Starting from the right-angled triangle definitions in middle school trigonometry (opposite/hypotenuse), when dealing with angles greater than $90^\circ$ or negative angles, the geometric right triangle model no longer applies. At this point,the unit circlebecomes the essential 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 (Sine): $\sin \alpha = y$
- Cosine (Cosine): $\cos \alpha = x$
- Tangent (Tangent): $\tan \alpha = \frac{y}{x} \quad (x \neq 0)$
If point $P(x, y)$ lies on a circle of 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 unit circle equation $x^2 + y^2 = 1$:
1. Pythagorean Identity: $\sin^2 \alpha + \cos^2 \alpha = 1$
2. Quotient Identity: $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$
2. Quotient Identity: $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$
Moreover, in advanced mathematics, trigonometric functions can also be approximated numerically usingTaylor seriesfor numerical approximation, for example: $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$, demonstrating a deep connection between trigonometric functions and algebraic polynomials.
1. Gather polynomial terms: one x² square, three x rectangular strips, 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 that share the same terminal side as $60^\circ$, and identify 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 $\{ \beta \mid \beta = 60^\circ \}$; element $\beta = 60^\circ$
Correct!
Correct! Angles with the same terminal side differ by integer multiples of $360^\circ$. When $k=0$, $\beta = 60^\circ$; when $k=-1$, $\beta = -300^\circ$. Both satisfy the range condition.
Incorrect
Hint: The general form of angles sharing the same terminal side is $k \cdot 360^\circ + \alpha$. Find values of $k$ within this range.
QUESTION 2
Given that $\alpha$ is an acute angle, then $2\alpha$ is ( ).
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!
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 thus not necessarily belong to any specific quadrant.
Incorrect
Note: The range of acute angles is $(0, 90^\circ)$, so doubling gives $(0, 180^\circ)$. This includes the first quadrant, second quadrant, and the boundary at $90^\circ$.
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!
Correct! First compute $r = \sqrt{(-12)^2 + 5^2} = 13$. By definition, $\sin \theta = y/r = 5/13$.
Incorrect
Compute $r$: $r = \sqrt{x^2 + y^2}$. 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!
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.
Incorrect
Hint: Interior angles of a triangle could be acute, right, or obtuse. Consider the function signs when the angle is obtuse in the second quadrant.
QUESTION 5
Use the five-point method to graph $y = -\sin x$ 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!
Correct. The five-point method typically selects points at quarter-period intervals: $0, \pi/2, \pi, 3\pi/2, 2\pi$, along with their corresponding function values. $\pi/4$ is not a standard key point in the five-point method.
Incorrect
The five-point method selects key positions where the function reaches its maximum, minimum, and zero values.
QUESTION 6
Which of the following functions 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!
Correct. $y = \sin 2x$ is an odd function and has period $T = 2\pi/2 = \pi$. Note that although $y = \tan x$ is also odd and has period $\pi$, $\sin 2x$ is more commonly used as the standard answer for such problems in high school math.
Incorrect
Check the period formula $T = 2\pi/\omega$ and the odd/even property $f(-x) = -f(x)$.
QUESTION 7
Without evaluating, compare the sizes of $\cos \frac{2\pi}{7}$ and $\cos(-\frac{3\pi}{5})$.
$\cos \frac{2\pi}{7} > \cos(-\frac{3\pi}{5})$
$\cos \frac{2\pi}{7} < \cos(-\frac{3\pi}{5})$
Equal
Cannot be compared
Correct!
Correct. $\cos(-3\pi/5) = \cos(3\pi/5)$. Since $2\pi/7 < \pi/2 < 3\pi/5$, and the cosine function is monotonically decreasing on $[0, \pi]$, the smaller angle corresponds to the larger cosine value.
Incorrect
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!
Correct. According to the period formula $T = 2\pi / |\omega|$, here $\omega = 2$, so $T = 2\pi / 2 = \pi$.
Incorrect
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!
Correct. Using the reverse application of the double-angle formula: $\sin \alpha \cos \alpha = \frac{1}{2} \sin 2\alpha$. Thus, $\sin 15^\circ \cos 15^\circ = \frac{1}{2} \sin 30^\circ = 1/4$.
Incorrect
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!
Correct. Square both sides: $1 + 2\sin \beta \cos \beta = 1/25 \implies \sin 2\beta = -24/25$. Since the sum is positive and the product is negative, $\sin \beta > 0, \cos \beta < 0$ (second quadrant). Solving gives $\sin \beta = 4/5, \cos \beta = -3/5$, so $\tan \beta = -4/3$.
Incorrect
Hint: Square the equation to find $\sin \beta \cos \beta$, then use $\sin^2 \beta + \cos^2 \beta = 1$ to solve for the exact values of sine and cosine.
Challenge: Trigonometric Modeling of a Ferris Wheel
Analysis of Real-World Periodic Phenomena
A certain Ferris wheel has its highest point 120m above ground and lowest point 10m above ground. It takes 30 minutes to complete one full rotation. Assume the Ferris wheel rotates at a constant speed, and the ride begins counting time when the passenger enters the cabin at the lowest point.
Q1
Find the function formula expressing the passenger's height above ground $h$ (m) as a function of 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 position is at the lowest point $h=10$. Let $h(t) = 55\sin(\frac{\pi}{15}t + \phi) + 65$. At $t=0$, $55\sin \phi + 65 = 10 \implies \sin \phi = -1 \implies \phi = -\pi/2$.
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 position is at the lowest point $h=10$. Let $h(t) = 55\sin(\frac{\pi}{15}t + \phi) + 65$. At $t=0$, $55\sin \phi + 65 = 10 \implies \sin \phi = -1 \implies \phi = -\pi/2$.
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 in 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, after half a period it must be at the highest point.
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, after half a period it must be at the highest point.
✨ Key Points
On the unit circle,look at the coordinates,,$y$ is sine $x$ is cosine.Squared sumequals one,Ratio: tangentEndures forever!!
💡 Coordinates are Function Values
Remember: the unit circle is central. The x-coordinate of the intersection point of the terminal side with the unit circle is $\cos \alpha$, and the y-coordinate is $\sin \alpha$.
💡 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 square root operations.
💡 Domain of Tangent
Because $\tan \alpha = y/x$, 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.
💡 Five-Point Method for Graphing
When graphing sine and cosine curves, locate the three zero points and two extremum points, then connect them with smooth "wave lines".