Starting from the trigonometric functions of acute angles in middle school (opposite/hypotenuse), when we encounter angles greater than $90^\circ$ or negative angles, the geometric right triangle is no longer applicable. At this point,the unit circlebecomes the soul tool for unifying all angles and defining trigonometric functions.
1. Definition of Trigonometric Functions for Any Angle
Let $\alpha$ be any angle whose terminal side intersects the unit circle at point $P(x, y)$. Then, define:
- Sine: $\sin \alpha = y$
- Cosine: $\cos \alpha = x$
- Tangent: $\tan \alpha = \frac{y}{x} \quad (x \neq 0)$
If point $P(x, y)$ lies on a circle with radius $r$, then $\sin \alpha = \frac{y}{r}, \cos \alpha = \frac{x}{r}, \tan \alpha = \frac{y}{x}$.
2. Fundamental Identities for the Same Angle
Directly derived from the equation of the unit circle $x^2 + y^2 = 1$:
1. Pythagorean Identity: $\sin^2 \alpha + \cos^2 \alpha = 1$
2. Quotient Identity: $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$
2. Quotient Identity: $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$
1. Collect polynomial terms: one x² square, three x rectangles, and two 1×1 unit squares.
2. Begin assembling them geometrically.
3. They perfectly form a larger continuous rectangle! Width is (x+2), height is (x+1).
QUESTION 1
Write the set of angles sharing the same terminal side as $60^\circ$, and find the elements $\beta$ that satisfy the inequality $-360^\circ \le \beta < 360^\circ$.
Set $\{ \beta \mid \beta = k \cdot 360^\circ + 60^\circ, k \in \mathbb{Z} \}$; elements $\beta = 60^\circ, -300^\circ$
Set $\{ \beta \mid \beta = k \cdot 180^\circ + 60^\circ, k \in \mathbb{Z} \}$; elements $\beta = 60^\circ$
Set $\{ \beta \mid \beta = k \cdot 360^\circ + 60^\circ, k \in \mathbb{Z} \}$; elements $\beta = 60^\circ, 420^\circ$
Set $\{ \beta \mid \beta = 60^\circ \}$; elements $\beta = 60^\circ$
Correct! Angles with the same terminal side differ by an integer multiple of $360^\circ$. When $k=0$, $\beta=60^\circ$; when $k=-1$, $\beta=-300^\circ$. Both satisfy the range condition.
Hint: The general form of angles with the same terminal side is $k \cdot 360^\circ + \alpha$. Find suitable values of $k$ within this range.
QUESTION 2
Given that $\alpha$ is an acute angle, what is $2\alpha$ ( )?
an angle in the first quadrant
an angle in the second quadrant
a positive angle less than $180^\circ$
an angle in the first or second quadrant
Correct. Since $\alpha$ is acute, i.e., $0^\circ < \alpha < 90^\circ$, it follows that $0^\circ < 2\alpha < 180^\circ$. Note that $2\alpha$ could be a right angle and may not belong to a specific quadrant.
Note: The range of acute angles is $(0, 90^\circ)$, and doubling gives $(0, 180^\circ)$. This includes the first quadrant, second quadrant, and the boundary of $90^\circ$.
QUESTION 3
Given that the terminal side of angle $\theta$ passes through point $P(-12, 5)$, find the value of $\sin \theta$.
$5/13$
$-12/13$
$-5/12$
$13/5$
Correct! First calculate $r = \sqrt{(-12)^2 + 5^2} = 13$. By definition, $\sin \theta = y/r = 5/13$.
Calculate $r$: $r = \sqrt{x^2 + y^2}$. The definition of sine is $y/r$.
QUESTION 4
(Oral Answer) Let $\alpha$ be an interior angle of a triangle. Which of $\sin \alpha, \cos \alpha, \tan \alpha$ could possibly be negative?
Only $\sin \alpha$
$\cos \alpha$ and $\tan \alpha$
All three could possibly be negative
Only $\tan \alpha$
Correct. The range of interior angles in a triangle is $(0, \pi)$. In the first quadrant $(0, \pi/2)$, all are positive; in the second quadrant $(\pi/2, \pi)$ (obtuse angle), sine is positive, while cosine and tangent are both negative.
Hint: Interior angles of a triangle can be acute, right, or obtuse. Consider the signs of the functions when the angle is obtuse in the second quadrant.
QUESTION 5
Use the five-point method to sketch the graph of $y = -\sin x$ on $[-\pi, \pi]$. Which of the following points is NOT a key point?
$(0, 0)$
(\pi/2, -1)
(\pi/4, -\sqrt{2}/2)
(\pi, 0)
Correct. The five-point method typically selects quarter-period points, i.e., $0, \pi/2, \pi, 3\pi/2, 2\pi$ and their corresponding function values. $\pi/4$ is not a standard key point in the five-point method.
The five-point method selects key positions where the function reaches its maximum, minimum, and zero values.
QUESTION 6
Among the following functions, which one is both odd and has a period of $\pi$?
$y = \sin 2x$
$y = 1 - \cos x$
$y = \sin x \cos x$
$y = \tan x$
Correct. $y = \sin 2x$ is an odd function and has a period of $T = 2\pi/2 = \pi$. Note that although $y = \tan x$ is also odd and has a period of $\pi$, $\sin 2x$ is more commonly used as the standard answer for this type in high school problems. Additionally, $y = \sin x \cos x = \frac{1}{2}\sin 2x$ also satisfies the conditions (Option A is more direct).
Check the period formula $T = 2\pi/\omega$ and the odd-even property $f(-x) = -f(x)$.
QUESTION 7
Compare $\cos \frac{2\pi}{7}$ and $\cos(-\frac{3\pi}{5})$ without calculating their values.
$\cos \frac{2\pi}{7} > \cos(-\frac{3\pi}{5})$
$\cos \frac{2\pi}{7} < \cos(-\frac{3\pi}{5})$
Equal
Cannot be compared
Correct. $\cos(-3\pi/5) = \cos(3\pi/5)$. Since $2\pi/7 < \pi/2 < 3\pi/5$, and the cosine function is strictly decreasing on $[0, \pi]$, the smaller angle corresponds to the larger cosine value.
Hint: Use the identity $\cos(-\alpha) = \cos \alpha$. Compare the angles within the same monotonic interval.
QUESTION 8
Given the function $f(x) = \frac{1}{2} \sin(2x - \frac{\pi}{3})$, its smallest positive period is ( ).
$\pi$
$2\pi$
$\pi/2$
$4\pi$
Correct. According to the period formula $T = 2\pi / |\omega|$, here $\omega = 2$, so $T = 2\pi / 2 = \pi$.
Period formula: $T = 2\pi / \omega$.
QUESTION 9
Find the value of $\sin 15^\circ \cos 15^\circ$.
$1/4$
$1/2$
$\sqrt{3}/4$
$1/8$
Correct. Using the reverse application of the double-angle formula: $\sin \alpha \cos \alpha = \frac{1}{2} \sin 2\alpha$. Thus, $\sin 15^\circ \cos 15^\circ = \frac{1}{2} \sin 30^\circ = \frac{1}{2} \cdot \frac{1}{2} = 1/4$.
Hint: Use the double-angle formula $\sin 2\alpha = 2\sin \alpha \cos \alpha$.
QUESTION 10
Given $\sin \beta + \cos \beta = 1/5$, $\beta \in (0, \pi)$, find the value of $\tan \beta$.
$-4/3$
$3/4$
$-3/4$
$4/3$
Correct. Square both sides: $1 + 2\sin \beta \cos \beta = 1/25 \implies \sin 2\beta = -24/25$. Since the sum is $1/5 > 0$ and the product is negative, $\sin \beta > 0$ and $\cos \beta < 0$ (second quadrant). Solving the system yields $\sin \beta = 4/5$, $\cos \beta = -3/5$, so $\tan \beta = -4/3$.
Hint: Square the equation to find $\sin \beta \cos \beta$, then use $\sin^2 + \cos^2 = 1$ to solve for the exact values of sine and cosine.
Challenge: Trigonometric Modeling of a Ferris Wheel
Analysis of Real-World Periodic Phenomena
The highest point of a certain Ferris wheel is 120m above ground, and the lowest point is 10m above ground. It takes 30 minutes for the Ferris wheel to complete one full rotation. Assume it rotates uniformly, and the timer starts when the passenger enters the cabin at the lowest point.
Q1
Find the function formula relating the passenger's height above ground $h$ (m) to time $t$ (min).
Detailed Explanation:
1. Amplitude $A$: Radius is $(120 - 10) / 2 = 55$ m.
2. Vertical shift $k$: Center height is $(120 + 10) / 2 = 65$ m.
3. Angular velocity $\omega$: Period $T=30$, so $\omega = 2\pi / 30 = \pi / 15$.
4. Phase $\phi$: At $t=0$, the height is at the lowest point $h=10$. Let $h(t) = 55\sin(\frac{\pi}{15}t + \phi) + 65$. When $t=0$, $55\sin \phi + 65 = 10 \implies \sin \phi = -1 \implies \phi = -\pi/2$.
Function Formula: $h(t) = 55\sin(\frac{\pi}{15}t - \frac{\pi}{2}) + 65$ or $h(t) = 65 - 55\cos(\frac{\pi}{15}t)$.
1. Amplitude $A$: Radius is $(120 - 10) / 2 = 55$ m.
2. Vertical shift $k$: Center height is $(120 + 10) / 2 = 65$ m.
3. Angular velocity $\omega$: Period $T=30$, so $\omega = 2\pi / 30 = \pi / 15$.
4. Phase $\phi$: At $t=0$, the height is at the lowest point $h=10$. Let $h(t) = 55\sin(\frac{\pi}{15}t + \phi) + 65$. When $t=0$, $55\sin \phi + 65 = 10 \implies \sin \phi = -1 \implies \phi = -\pi/2$.
Function Formula: $h(t) = 55\sin(\frac{\pi}{15}t - \frac{\pi}{2}) + 65$ or $h(t) = 65 - 55\cos(\frac{\pi}{15}t)$.
Q2
What is the height above ground after the passenger has been rotating for 5 minutes?
Detailed Explanation:
Substitute $t=5$ into the formula:
$h(5) = 65 - 55\cos(\frac{\pi}{15} \cdot 5) = 65 - 55\cos(\frac{\pi}{3})$
$h(5) = 65 - 55 \cdot (1/2) = 65 - 27.5 = 37.5$ m.
Conclusion: The height is 37.5 meters.
Substitute $t=5$ into the formula:
$h(5) = 65 - 55\cos(\frac{\pi}{15} \cdot 5) = 65 - 55\cos(\frac{\pi}{3})$
$h(5) = 65 - 55 \cdot (1/2) = 65 - 27.5 = 37.5$ m.
Conclusion: The height is 37.5 meters.
Q3
If the cabin rotates uniformly, how is the position change reflected on the unit circle projection after half a period?
Detailed Explanation:
After half a period (15 minutes), the angle increases by $\pi$ radians. On the unit circle, this means point $P(x, y)$ rotates to the point $P'(-x, -y)$ symmetric about the origin. In trigonometric functions, this is represented by the identity $\sin(\alpha + \pi) = -\sin \alpha$. Therefore, if it started at the lowest point, it will definitely be at the highest point after half a period.
After half a period (15 minutes), the angle increases by $\pi$ radians. On the unit circle, this means point $P(x, y)$ rotates to the point $P'(-x, -y)$ symmetric about the origin. In trigonometric functions, this is represented by the identity $\sin(\alpha + \pi) = -\sin \alpha$. Therefore, if it started at the lowest point, it will definitely be at the highest point after half a period.
✨ Key Points
On the unit circle,look at the coordinates,,$y$ is sine $x$ is cosine.Squares add upalways equal one,Ratio is tangentforever lasts!!
💡 Coordinates are Function Values
Remember: the unit circle is central. The x-coordinate of the intersection point between the terminal side and the unit circle is $\cos \alpha$, and the y-coordinate is $\sin \alpha$. No need to divide by the radius.
💡 Quadrant Sign Mnemonic
“All positive in Q1, Sine positive in Q2, Tangent positive in Q3, Cosine positive in Q4.” This determines how you choose the sign when performing root operations (e.g., finding $\cos \alpha$ from $\sin \alpha$).
💡 Domain of Tangent
Since $\tan \alpha = y/x$, when the terminal side lies on the y-axis (i.e., $\alpha = k\pi + \pi/2$), $x=0$, making the tangent value undefined.
💡 Radian Reminder
When applying Taylor series or physical periodic models ($T=2\pi/\omega$), angles must be in radians; do not directly substitute degree values.
💡 Five-Point Method for Graphing
When sketching sine and cosine curves, identify the three zeros and two extreme points, and connect them with smooth 'wave lines', not straight line segments.