Unit 4 · Trigonometric Functions　三角函数

By the end of this chapter you can:

1. Convert freely between degrees and radians and place any angle on the unit circle
2. Evaluate $\sin$, $\cos$, $\tan$ for any angle using quadrant signs and reference angles
3. Apply the Pythagorean identity and its derived forms to simplify or evaluate expressions
4. Use reduction formulae ("奇变偶不变，符号看象限") to rewrite trig functions of $k\cdot 90° \pm \alpha$
5. Compute exact values using sum/difference, double-angle, and half-angle formulas, and read off period, amplitude, and phase shift from a graph equation

Exam weight on past CSCA papers: ~15% (7–8 of 48 MCQs — the highest of any topic).

4.1　Angles and Radian Measure　角度与弧度

What is an angle?

In Chinese high-school math (and on the CSCA), angles are defined as rotation angles: a ray starting from the positive $x$-axis rotates to form an angle $\alpha$. Counter-clockwise rotation gives a positive angle; clockwise gives a negative angle. This extends the familiar $0°$–$360°$ range to all of $\mathbb{R}$.

Radians　弧度

The radian is the natural unit: one radian is the angle subtended by an arc equal in length to the radius.

For a circle of radius $r$, arc length $l$, and central angle $\alpha$ (in radians):

$$\boxed{l = r\alpha}, \qquad \alpha = \frac{l}{r}$$

Since the full circumference is $2\pi r$, one full revolution $= 2\pi$ radians $= 360°$.

Degree–radian conversion

$$\text{radians} = \text{degrees} \times \frac{\pi}{180°}, \qquad \text{degrees} = \text{radians} \times \frac{180°}{\pi}$$

Key angles to memorise (both units)

🔑 Quick conversion shortcut: $180° = \pi$. So $1° = \dfrac{\pi}{180}$ and $1\,\text{rad} = \dfrac{180°}{\pi} \approx 57.3°$.

⚠️ Common exam trap: An angle of $\dfrac{5\pi}{4}$ is not in the first quadrant. Divide: $\dfrac{5\pi}{4} \div \dfrac{\pi}{2} = 2.5$ — it is between $2$ and $3$ half-turns, so it lies in Quadrant III ($180°$ to $270°$).

Worked Example 4.1.A

Convert $-\dfrac{7\pi}{6}$ to degrees, and state which quadrant it lies in.

Solution.

$$-\frac{7\pi}{6} \times \frac{180°}{\pi} = -\frac{7 \times 180°}{6} = -210°$$

A negative angle means clockwise rotation. $-210°$ clockwise = $360° - 210° = 150°$ measured counter-clockwise, which lands in Quadrant II (between $90°$ and $180°$).

$$\boxed{-210°,\text{ Quadrant II}}$$

4.2　Trig Function Definitions　三角函数定义

Unit circle definition

Place angle $\alpha$ in standard position. The terminal ray intersects the unit circle (radius $1$) at point $P = (x, y)$. Define:

$$\sin\alpha = y, \qquad \cos\alpha = x, \qquad \tan\alpha = \frac{y}{x}\ (x \neq 0)$$

This definition works for any angle — positive, negative, or greater than $360°$.

Signs by quadrant

The mnemonic "All Students Take Calculus" (or in Chinese "一全二正弦，三正切，四余弦") tells you which functions are positive in each quadrant:

Special angle values

🔑 Memory trick for $\sin$ row: $0, \tfrac{1}{2}, \tfrac{\sqrt{2}}{2}, \tfrac{\sqrt{3}}{2}, 1$ — the numerators go $\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}$, all divided by $2$. The $\cos$ row is the reverse.

⚠️ Sign quadrant trap: $\sin(210°) = \sin(180° + 30°)$. Since $210°$ is in Quadrant III, both $\sin$ and $\cos$ are negative. $\sin(210°) = -\sin(30°) = -\tfrac{1}{2}$.

Worked Example 4.2.A

Evaluate $\cos\!\left(-\dfrac{5\pi}{6}\right)$.

Solution.

The angle $-\dfrac{5\pi}{6} = -150°$ is the same as $360° - 150° = 210°$ (or equivalently, the reference angle is $30°$). The angle $210°$ is in Quadrant III, where $\cos < 0$.

$$\cos(210°) = -\cos(30°) = -\frac{\sqrt{3}}{2}$$

$$\boxed{\cos\!\left(-\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}}$$

4.3　Pythagorean Identity　同角三角函数关系

The three fundamental identities

From $x^2 + y^2 = 1$ on the unit circle (where $x = \cos\alpha$, $y = \sin\alpha$):

$$\boxed{\sin^2\alpha + \cos^2\alpha = 1}$$

Dividing by $\cos^2\alpha$ (where $\cos\alpha \neq 0$):

$$\boxed{1 + \tan^2\alpha = \sec^2\alpha}$$

Dividing the original by $\sin^2\alpha$:

$$\boxed{1 + \cot^2\alpha = \csc^2\alpha}$$

The quotient relation:

$$\tan\alpha = \frac{\sin\alpha}{\cos\alpha}$$

Common manipulation: given one value, find others

If given $\sin\alpha$ and the quadrant, find $\cos\alpha$ and $\tan\alpha$:

1. Use $\cos^2\alpha = 1 - \sin^2\alpha \Rightarrow \cos\alpha = \pm\sqrt{1 - \sin^2\alpha}$
2. Use the quadrant to choose the correct sign
3. Then $\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha}$

⚠️ Biggest trap on the CSCA: After computing $\cos^2\alpha$, students forget to take the square root and check the sign. Always determine the quadrant first, then assign the sign to $\cos\alpha$.

Worked Example 4.3.A

Given $\sin\alpha = \dfrac{3}{5}$ and $\alpha \in \left(\dfrac{\pi}{2}, \pi\right)$ (Quadrant II), find $\cos\alpha$ and $\tan\alpha$.

Solution.

Step 1: $\cos^2\alpha = 1 - \sin^2\alpha = 1 - \dfrac{9}{25} = \dfrac{16}{25}$

Step 2: Quadrant II → $\cos\alpha < 0$, so $\cos\alpha = -\dfrac{4}{5}$

Step 3: $\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}$

$$\boxed{\cos\alpha = -\frac{4}{5},\quad \tan\alpha = -\frac{3}{4}}$$

Worked Example 4.3.B — Simplification

Simplify $\dfrac{\sin^2\alpha - 1}{\sin\alpha \cdot \cos\alpha}$.

Solution.

$$\frac{\sin^2\alpha - 1}{\sin\alpha \cdot \cos\alpha} = \frac{-\cos^2\alpha}{\sin\alpha \cdot \cos\alpha} = \frac{-\cos\alpha}{\sin\alpha} = -\cot\alpha$$

$$\boxed{-\cot\alpha}$$

4.4　Reduction Formulae　诱导公式

Reduction formulae (诱导公式) convert trig functions of "awkward" angles like $\pi - \alpha$ or $\dfrac{3\pi}{2} + \alpha$ into functions of the "nice" reference angle $\alpha$.

The master mnemonic　奇变偶不变，符号看象限

🔑 "奇变偶不变，符号看象限"
("Odd multiples change the function name; even multiples keep it. The sign depends on which quadrant $\alpha$ would land in.")

• Count how many times $\dfrac{\pi}{2}$ appears: odd multiple of $\dfrac{\pi}{2}$ → change $\sin \leftrightarrow \cos$ (and $\tan \leftrightarrow \cot$); even multiple → keep the function name.
• Then determine the sign by treating $\alpha$ as acute (first quadrant) and checking which quadrant $k\cdot\dfrac{\pi}{2} \pm \alpha$ falls in.

Complete reference table

⚠️ The #1 sign error: Students get the function-name change right but forget to check the sign. Always finish by asking: "If $\alpha$ is acute (Q I), which quadrant does $k\cdot\tfrac{\pi}{2} \pm \alpha$ land in? Is the original function positive or negative there?"

Worked Example 4.4.A

Evaluate $\sin\!\left(\dfrac{7\pi}{6}\right)$ using reduction.

Solution.

$\dfrac{7\pi}{6} = \pi + \dfrac{\pi}{6}$. This is in the form $\pi + \alpha$ with $\alpha = \dfrac{\pi}{6}$.

Rule: $\sin(\pi + \alpha) = -\sin\alpha$ (even multiple of $\tfrac{\pi}{2}$: keep $\sin$; Q III: negative).

$$\sin\!\left(\frac{7\pi}{6}\right) = -\sin\!\left(\frac{\pi}{6}\right) = -\frac{1}{2}$$

$$\boxed{-\dfrac{1}{2}}$$

Worked Example 4.4.B

Simplify $\cos\!\left(\dfrac{3\pi}{2} + \alpha\right) \cdot \tan(\pi - \alpha)$.

Solution.

Apply the reduction formulae term by term:

• $\cos\!\left(\dfrac{3\pi}{2} + \alpha\right) = \sin\alpha$ (odd: $\cos \to \sin$; Q IV: $\cos > 0$ → positive)
• $\tan(\pi - \alpha) = -\tan\alpha$

Product: $\sin\alpha \cdot (-\tan\alpha) = -\sin\alpha \cdot \dfrac{\sin\alpha}{\cos\alpha} = -\dfrac{\sin^2\alpha}{\cos\alpha}$

$$\boxed{-\dfrac{\sin^2\alpha}{\cos\alpha}}$$

4.5　Sum and Difference Formulas　和差公式

Complete reference table

🔑 Memory pattern for $\cos$: "cos–cos minus sin–sin" (both same angle combinations, but subtracted). For $\sin$: "sin–cos plus cos–sin" (cross-paired, added for $+$, subtracted for $-$).

⚠️ Trap: Students mix up the sign inside $\cos(\alpha + \beta)$. The cosine-sum formula has a minus sign between its terms even when you add the angles. Do not write $\cos(\alpha + \beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta$ — that is $\cos(\alpha - \beta)$.

Worked Example 4.5.A

Compute $\cos 75°$ exactly.

Solution.

Write $75° = 45° + 30°$ and apply the cosine addition formula:

$$\cos 75° = \cos(45° + 30°) = \cos 45°\cos 30° - \sin 45°\sin 30°$$
$$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$$

$$\boxed{\cos 75° = \dfrac{\sqrt{6} - \sqrt{2}}{4}}$$

Worked Example 4.5.B

If $\sin\alpha = \dfrac{1}{2}$, $\alpha \in \left(\dfrac{\pi}{2}, \pi\right)$, $\cos\beta = -\dfrac{\sqrt{2}}{2}$, $\beta \in (\pi, 2\pi)$, find $\sin(\alpha + \beta)$.

Solution.

First find the missing values using quadrant information and the Pythagorean identity.

$\alpha$ in Q II, $\sin\alpha = \tfrac{1}{2}$: $\cos\alpha = -\dfrac{\sqrt{3}}{2}$.

$\beta$ in Q III or IV with $\cos\beta < 0$: since $\cos\beta = -\dfrac{\sqrt{2}}{2}$ and $\beta \in (\pi, 2\pi)$, and $\cos < 0$ in Q III, $\beta \in (\pi, \tfrac{3\pi}{2})$, so $\sin\beta < 0$: $\sin\beta = -\dfrac{\sqrt{2}}{2}$.

Now:
$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{1}{2}\cdot\!\left(-\frac{\sqrt{2}}{2}\right) + \left(-\frac{\sqrt{3}}{2}\right)\cdot\!\left(-\frac{\sqrt{2}}{2}\right)$$
$$= -\frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$$

$$\boxed{\dfrac{\sqrt{6} - \sqrt{2}}{4}}$$

4.6　Double Angle Formulas　倍角公式

Complete reference table

🔑 Power-lowering trick: These rearrange to eliminate squares:
$$\sin^2\alpha = \frac{1 - \cos 2\alpha}{2}, \qquad \cos^2\alpha = \frac{1 + \cos 2\alpha}{2}$$
These appear constantly in integration (later courses) and in CSCA simplification questions.

⚠️ Three forms of $\cos 2\alpha$: The examiners love to force you into one specific form. If the question involves $\sin^2\alpha$, use $\cos 2\alpha = 1 - 2\sin^2\alpha$. If it involves $\cos^2\alpha$, use $\cos 2\alpha = 2\cos^2\alpha - 1$.

Worked Example 4.6.A

Given $\sin\alpha + \cos\alpha = \dfrac{\sqrt{2}}{2}$, find $\sin 2\alpha$.

Solution.

Square both sides:

$$(\sin\alpha + \cos\alpha)^2 = \frac{1}{2}$$
$$\sin^2\alpha + 2\sin\alpha\cos\alpha + \cos^2\alpha = \frac{1}{2}$$
$$1 + \sin 2\alpha = \frac{1}{2}$$
$$\sin 2\alpha = -\frac{1}{2}$$

$$\boxed{\sin 2\alpha = -\dfrac{1}{2}}$$

Worked Example 4.6.B

Simplify $\dfrac{1 - \cos 2\alpha}{\sin 2\alpha}$.

Solution.

$$\frac{1 - \cos 2\alpha}{\sin 2\alpha} = \frac{2\sin^2\alpha}{2\sin\alpha\cos\alpha} = \frac{\sin\alpha}{\cos\alpha} = \tan\alpha$$

$$\boxed{\tan\alpha}$$

4.7　Half Angle Formulas　半角公式

Complete reference table

⚠️ The $\pm$ trap: The square-root forms of the half-angle formulas require you to determine the sign of $\dfrac{\alpha}{2}$ separately. The two equivalent forms of $\tan\dfrac{\alpha}{2}$ that avoid the $\pm$ are much safer for calculation — prefer them.

🔑 Connection to double-angle: Half-angle formulas are just the double-angle/power-lowering formulas run in reverse. If $\beta = \dfrac{\alpha}{2}$, then $\alpha = 2\beta$ and $\cos\alpha = 1 - 2\sin^2\!\dfrac{\alpha}{2}$ rearranges to the $\sin\dfrac{\alpha}{2}$ formula.

Worked Example 4.7.A

Given $\cos\alpha = \dfrac{3}{5}$ with $0 < \alpha < \dfrac{\pi}{2}$, find $\sin\dfrac{\alpha}{2}$, $\cos\dfrac{\alpha}{2}$, and $\tan\dfrac{\alpha}{2}$.

Solution.

Since $0 < \alpha < \dfrac{\pi}{2}$, we have $0 < \dfrac{\alpha}{2} < \dfrac{\pi}{4} \subset$ Quadrant I — all trig values are positive.

$$\sin\frac{\alpha}{2} = \sqrt{\frac{1 - \cos\alpha}{2}} = \sqrt{\frac{1 - 3/5}{2}} = \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$$

$$\cos\frac{\alpha}{2} = \sqrt{\frac{1 + \cos\alpha}{2}} = \sqrt{\frac{1 + 3/5}{2}} = \sqrt{\frac{8/5}{2}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$

$$\tan\frac{\alpha}{2} = \frac{\sin\alpha}{1 + \cos\alpha} = \frac{4/5}{1 + 3/5} = \frac{4/5}{8/5} = \frac{1}{2}$$

$$\boxed{\sin\frac{\alpha}{2} = \frac{\sqrt{5}}{5},\quad \cos\frac{\alpha}{2} = \frac{2\sqrt{5}}{5},\quad \tan\frac{\alpha}{2} = \frac{1}{2}}$$

Verification: $\sin^2\!\dfrac{\alpha}{2} + \cos^2\!\dfrac{\alpha}{2} = \dfrac{1}{5} + \dfrac{4}{5} = 1$ ✓

4.8　Trigonometric Function Graphs　三角函数图像

The standard sine graph $y = A\sin(\omega x + \varphi) + b$

The same template applies to $y = A\cos(\omega x + \varphi) + b$, with period $T = \dfrac{2\pi}{|\omega|}$.

For tangent: $y = A\tan(\omega x + \varphi)$ has period $T = \dfrac{\pi}{|\omega|}$.

Key graph features

🔑 Read the period from the equation: Given $y = 3\sin(2x - \pi)$, the period is $\dfrac{2\pi}{2} = \pi$. The phase shift is $\dfrac{\pi}{2}$ to the right (because $2x - \pi = 0 \Rightarrow x = \dfrac{\pi}{2}$). The amplitude is $3$.

⚠️ Phase-shift direction trap: For $y = \sin\!\left(x - \dfrac{\pi}{3}\right)$, the graph shifts $\dfrac{\pi}{3}$ to the right (positive direction), NOT left. The rule: $y = \sin(x - c)$ shifts right by $c$ when $c > 0$.

⚠️ Tangent period is $\pi$, not $2\pi$: This catches students every exam. $y = \tan x$ has period $\pi$ because tan repeats after half a revolution, not a full one.

Worked Example 4.8.A

For $y = 2\sin\!\left(3x + \dfrac{\pi}{6}\right)$, find the amplitude, period, and the horizontal shift of the graph relative to $y = 2\sin(3x)$.

Solution.

• Amplitude: $|A| = |2| = 2$
• Period: $T = \dfrac{2\pi}{|\omega|} = \dfrac{2\pi}{3}$
• Phase shift: Set $3x + \dfrac{\pi}{6} = 0 \Rightarrow x = -\dfrac{\pi}{18}$. The graph shifts $\dfrac{\pi}{18}$ to the left relative to $y = 2\sin(3x)$.

$$\boxed{A = 2,\quad T = \dfrac{2\pi}{3},\quad \text{shift: } \dfrac{\pi}{18}\text{ left}}$$

Worked Example 4.8.B — Reading the equation from a graph

A sine graph has maximum value $4$, minimum value $-2$, and passes through its first maximum at $x = \dfrac{\pi}{4}$. The period is $2\pi$. Find the equation.

Solution.

• Midline: $b = \dfrac{4 + (-2)}{2} = 1$
• Amplitude: $A = \dfrac{4 - (-2)}{2} = 3$
• Period $= 2\pi$, so $\omega = 1$
• Phase shift: max occurs at $x = \dfrac{\pi}{4}$. For $y = 3\sin(x + \varphi) + 1$, max when $x + \varphi = \dfrac{\pi}{2}$, so $\varphi = \dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}$.

$$\boxed{y = 3\sin\!\left(x + \dfrac{\pi}{4}\right) + 1}$$

Try it!　自测练习

Q1. Convert $225°$ to radians. Which quadrant?

Q2. Given $\tan\alpha = -\sqrt{3}$ and $\cos\alpha > 0$, find $\sin\alpha$.

Q3. Simplify $\sin\!\left(\dfrac{3\pi}{2} - \alpha\right) \cdot \cos(\pi + \alpha) - \cos^2\!\alpha$.

Q4. If $\cos 2\alpha = \dfrac{1}{3}$ and $\alpha \in \left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$, find $\sin\alpha$.

Q5. For $y = -\cos\!\left(2x - \dfrac{\pi}{3}\right)$, state the period, amplitude, and the $x$-coordinate of the first maximum for $x > 0$.

📌 Chapter summary

Topic	Key formula / mnemonic
4.1 Radians	$l = r\alpha$; $180° = \pi$; memorise 6 key angles
4.2 Signs	ASTC: All / Sin / Tan / Cos positive in Q I/II/III/IV
4.3 Pythagorean	$\sin^2\alpha + \cos^2\alpha = 1$; always determine quadrant before taking $\pm$
4.4 Reduction	奇变偶不变，符号看象限: odd multiple of $\tfrac{\pi}{2}$ → swap name; even → keep name; sign from quadrant
4.5 Sum/Diff	$\cos(\alpha\pm\beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta$ (note the sign flip for cos)
4.6 Double angle	$\sin 2\alpha = 2\sin\alpha\cos\alpha$; three forms of $\cos 2\alpha$ — pick the one matching what's given
4.7 Half angle	Use $\tan\dfrac{\alpha}{2} = \dfrac{\sin\alpha}{1+\cos\alpha}$ to avoid $\pm$ ambiguity
4.8 Graphs	$T = \dfrac{2\pi}{|\omega|}$ for sin/cos; $T = \dfrac{\pi}{|\omega|}$ for tan; phase shift = $-\dfrac{\varphi}{\omega}$

What's next → Unit 5 (Sequences) and Unit 6 (Plane Vectors) are self-contained but the trig identities from this chapter reappear in dot products (Unit 6) and conic sections (Unit 9).