Topic drillAbout 15 minutesTested in all 5 recent real CSCA papers (7 questions)

Trig Reduction (Induction) Formulas

Stop memorising six formula groups: see the symmetry on the unit circle and derive "odd multiple: swap the name; sign from the quadrant" yourself. With CSCA exam examples, the three most common mistakes and a 5-question mastery check.

This page is written in Chinese and English (the two CSCA exam languages); you are reading the English version.

1Warm-up

Before any formula, answer this on instinct:

MockSource: Math Mock Exam 12

Given

\cos\alpha=\frac{\sqrt2}{3}

, find

\cos(\pi-\alpha)

2The idea: reduction formulas are symmetries of the unit circle

The terminal side of $\alpha$ meets the unit circle at $P(\cos\alpha,\ \sin\alpha)$ — the x-coordinate is cosine, the y-coordinate is sine. A reduction formula just reflects $P$ and reads off the new coordinates.

Drag the slider to change

\alpha

, then pick a transformation and watch the reflected point's coordinates.

\alpha = 30^\circ

reflect across the y-axis

P: (0.866, 0.500) = (cos α, sin α)
P′: (-0.866, 0.500)

$\sin\left(\pi-\alpha\right)=\sin\alpha$

$\cos\left(\pi-\alpha\right)=-\cos\alpha$

k = 2 (even) → name stays

$\pi-\alpha$ , $\pi+\alpha$ , $-\alpha$ and $2\pi-\alpha$ reflect $P$ across the y-axis, the origin or the x-axis: the coordinates are the same two numbers, only their signs can change — so the function name stays and you only decide the sign.

$\frac{\pi}{2}-\alpha$ reflects $P$ across the line $y=x$ : the two coordinates swap — sine becomes cosine and cosine becomes sine, so the name changes. $\frac{\pi}{2}+\alpha$ is that swap followed by a reflection across the y-axis: the swap is what changes the name; the extra flip only affects the sign.

That is where the rule "odd multiple: swap the name; even: keep it; sign from the quadrant" comes from. Write the angle as $k\cdot\frac{\pi}{2}\pm\alpha$ : an odd $k$ is a swap (name changes), an even $k$ is a flip (name stays). The sign always comes from treating $\alpha$ as acute and asking which quadrant the original angle lands in.

34 steps

1. Remove full turns
Subtract whole multiples of $2\pi$ ( $360^\circ$ ); the value does not change.
2. Split the angle
Write what is left as $k\cdot\frac{\pi}{2}\pm\alpha$ , with $\alpha$ treated as acute.
3. Decide the name
Odd $k$ : $\sin\leftrightarrow\cos$ ( $\tan$ becomes $\cot$ ). Even $k$ : the name stays.
4. Decide the sign
Treat $\alpha$ as acute, find the quadrant of $k\cdot\frac{\pi}{2}\pm\alpha$ , and use the sign of the original function there.

Four-step reducer

Pick an angle. At each step, choose first, then check.

Step 1After removing whole multiples of $360^\circ$ ( $2\pi$ ), what angle is left?

Read all 5 worked examples in full

$\sin 210^\circ$

Step 1After removing whole multiples of $360^\circ$ ( $2\pi$ ), what angle is left?
$210^\circ<360^\circ$ , so this step is skipped.
✗ $-150^\circ$ : $210^\circ$ is less than a full turn, so nothing to remove; a negative angle only makes it harder.
Step 2Write it as $k\cdot 90^\circ\pm\alpha$ ( $\alpha$ acute):
$180^\circ=2\cdot 90^\circ$ , so $k=2$ and $\alpha=30^\circ$ .
✗ $90^\circ+120^\circ$ :That $\alpha$ is not acute; the method needs $\alpha$ to be acute.
✗ $360^\circ-150^\circ$ :That $\alpha$ is not acute; the method needs $\alpha$ to be acute.
Step 3Does the function name change?
$k$ is even, so the name stays.
✗ name swaps: $k$ is even, so the name stays.
Step 4With $\alpha$ acute, which sign?
$\sin<0$ in quadrant III: minus.
✗ $+$ : $180^\circ+30^\circ$ is in quadrant III, where $\sin$ is negative.

Result: $\sin 210^\circ=-\sin 30^\circ=-\frac{1}{2}$

$\cos 600^\circ$

Step 1After removing whole multiples of $360^\circ$ ( $2\pi$ ), what angle is left?
$600^\circ-360^\circ=240^\circ$ .
✗ $600^\circ$ : $600^\circ$ is more than a full turn: subtract $360^\circ$ first.
✗ $300^\circ$ : $600^\circ-360^\circ=240^\circ$ — subtract, do not halve.
Step 2Write it as $k\cdot 90^\circ\pm\alpha$ ( $\alpha$ acute):
$k=2$ , $\alpha=60^\circ$ .
✗ $90^\circ+150^\circ$ :That $\alpha$ is not acute; the method needs $\alpha$ to be acute.
Step 3Does the function name change?
$k$ is even, so the name stays.
✗ name swaps: $k$ is even, so the name stays.
Step 4With $\alpha$ acute, which sign?
$\cos<0$ in quadrant III: minus.
✗ $+$ : $240^\circ$ is in quadrant III, where $\cos$ is negative.

Result: $\cos 600^\circ=\cos 240^\circ=-\cos 60^\circ=-\frac{1}{2}$

$\sin\frac{2023\pi}{6}$

Step 1After removing whole multiples of $360^\circ$ ( $2\pi$ ), what angle is left?
$\frac{2023\pi}{6}=168\cdot 2\pi+\frac{7\pi}{6}$ .
✗ $\frac{\pi}{6}$ : $\frac{2023\pi}{6}=336\pi+\frac{7\pi}{6}$ ; only multiples of $2\pi$ can go — a single $\pi$ cannot.
✗ $\frac{5\pi}{6}$ : $2023=6\times 336+7$ , which leaves $\frac{7\pi}{6}$ .
Step 2Write it as $k\cdot\frac{\pi}{2}\pm\alpha$ ( $\alpha$ acute):
$\pi=2\cdot\frac{\pi}{2}$ , so $k=2$ and $\alpha=\frac{\pi}{6}$ .
✗ $\frac{\pi}{2}+\frac{2\pi}{3}$ :That $\alpha$ is not acute; the method needs $\alpha$ to be acute.
Step 3Does the function name change?
$k$ is even, so the name stays.
✗ name swaps: $k$ is even, so the name stays.
Step 4With $\alpha$ acute, which sign?
$\sin<0$ in quadrant III: minus.
✗ $+$ : $\pi+\frac{\pi}{6}$ is in quadrant III, where $\sin$ is negative.

Result: $\sin\frac{2023\pi}{6}=\sin\frac{7\pi}{6}=-\sin\frac{\pi}{6}=-\frac{1}{2}$

$\cos\left(\frac{\pi}{2}+\alpha\right)$

Step 1After removing whole multiples of $360^\circ$ ( $2\pi$ ), what angle is left?
No full turn: skip.
✗ $\alpha-\frac{3\pi}{2}$ :Less than a full turn: nothing to subtract.
Step 2Write it as $k\cdot\frac{\pi}{2}\pm\alpha$ ( $\alpha$ acute):
$k=1$ .
✗ $k=2$ : $\frac{\pi}{2}=1\cdot\frac{\pi}{2}$ , so $k=1$ .
Step 3Does the function name change?
$k$ is odd: $\cos$ becomes $\sin$ .
✗ name stays: $k$ is odd, so the name swaps.
Step 4With $\alpha$ acute, which sign?
Use the original $\cos$ in quadrant II: negative.
✗ $+$ : $\frac{\pi}{2}+\alpha$ is in quadrant II; use the original function $\cos$ , which is negative there.

Result: $\cos\left(\frac{\pi}{2}+\alpha\right)=-\sin\alpha$

$\tan(3\pi-x)$

Step 1After removing whole multiples of $360^\circ$ ( $2\pi$ ), what angle is left?
$3\pi-x=2\pi+(\pi-x)$ .
✗ $3\pi-x$ : $3\pi=2\pi+\pi$ : remove one $2\pi$ first.
Step 2Write it as $k\cdot\frac{\pi}{2}\pm\alpha$ ( $\alpha$ acute):
$k=2$ .
✗ $k=1$ : $\pi=2\cdot\frac{\pi}{2}$ , so $k=2$ .
Step 3Does the function name change?
$k$ is even, so the name stays.
✗ name swaps: $k$ is even, so the name stays.
Step 4With $\alpha$ acute, which sign?
$\tan<0$ in quadrant II: minus.
✗ $+$ : $\pi-x$ is in quadrant II, where $\tan$ is negative.

Result: $\tan(3\pi-x)=\tan(\pi-x)=-\tan x$

4Worked examples

Try it yourself, then open the steps

Real examSource: CSCA Math April 2026 Past Paper

\sin 120 ^ { \circ } =

Step 1 $120^\circ=180^\circ-60^\circ=2\cdot 90^\circ-60^\circ$ , so $k=2$ is even → the name stays $\sin$ .
Step 2With $60^\circ$ acute, $120^\circ$ is in quadrant II, where $\sin$ is positive → plus sign.
Step 3 $\sin 120^\circ=\sin 60^\circ=\frac{\sqrt3}{2}$ : answer A. You could also split it as $90^\circ+30^\circ$ : $k=1$ swaps the name, $\sin$ is positive in quadrant II, giving $\cos 30^\circ=\frac{\sqrt3}{2}$ — any split with an acute $\alpha$ gives the same answer.

Try it yourself, then open the steps

MockSource: Math Mock Exam 1

If the terminal side of angle

a

passes through

P(1,-\sqrt2)

, find

\sin(a+\frac\pi2)

Step 1From the point on the terminal side: $r=\sqrt{1^2+(-\sqrt2)^2}=\sqrt3$ , so $\cos a=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$ .
Step 2In $a+\frac{\pi}{2}$ , $k=1$ is odd → swap the name: $\sin$ becomes $\cos$ .
Step 3With $a$ acute, $a+\frac{\pi}{2}$ is in quadrant II, where $\sin$ is positive → $\sin\left(a+\frac{\pi}{2}\right)=\cos a=\frac{\sqrt3}{3}$ : answer A.

Try it yourself, then open the steps

MockSource: Math Mock Exam 1

Given that

a

is an angle in the fourth quadrant and

\sin(\frac{7\pi}{2}+a)=m

, find

\sin a

Step 1Remove a full turn: $\frac{7\pi}{2}=2\pi+\frac{3\pi}{2}$ , so $\sin\left(\frac{7\pi}{2}+a\right)=\sin\left(\frac{3\pi}{2}+a\right)$ .
Step 2 $\frac{3\pi}{2}=3\cdot\frac{\pi}{2}$ , $k=3$ is odd → swap the name; with $a$ acute, $\frac{3\pi}{2}+a$ is in quadrant IV, where $\sin$ is negative → $\sin\left(\frac{3\pi}{2}+a\right)=-\cos a$ , hence $\cos a=-m$ .
Step 3Then the Pythagorean identity: $\sin^2a=1-\cos^2a=1-m^2$ . The question puts $a$ in quadrant IV, so $\sin a<0$ and $\sin a=-\sqrt{1-m^2}$ : answer D.

5Common mistakes

Answer data: CSCA.app students' answers over the 90 days before 2026-10-21.

1. Swapping the name when you shouldn't — or not when you should

What the data shows:On the June 2026 real paper, about one in six students took $\tan\left(\frac{\pi}{2}-\alpha\right)=\tan\alpha$ ; on mock questions, "name not swapped" is one of the most common wrong answers.

How to avoid it:Look only at $k$ in $k\cdot\frac{\pi}{2}\pm\alpha$ : odd → swap, even → keep. $\pi$ , $2\pi$ and $-\alpha$ all have an even $k$ .

Check yourself

Real examSource: CSCA Math June 2026 Past Paper

About reduction formulae, which of the following statements is correct?

2. Taking the sign from the wrong quadrant

What the data shows:On the same real-exam question, about one in five students took $\sin(\pi+\alpha)=\sin\alpha$ ; on a mock question, nearly a third forgot that $\sin$ is negative in quadrant IV.

How to avoid it:For the sign, always assume $\alpha$ is acute and use the original function (the one before any swap) in the quadrant of $k\cdot\frac{\pi}{2}\pm\alpha$ — not the swapped one.

Check yourself

MockSource: Math Mock Exam 2

Given that

\cos \left( \frac { \pi } { 2 } + \varphi \right) = \frac { \sqrt { 3 } } { 2 }

, then

\sin \varphi = ( \quad )

3. Not removing full turns from a large angle

What the data shows:The mock question $\tan(3\pi-x)$ is among the lowest-scoring reduction-formula questions, and every option draws a fair share of picks — a sign that $3\pi$ alone shakes many students' confidence.

How to avoid it:Subtract multiples of $2\pi$ first: $3\pi-x=2\pi+(\pi-x)$ , $\frac{7\pi}{2}=2\pi+\frac{3\pi}{2}$ , $600^\circ=360^\circ+240^\circ$ . Then split, name and sign what is left.

Check yourself

MockSource: Math Mock Exam 1

Which of the following trigonometric identity transformations is incorrect?

6Practice

Question 1 of 6

MockSource: Math Mock Exam 3

Given that

\cos a = \frac { 1 } { 3 }

, then

\cos ( \pi - a ) = ( \quad )

0 of 6 right in this set

7Mastery check

5 questions; 4 right means you have it. No solutions until the end.

Learn first

Fundamental Trigonometric Identities

Concept and formulas

Trigonometric Induction Formulas

Learn next

Double-Angle Formulas Half-Angle Formulas

Chapter practice

Want more? Go to practice

1Warm-up

2The idea: reduction formulas are symmetries of the unit circle

34 steps

Four-step reducer

sin⁡210∘\sin 210^\circsin210∘

cos⁡600∘\cos 600^\circcos600∘

sin⁡2023π6\sin\frac{2023\pi}{6}sin62023π​

cos⁡(π2+α)\cos\left(\frac{\pi}{2}+\alpha\right)cos(2π​+α)

tan⁡(3π−x)\tan(3\pi-x)tan(3π−x)

4Worked examples

5Common mistakes

1. Swapping the name when you shouldn't — or not when you should

2. Taking the sign from the wrong quadrant

3. Not removing full turns from a large angle

6Practice

7Mastery check

Learn first

Concept and formulas

Learn next

Chapter practice

$\sin 210^\circ$

$\cos 600^\circ$

$\sin\frac{2023\pi}{6}$

$\cos\left(\frac{\pi}{2}+\alpha\right)$

$\tan(3\pi-x)$