Topic drillAbout 15 minutesTested in 4 of the last 5 real CSCA papers (4 questions)

Trig Values from a Point on the Terminal Side

Nothing to memorise: with a point P(x,y) on the terminal side, find r and the three ratios y/r, x/r, y/x are sin, cos, tan. See how r pulls the point back onto the unit circle, then learn to fix the quadrant and the signs. 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

Start with a real exam question, on instinct:

Real examSource: CSCA Math April 2026 Past Paper

If point

P ( x , y )

lies on the terminal side of angle

\alpha

with

r = \sqrt { x ^ { 2 } + y ^ { 2 } }

, then

\cos \alpha =

2The idea: one point, three ratios

Take any point $P(x,y)$ (not the origin) on the terminal side and the trig values follow: first $r=\sqrt{x^2+y^2}$ , then $\sin\alpha=\dfrac yr$ , $\cos\alpha=\dfrac xr$ , $\tan\alpha=\dfrac yx$ . It does not matter where on the terminal side you take the point: the ratios do not change as it slides along.

Pick a point or drag $x$ and $y$ : watch the triangle, $r$ , and how sin, cos and tan are worked out. The hollow dot is $P$ pulled back onto the unit circle.

Filled: P. Hollow: P pulled back onto the unit circle; its coordinates are (cos α, sin α).

x = 1y = 2

$r=\sqrt{1^2+2^2}=\sqrt{5}$
$\sin\alpha=\dfrac{y}{r}=\dfrac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$
$\cos\alpha=\dfrac{x}{r}=\dfrac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$
$\tan\alpha=\dfrac{y}{x}=\dfrac{2}{1}=2$
quadrant I: sin +, cos +, tan +

One point → three ratios

a point $P(x,y)$ on the terminal side

find $r=\sqrt{x^2+y^2}$

$\sin\alpha=\frac yr$ 　 $\cos\alpha=\frac xr$ 　 $\tan\alpha=\frac yx$

Sine and cosine divide by $r$ ; tangent is the ratio of the coordinates and needs no $r$ .

Signs by quadrant

Quadrant	$\sin\alpha$	$\cos\alpha$	$\tan\alpha$
I	+	+	+
II	+	−	−
III	−	−	+
IV	−	+	−

Chant: all in I, sine in II, tangent in III, cosine in IV.

Faced with $\alpha+\frac\pi2$ : reduce first, then use the definition

$\sin\left(\alpha+\frac\pi2\right)$

reduces to $\cos\alpha$

$=\dfrac xr$

$\cos\left(\alpha+\frac\pi2\right)$

reduces to $-\sin\alpha$

$=-\dfrac yr$

Keep the order: the $\frac\pi2$ inside the bracket cannot be fed straight into the point's coordinates.

Read it as text

$r$ is the distance from the point to the origin. Dividing by $r$ pulls $P$ back along the terminal side onto the unit circle: the coordinates of that point, $(\cos\alpha,\sin\alpha)$ , are exactly $\cos\alpha$ and $\sin\alpha$ . So sine and cosine divide by $r$ ; tangent, $\dfrac yx$ , is a ratio of the two coordinates and needs no $r$ .

Signs come from the quadrant: all positive in I, sine in II, tangent in III, cosine in IV. If the point is on the $y$ -axis ( $x=0$ ), $\tan\alpha$ is undefined.

To find a coordinate from a trig value, write the definition as an equation, e.g. $\dfrac{y}{\sqrt{8+y^2}}=\dfrac{\sqrt3}{3}$ , square it to get $y^2$ , and let the quadrant fix the sign. For something like $\sin(\alpha+\frac\pi2)$ , first reduce it to a function of $\alpha$ : $\sin(\alpha+\frac\pi2)=\cos\alpha$ and $\cos(\alpha+\frac\pi2)=-\sin\alpha$ .

34 steps

Walking through $P(1,2)$ and $\sin\alpha$

1
Plot the point, find the quadrant
$P(1,2)$ : $x>0,\ y>0$ , quadrant I
In quadrant I all three ratios are positive; if unsure, write the signs of $x$ and $y$ first.
2
Find $r$
$r=\sqrt{1^2+2^2}=\sqrt5$
$r$ is always positive.
3
Use the definition
$\sin\alpha=\dfrac{2}{\sqrt5}=\dfrac{2\sqrt5}{5}$
- $\sin\alpha=\dfrac yr$
- $\cos\alpha=\dfrac xr$
- $\tan\alpha=\dfrac yx$ (no $r$ )
No root in the denominator: multiply top and bottom by $\sqrt5$ .
4
Reduce first if there is an $\alpha\pm\frac\pi2$
- $\sin(\alpha+\frac\pi2)=\cos\alpha$
- $\cos(\alpha+\frac\pi2)=-\sin\alpha$
If you cannot recall them, go back to the reduction-formula drill.

Read it as text

1. Plot the point, find the quadrant
Write the signs of $x$ and $y$ : the quadrant fixes the signs of the three ratios.
2. Find $r$
$r=\sqrt{x^2+y^2}$ , always positive.
3. Use the definition
$\sin\alpha=\frac yr$ , $\cos\alpha=\frac xr$ , $\tan\alpha=\frac yx$ ; no root in the denominator.
4. Reduce first if there is an $\alpha\pm\frac\pi2$
Reduce to a function of $\alpha$ with the reduction formulas, then use the definition.

Choose first, then check: 4 points

Commit to each step first; a wrong pick tells you why.

Step 1Step 1, find $r$ :

Read all 4 worked examples in full

$P(1,2),\ \sin\alpha$

Step 1Step 1, find $r$ :
$r=\sqrt{1+4}=\sqrt5$ .
✗ $3$ : $1+2=3$ is not $r$ . $r=\sqrt{x^2+y^2}$ : square, add, then take the root.
✗ $\sqrt3$ : $x^2+y^2=1+4=5$ , not $3$ : $2^2=4$ .
Step 2Step 2, $\sin\alpha=$ ?
$\sin\alpha=\dfrac yr=\dfrac2{\sqrt5}$ .
✗ $\dfrac{x}{r}=\dfrac{1}{\sqrt5}$ :That is $\cos\alpha$ . Sine uses $y$ , the vertical coordinate.
✗ $\dfrac{x}{y}=\dfrac12$ :That is the cotangent $\cot\alpha=\frac xy=\frac12$ , not the sine: sine divides by $r$ , $\sin\alpha=\frac yr=\frac{2}{\sqrt5}$ .
Step 3Step 3, rationalise the denominator:
Multiply top and bottom by $\sqrt5$ .
✗ $\dfrac{\sqrt5}{5}$ :The $2$ on top got lost: $\frac{2}{\sqrt5}\cdot\frac{\sqrt5}{\sqrt5}=\frac{2\sqrt5}{5}$ .

Result: $\sin\alpha=\frac{2\sqrt5}{5}$

$P(-5,-12),\ \cos\alpha,\ \tan\alpha$

Step 1Step 1, find $r$ :
$r=13$ .
✗ $17$ : $5+12=17$ is not $r$ . $r=\sqrt{25+144}=\sqrt{169}=13$ .
✗ $\sqrt{119}$ : $r$ uses $x^2+y^2$ , a sum, not $144-25$ .
Step 2Step 2, $\cos\alpha=$ ?
$\cos\alpha=\dfrac xr=-\dfrac5{13}$ .
✗ $\dfrac{5}{13}$ : $x=-5$ is negative, so the sign comes along: $\frac{-5}{13}$ . In quadrant III cosine is negative anyway.
✗ $-\dfrac{12}{13}$ :That is $\sin\alpha=\frac yr$ . Cosine uses $x$ .
Step 3Step 3, $\tan\alpha=$ ?
$\tan\alpha=\dfrac yx=\dfrac{-12}{-5}=\dfrac{12}{5}$ .
✗ $-\dfrac{12}{5}$ : $\frac{-12}{-5}$ : two negatives make a positive. In quadrant III tangent is positive.
✗ $\dfrac{5}{12}$ : $\tan\alpha=\frac yx$ with $y$ on top; $\frac xy$ is the cotangent.

Result: $\cos\alpha=-\frac{5}{13},\ \tan\alpha=\frac{12}{5}$

$P(3,-1),\ \tan\alpha,\ \sin\alpha$

Step 1Step 1, the quadrant:
Quadrant IV: only cosine is positive; sine and tangent are negative.
✗ quadrant II:Quadrant II has $x<0,\ y>0$ . Here $x=3>0$ and $y=-1<0$ .
Step 2Step 2, $\tan\alpha=$ ?
$\tan\alpha=\dfrac yx=-\dfrac13$ .
✗ $-3$ : $\tan\alpha=\frac yx=\frac{-1}{3}$ , not $\frac xy$ .
✗ $-\dfrac{\sqrt{10}}{10}$ :That is $\sin\alpha=\frac yr$ . Tangent does not use $r$ .
Step 3Step 3, $\sin\alpha=$ ? ( $r=\sqrt{10}$ )
$\sin\alpha=\dfrac{-1}{\sqrt{10}}=-\dfrac{\sqrt{10}}{10}$ .
✗ $\dfrac{\sqrt{10}}{10}$ : $y=-1$ is negative; in quadrant IV sine is negative.
✗ $-\dfrac13$ :That is $\tan\alpha$ . Sine divides by $r$ : $\frac{-1}{\sqrt{10}}$ .

Result: $\tan\alpha=-\frac13,\ \sin\alpha=-\frac{\sqrt{10}}{10}$

$P(x,4),\ \cos\alpha=\frac{x}{5},\ \alpha\in\mathrm{II}$

Step 1Step 1, the sign of $x$ in quadrant II:
$x<0$ .
✗ positive: $x>0$ :Quadrant II is left of the $y$ -axis and above the $x$ -axis: $x<0$ , so $\cos\alpha<0$ .
Step 2Step 2, solve $\dfrac{x}{\sqrt{x^2+16}}=\dfrac x5$ for $x$ :
$\sqrt{x^2+16}=5$ , so $x^2=9$ , and $x<0$ gives $x=-3$ .
✗ $x=3$ : $x^2=9$ has two solutions $\pm3$ ; the $x<0$ from step 1 picks $-3$ .
✗ $x=-5$ : $5$ is $r$ , not $x$ : $\sqrt{x^2+16}=5$ gives $x^2=9$ .
Step 3Step 3, $\tan\alpha=$ ?
$\tan\alpha=\dfrac{4}{-3}=-\dfrac43$ .
✗ $\dfrac43$ : $x=-3$ is negative: $\frac{4}{-3}$ . In quadrant II tangent is negative.
✗ $-\dfrac34$ : $\tan\alpha=\frac yx$ with $y=4$ on top.

Result: $x=-3,\ \tan\alpha=-\frac43$

4Worked examples

Try it yourself, then open the steps

Real examSource: CSCA Math January 2026 Past Paper

If the terminal side of angle

\alpha

passes through

P(1,3)

, then

\sin\alpha=

( ).

Step 1 $P(1,3)$ : $r=\sqrt{1^2+3^2}=\sqrt{10}$ .
Step 2 $\sin\alpha=\dfrac yr=\dfrac{3}{\sqrt{10}}$ .
Step 3Rationalise: $\dfrac{3\sqrt{10}}{10}$ , answer B.

Try it yourself, then open the steps

MockSource: Math Sprint · Batch 2 · Paper 2

Let

a

be an angle in the second quadrant,

P(x,4)

be a point on its terminal side, and

\cos a=\dfrac{1}{5}x

, then

\tan a

equals ( )

Step 1Quadrant II means $x<0$ , consistent with $\cos\alpha=\dfrac x5<0$ .
Step 2By definition $\cos\alpha=\dfrac{x}{\sqrt{x^2+16}}=\dfrac x5$ , so $\sqrt{x^2+16}=5$ and $x^2=9$ .
Step 3 $x<0$ gives $x=-3$ ; $\tan\alpha=\dfrac yx=\dfrac{4}{-3}=-\dfrac43$ , answer D.

Try it yourself, then open the steps

MockSource: Math Sprint · Batch 2 · Paper 1

Given

\cos a=-\dfrac{4}{5}

\tan a<0

\sin a=

( )

Step 1 $\cos\alpha=-\frac45<0$ : quadrants II or III; $\tan\alpha<0$ : II or IV. The overlap: quadrant II.
Step 2 $\sin^2\alpha=1-\cos^2\alpha=1-\frac{16}{25}=\frac9{25}$ , so $\sin\alpha=\pm\frac35$ .
Step 3In quadrant II $\sin\alpha>0$ , so $\sin\alpha=\frac35$ : answer B.

5Common mistakes

Answer data: CSCA.app students' answers over the 90 days before 2026-10-23.

1. Using bare coordinates, guessing a familiar value, or swapping sin and cos

What the data shows:On the June 2026 real exam, for $P(1,2)$ and $\sin\alpha$ , fewer than half get it right. About one in five choose $\frac12$ , the cotangent $\frac xy$ , with no $r$ in it; about one in five choose $\frac{\sqrt3}{2}$ , a familiar special-angle value unrelated to the point; about one in ten choose $\frac{\sqrt5}{5}$ , which is $\cos\alpha$ .

How to avoid it:Write down $x=1,\ y=2,\ r=\sqrt5$ first, then ask which ratio you need: sine is $\frac yr$ (the vertical $y$ on top) and cosine is $\frac xr$ , both divided by $r$ . When a familiar value like $\frac{\sqrt3}{2}$ shows up, ask whether it has anything to do with this point: $r=\sqrt5$ , so no $\sqrt3$ can appear.

Check yourself

Real examSource: CSCA Math June 2026 Past Paper

If the point

P(1, 2)

is on the terminal side of the angle

\alpha

, then

\sin \alpha =

2. Not reducing $\sin(\alpha+\frac\pi2)$ to a function of $\alpha$ first

What the data shows:For $P(1,-\sqrt2)$ and $\sin(\alpha+\frac\pi2)$ , about a third of students get it right. More than a third choose $\pm\frac{\sqrt6}{3}$ , which is $\sin\alpha$ itself or its opposite, as if the $+\frac\pi2$ in the bracket were not there; nearly three in ten choose $-\frac{\sqrt3}{3}$ , the right size with the wrong sign.

How to avoid it:The $\alpha+\frac\pi2$ in the bracket cannot be fed straight into the point's coordinates: reduce it to a function of $\alpha$ first, $\sin(\alpha+\frac\pi2)=\cos\alpha$ , then use $\cos\alpha=\frac xr$ . Here $x=1>0$ , so the result is positive.

Check yourself

MockSource: Math Mock Exam 1

If the terminal side of angle

a

passes through

P(1,-\sqrt2)

, find

\sin(a+\frac\pi2)

3. Solving for a coordinate: no reduction first, no sign from the quadrant

What the data shows:For $P(-2\sqrt2,y)$ with $\cos(\alpha+\frac\pi2)=-\frac{\sqrt3}{3}$ , find $y$ : only about a quarter of students get it right ( $y=2$ ); about four in ten choose $\sqrt2$ and another quarter choose $-\sqrt2$ .

How to avoid it:Three steps: ① reduce $\cos(\alpha+\frac\pi2)=-\sin\alpha$ to get $\sin\alpha=\frac{\sqrt3}{3}>0$ , so $y>0$ ; ② write the definition $\dfrac{y}{\sqrt{8+y^2}}=\dfrac{\sqrt3}{3}$ ; ③ square: $3y^2=8+y^2$ , $y^2=4$ , $y=2$ . Then check by substituting back.

Check yourself

MockSource: Math Mock Exam 1

If point

P(-2\sqrt{2},y)

lies on the terminal side of angle

a

and

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

, then the value of

y

is:

6Practice

Question 1 of 4

MockSource: Math Mock Exam 15

The terminal side of

\alpha

passes through

(3,-1)

. Find

\tan\alpha

0 of 4 right in this set

7Mastery check

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

Learn first

Quadrants of the Coordinate Plane

Concept and formulas

Fundamental Trigonometric Identities

Learn next

Trigonometric Induction Formulas Double-Angle Formulas

Chapter practice

Want more? Go to practice

1Warm-up

2The idea: one point, three ratios

One point → three ratios

Signs by quadrant

Faced with α+π2\alpha+\frac\pi2α+2π​: reduce first, then use the definition

34 steps

Walking through P(1,2)P(1,2)P(1,2) and sin⁡α\sin\alphasinα

Choose first, then check: 4 points

P(1,2), sin⁡αP(1,2),\ \sin\alphaP(1,2), sinα

P(−5,−12), cos⁡α, tan⁡αP(-5,-12),\ \cos\alpha,\ \tan\alphaP(−5,−12), cosα, tanα

P(3,−1), tan⁡α, sin⁡αP(3,-1),\ \tan\alpha,\ \sin\alphaP(3,−1), tanα, sinα

P(x,4), cos⁡α=x5, α∈IIP(x,4),\ \cos\alpha=\frac{x}{5},\ \alpha\in\mathrm{II}P(x,4), cosα=5x​, α∈II

4Worked examples

5Common mistakes

1. Using bare coordinates, guessing a familiar value, or swapping sin and cos

2. Not reducing sin⁡(α+π2)\sin(\alpha+\frac\pi2)sin(α+2π​) to a function of α\alphaα first

3. Solving for a coordinate: no reduction first, no sign from the quadrant

6Practice

7Mastery check

Learn first

Concept and formulas

Learn next

Chapter practice

Faced with $\alpha+\frac\pi2$ : reduce first, then use the definition

Walking through $P(1,2)$ and $\sin\alpha$

$P(1,2),\ \sin\alpha$

$P(-5,-12),\ \cos\alpha,\ \tan\alpha$

$P(3,-1),\ \tan\alpha,\ \sin\alpha$

$P(x,4),\ \cos\alpha=\frac{x}{5},\ \alpha\in\mathrm{II}$

2. Not reducing $\sin(\alpha+\frac\pi2)$ to a function of $\alpha$ first