心 (Mid): $(\pi, 0)$

4. 波谷 (Trough): $(\frac{3\pi}{2}, -1)$

5. 终点 (End): $(2\pi, 0)$

4. 变换公式 | General Transformation

对于函数 $y = A \sin(\omega x + \phi) + B$:

1 ) Draw the graph of $y = \sin x$, $y = \cos x$, $y = \tan x$. Find the period.

3 ) What is the symmetry relationship of $\sin x$, $\cos x$, $\tan x$?

What is the asymptote of $\tan x$?

4 )What is the number of solutions of the equation $\dfrac{x}{100} = \sin x$?

5) At how many points with $x$-coordinate between $0$ and $100$ do the graphs of $y = \sin x$ and $y = \cos x$ intersect?

To find the number of intersections for $\sin x = \cos x$ (or $\tan x = 1$) in the interval $(0, 100)$:

1. General Solution

The graphs intersect when:

$$x = \frac{\pi}{4} + k\pi, \quad k \in \mathbb{Z}$$

2. Range Analysis

We need to find the number of integers $k$ such that:

$$0 < \frac{\pi}{4} + k\pi < 100$$

Subtracting $\frac{\pi}{4}$:

$$-\frac{\pi}{4} < k\pi < 100 - \frac{\pi}{4}$$

Dividing by $\pi$ ($\approx 3.14159$):

$$-0.25 < k < \frac{100}{\pi} - 0.25$$

$$-0.25 < k < 31.83 - 0.25$$

$$-0.25 < k < 31.58$$

3. Conclusion

The possible values for $k$ are $\{0, 1, 2, \dots, 31\}$.

The total number of points is 32.

6 ) Draw the graph of $y = 2\sin(2x + 1) + 2$, what is the period and phase shift?

7 ) Draw the graph of $y = -2\cos\left(3x - \dfrac{1}{4}\right) - 1$, what is the period and phase shift?

The Vertical Line Test is a simple, visual method used to determine whether a graph (typically a curve in the Cartesian coordinate system) represents a function.

1. Core Principle

According to the definition of a function, every input ($x$-value) must correspond to exactly one output ($y$-value).

2. Comparison of Common Examples

Type of GraphIntersectionsIs it a Function?Example
Non-vertical LineAlways 1Yes$y = 2x + 1$
Upward ParabolaAlways 1Yes$y = x^2$
Circle2 (in most areas)No$x^2 + y^2 = 25$
Sideways Parabola2 (in most areas)No$x = y^2$

3. Summary

The Vertical Line Test is essentially checking for "Single-valuedness."

Note: This is exactly why we restrict the domain and range of inverse trigonometric functions. By limiting the interval, we ensure the curve passes the vertical line test, allowing it to be defined as a function.

2. Inversion Function

The inverse of $y = \sin x$ is $y = \arcsin(x)$ (or $\sin^{-1}x$).

The inverse of $y = \cos x$ is $y = \arccos(x)$ (or $\cos^{-1}x$).

Featurey=cosx (Restricted)y=arccosx
InputAngle ($\theta$)Ratio ($x$)
OutputRatio ($x$)Angle ($\theta$)
Domain$[0, \pi]$$[-1, 1]$
Range$[-1, 1]$$[0, \pi]$
Featurey=tanx (Restricted)y=arctanx
Domain$(-\dfrac{\pi}{2}, \dfrac{\pi}{2})$$(-\infty, \infty)$
Range$(-\infty, \infty)$$(-\dfrac{\pi}{2}, \dfrac{\pi}{2})$
AsymptotesVertical ($x = \pm \dfrac{\pi}{2}$)Horizontal ($y = \pm \dfrac{\pi}{2}$)

1 ) Evaluate $\arcsin \left( \dfrac{\sqrt{2}}{2} \right)$

2 ) Evaluate $\arcsin\left(\sin\pi\right)$

8 ) What is the value of $\arcsin\left(\sin \dfrac{5\pi}{3}\right)$?

4 ) Evaluate $\sin\left(\arccos \dfrac{4}{5}\right)$

9 ) $\tan\left(\arccos \dfrac{2}{3}\right) =$

3. Parameterization

A parametrization (represented as $x = x(t), y = y(t)$) is a set of functions where each coordinate depends on an independent parameter, typically time $t$.

Is a Parametrization a Function?

2) Find the graph of the parametric equations

$x = 3 + 2\cos t$, $y = 4 + 2\sin t$.

3 ) Find the graph of the parametric equations $x = \sin t$, $y = 3 - 2 \cos 2t$

1:56:55 A pitfall

6 ) Find parametric equations for $x$ and $y$ such that $x^2 - 3y^2 = 1$

What is the graph of the parametric equations

$x = \dfrac{4}{1 + t^2}$,

$y = \dfrac{4t}{1 + t^2}$

One important detail: As $t \to \infty$ or $t \to -\infty$, both $x$ and $y$ approach $0$. However, looking at the original equation $x = \frac{4}{1+t^2}$, $x$ can never actually be $0$ for any real value of $t$. Therefore, the graph is the entire circle except for the point $(0, 0)$.