The Basics
SOHCAHTOA · Right-angled triangle · Six trig functions · Exact values · Graphs
Right-angled triangle & SOHCAHTOA
sin θ
\[\dfrac{\text{opp}}{\text{hyp}}\]
cos θ
\[\dfrac{\text{adj}}{\text{hyp}}\]
tan θ
\[\dfrac{\text{opp}}{\text{adj}}\]
Six trig functions
sin θ
\(\dfrac{\text{opp}}{\text{hyp}}\)
recip → cosec θ
cos θ
\(\dfrac{\text{adj}}{\text{hyp}}\)
recip → sec θ
tan θ
\(\dfrac{\sin\theta}{\cos\theta}\)
recip → cot θ
cosec θ
\(\dfrac{1}{\sin\theta}\)
sec θ
\(\dfrac{1}{\cos\theta}\)
cot θ
\(\dfrac{\cos\theta}{\sin\theta}\)
Exact values
| θ° | θ rad | sin θ | cos θ | tan θ | cosec θ | sec θ | cot θ |
|---|---|---|---|---|---|---|---|
| 0° | \(0\) | \(0\) | \(1\) | \(0\) | ∞ | \(1\) | ∞ |
| 30° | \(\dfrac{\pi}{6}\) | \(\dfrac{1}{2}\) | \(\dfrac{\sqrt{3}}{2}\) | \(\dfrac{1}{\sqrt{3}}\) | \(2\) | \(\dfrac{2\sqrt{3}}{3}\) | \(\sqrt{3}\) |
| 45° | \(\dfrac{\pi}{4}\) | \(\dfrac{1}{\sqrt{2}}\) | \(\dfrac{1}{\sqrt{2}}\) | \(1\) | \(\sqrt{2}\) | \(\sqrt{2}\) | \(1\) |
| 60° | \(\dfrac{\pi}{3}\) | \(\dfrac{\sqrt{3}}{2}\) | \(\dfrac{1}{2}\) | \(\sqrt{3}\) | \(\dfrac{2\sqrt{3}}{3}\) | \(2\) | \(\dfrac{1}{\sqrt{3}}\) |
| 90° | \(\dfrac{\pi}{2}\) | \(1\) | \(0\) | ∞ | \(1\) | ∞ | \(0\) |
y = sin x | period 360° | amp 1
y = cos x | period 360° | amp 1
y = tan x | period 180° | no amp
Graph transformations ∞ \(y = a\cdot f(bx+c)+d\)
a
Vertical stretch. Amplitude = |a|. Negative a flips the graph.
b
Horizontal stretch. Period = \(\frac{360°}{b}\) for sin/cos, \(\frac{180°}{b}\) for tan.
c
Phase shift ∞ slides graph left/right. Careful with direction.
d
Vertical shift ∞ moves the midline. Max = a+d, Min = −a+d.
Interactive graph explorer ∞ drag the sliders
y = 1·sin(1·x + 0°) + 0
Amplitude
1
Period
360°
Max
1
Min
−1
a ∞ amplitude
1
Try negative a ∞ the graph flips
b ∞ period
1
b=2 → period halves. b=0.5 → period doubles
c ∞ phase shift (°)
0°
Positive c shifts graph left
d ∞ vertical shift
0
Moves the midline up or down
Degrees & Radians
deg → rad
\[\theta_{\text{rad}} = \theta° \times \dfrac{\pi}{180}\]
rad → deg
\[\theta° = \theta_{\text{rad}} \times \dfrac{180}{\pi}\]
Key conversions to memorise
| 30° | \(\dfrac{\pi}{6}\) | 90° | \(\dfrac{\pi}{2}\) |
| 45° | \(\dfrac{\pi}{4}\) | 180° | \(\pi\) |
| 60° | \(\dfrac{\pi}{3}\) | 360° | \(2\pi\) |
⚠ CALCULATOR MODE ∞ CHECK BEFORE EVERY EXAM
Wrong calculator mode = wrong answer even if working is perfect. Always check before you start.
How to switch mode ∞ Casio fx-991 (most common)
- Press SHIFT then SETUP (or MODE on older models)
- Look for Angle Unit ∞ usually option 3 or 4
- Select 1: Deg for degrees | 2: Rad for radians
- Check the top of your screen ∞ it should show D or R as confirmation
Sine rule · Cosine rule · Area of triangle
SINE RULE
\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\]
Use when you have a side and its opposite angle. Works for any triangle.
⚠ Ambiguous case ∞ when given two sides and a non-included angle, there may be two possible triangles. Always check if a 2nd solution exists.
COSINE RULE
\[a^2 = b^2 + c^2 - 2bc\cos A\]
rearranged to find angle:
\[\cos A = \frac{b^2 + c^2 - a^2}{2bc}\]
Use when you have all 3 sides, or 2 sides and the included angle.
AREA OF TRIANGLE
\[\text{Area} = \frac{1}{2}ab\sin C\]
Use when you know 2 sides and the included angle. Works for any triangle ∞ not just right-angled.
C is the angle between sides a and b.
WHICH RULE TO USE?
Sine
Given a side + opposite angle. Finding missing side or angle.
Cosine
Given 3 sides, or 2 sides + included angle.
Area
Given 2 sides + included angle C.
Trig Identities
The three IGCSE 0606 identities ∞ all derived from Pythagoras
Identity 1 ∞ Pythagorean
\[\sin^2 A + \cos^2 A = 1\]
Comes from Pythagoras on the unit circle. The foundation.
Identity 2
\[1 + \tan^2 A = \sec^2 A\]
Divide identity 1 through by \(\cos^2 A\)
Identity 3
\[1 + \cot^2 A = \cosec^2 A\]
Divide identity 1 through by \(\sin^2 A\)
Deriving identity 2 ∞ live
- Start: \(\sin^2 A + \cos^2 A = 1\)
- Divide every term by \(\cos^2 A\)
- \(\dfrac{\sin^2 A}{\cos^2 A} + \dfrac{\cos^2 A}{\cos^2 A} = \dfrac{1}{\cos^2 A}\)
- Since \(\frac{\sin}{\cos} = \tan\) and \(\frac{1}{\cos} = \sec\):
- \(\tan^2 A + 1 = \sec^2 A\) ✓
Deriving identity 3 ∞ live
- Start: \(\sin^2 A + \cos^2 A = 1\)
- Divide every term by \(\sin^2 A\)
- \(\dfrac{\sin^2 A}{\sin^2 A} + \dfrac{\cos^2 A}{\sin^2 A} = \dfrac{1}{\sin^2 A}\)
- Since \(\frac{\cos}{\sin} = \cot\) and \(\frac{1}{\sin} = \cosec\):
- \(1 + \cot^2 A = \cosec^2 A\) ✓
Key rearrangements to memorise
\(\sin^2 A = 1 - \cos^2 A\)
Rearrange identity 1
\(\cos^2 A = 1 - \sin^2 A\)
Rearrange identity 1
\(\tan^2 A = \sec^2 A - 1\)
Rearrange identity 2
\(\cot^2 A = \cosec^2 A - 1\)
Rearrange identity 3
Solving Trig Equations
CAST diagram · All solutions · Domain adjustment
CAST diagram
Positivity in each quadrant
A (0°—90°)
All positive
S (90°—180°)
Sin only positive
T (180°—270°)
Tan only positive
C (270°—360°)
Cos only positive
General method
1
Isolate one trig function: \(\sin x = k\), \(\cos x = k\), or \(\tan x = k\)
2
Find principal value: \(x_0 = \sin^{-1}(k)\) ∞ always between −90° and 90°
3
Use CAST to identify which quadrants give that sign → find 2nd angle
4
List ALL solutions in the domain ∞ never stop at one answer
5
For \(\sin(2x)\) or \(\cos(x+30°)\): expand the domain first, solve, then convert back
Finding the 2nd angle
sin
\(180° - x_0\)
cos
\(360° - x_0\)
tan
\(180° + x_0\)
rad
Replace 180° → \(\pi\), 360° → \(2\pi\)
Example ∞ \(\sin x = \dfrac{\sqrt{3}}{2},\;\; 0° \leq x \leq 360°\)
1
Principal value: \(x_0 = 60°\)
2
sin positive → quadrants S and A
3
Q1: \(x=60°\) Q2: \(x = 180°-60° = 120°\)
✓
Answer: \(x = 60°,\; 120°\)
Practice Questions
3 Easy · 3 Medium · 3 Hard — Cambridge 0606 Trigonometry
Easy 1
[2]
Write down the exact value of \(\sin 60° + \cos 30°\).
STEP-BY-STEP SOLUTION
Recall
From exact values table:
\(\sin 60° = \dfrac{\sqrt3}{2}\) \(\cos 30° = \dfrac{\sqrt3}{2}\)
\(\sin 60° = \dfrac{\sqrt3}{2}\) \(\cos 30° = \dfrac{\sqrt3}{2}\)
Add
\(\dfrac{\sqrt3}{2} + \dfrac{\sqrt3}{2} = \dfrac{2\sqrt3}{2}\)
Answer
\(\boldsymbol{\sqrt{3}}\)
Easy 2
[3]
Solve \(\sin\theta = 0.5\) for \(0° \leq \theta \leq 360°\).
STEP-BY-STEP — CAST DIAGRAM
Principal
\(\theta = \sin^{-1}(0.5) = 30°\)
CAST
\(\sin\) is positive in Q1 and Q2.
Q1: \(\theta = 30°\)
Q2: \(\theta = 180° - 30° = 150°\)
Q1: \(\theta = 30°\)
Q2: \(\theta = 180° - 30° = 150°\)
Answer
\(\boldsymbol{\theta = 30°}\) and \(\boldsymbol{\theta = 150°}\)
Always check both quadrants where the trig ratio is positive/negative using CAST.
Easy 3
[2]
Show that \(\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{1}{\sin\theta\cos\theta}\) is an identity.
STEP-BY-STEP SOLUTION
LHS
Combine over common denominator \(\sin\theta\cos\theta\):
\(\dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}\)
\(\dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta}\)
Identity
Apply \(\sin^2\theta + \cos^2\theta = 1\):
\(= \dfrac{1}{\sin\theta\cos\theta}\) = RHS ✓
\(= \dfrac{1}{\sin\theta\cos\theta}\) = RHS ✓
Medium 1
[4]
Solve \(2\cos^2\theta - \cos\theta - 1 = 0\) for \(0° \leq \theta \leq 360°\).
STEP-BY-STEP — QUADRATIC IN \(\cos\theta\)
Factorise
Let \(c = \cos\theta\):
\(2c^2 - c - 1 = (2c+1)(c-1) = 0\)
\(2c^2 - c - 1 = (2c+1)(c-1) = 0\)
Solutions
\(c = 1\) or \(c = -\tfrac{1}{2}\)
\(\cos\theta=1\)
\(\theta = 0°\) and \(\theta = 360°\)
\(\cos\theta=-\frac12\)
Principal: \(\cos^{-1}(\tfrac12)=60°\). Negative ⇒ Q2 and Q3:
\(\theta = 180°-60°=120°\) and \(\theta = 180°+60°=240°\)
\(\theta = 180°-60°=120°\) and \(\theta = 180°+60°=240°\)
Answer
\(\boldsymbol{\theta = 0°, 120°, 240°, 360°}\)
Medium 2
[4]
Prove that \(\dfrac{1-\cos^2\theta}{1+\tan^2\theta} = \sin^2\theta\cos^2\theta\).
STEP-BY-STEP — PROVING IDENTITIES
Numerator
\(1-\cos^2\theta = \sin^2\theta\) (Pythagorean identity)
Denominator
\(1 + \tan^2\theta = \sec^2\theta = \dfrac{1}{\cos^2\theta}\)
Combine
LHS \(= \dfrac{\sin^2\theta}{\frac{1}{\cos^2\theta}} = \sin^2\theta \cdot \cos^2\theta\) = RHS ✓
Medium 3
[4]
Solve \(\tan 2\theta = \sqrt{3}\) for \(0° \leq \theta \leq 180°\).
STEP-BY-STEP — DOUBLE ANGLE
Expand range
If \(0°\leq\theta\leq180°\), then \(0°\leq 2\theta\leq360°\)
Solve for \(2\theta\)
Principal: \(\tan^{-1}(\sqrt3) = 60°\)
tan is positive in Q1 and Q3:
\(2\theta = 60°\) or \(2\theta = 60°+180° = 240°\)
tan is positive in Q1 and Q3:
\(2\theta = 60°\) or \(2\theta = 60°+180° = 240°\)
Find \(\theta\)
\(\theta = 30°\) or \(\theta = 120°\)
Answer
\(\boldsymbol{\theta = 30°}\) and \(\boldsymbol{\theta = 120°}\)
⚠ Always expand the range FIRST when solving \(\tan(n\theta)\), before finding solutions.
Hard 1
[5]
Solve \(3\sin^2\theta = 2 + \cos\theta\) for \(0° \leq \theta \leq 360°\).
STEP-BY-STEP — CONVERT USING \(\sin^2\theta = 1-\cos^2\theta\)
Substitute
\(3(1-\cos^2\theta) = 2 + \cos\theta\)
\(3 - 3\cos^2\theta = 2 + \cos\theta\)
\(3\cos^2\theta + \cos\theta - 1 = 0\)
\(3 - 3\cos^2\theta = 2 + \cos\theta\)
\(3\cos^2\theta + \cos\theta - 1 = 0\)
Quadratic
Let \(c=\cos\theta\): \(3c^2+c-1=0\)
\(c = \dfrac{-1\pm\sqrt{1+12}}{6} = \dfrac{-1\pm\sqrt{13}}{6}\)
\(c = \dfrac{-1\pm\sqrt{1+12}}{6} = \dfrac{-1\pm\sqrt{13}}{6}\)
Values
\(c = \dfrac{-1+\sqrt{13}}{6} \approx 0.434\) or \(c = \dfrac{-1-\sqrt{13}}{6} \approx -0.768\)
Solve
For \(c\approx0.434\): \(\theta\approx64.3°\) (Q1) and \(\theta\approx295.7°\) (Q4)
For \(c\approx-0.768\): \(\theta\approx140.2°\) (Q2) and \(\theta\approx219.8°\) (Q3)
For \(c\approx-0.768\): \(\theta\approx140.2°\) (Q2) and \(\theta\approx219.8°\) (Q3)
Answer
\(\boldsymbol{\theta \approx 64.3°, 140.2°, 219.8°, 295.7°}\)
Hard 2
[5]
Given that \(f(\theta) = a\sin(b\theta + c)\) has amplitude 3, period 120°, and passes through \((0°, 0)\) with a positive gradient, state the values of \(a\), \(b\), \(c\).
STEP-BY-STEP — READING TRANSFORMATIONS
Amplitude
Amplitude \(= |a| = 3 \Rightarrow a = 3\) (positive since positive gradient at 0)
Period
Period \(= \dfrac{360°}{b} = 120° \Rightarrow b = 3\)
Phase shift
At \(\theta=0\): \(f(0)=3\sin(c)=0 \Rightarrow c=0\)
(and gradient positive since \(f'(0)=3b\cos(0)=9>0\) ✓)
(and gradient positive since \(f'(0)=3b\cos(0)=9>0\) ✓)
Answer
\(\boldsymbol{a=3,\; b=3,\; c=0°}\)
Hard 3
[6]
Prove the identity: \(\dfrac{\tan\theta + \sin\theta}{\tan\theta - \sin\theta} \equiv \dfrac{\sec\theta + 1}{\sec\theta - 1}\).
STEP-BY-STEP — PROOF FROM LHS
Rewrite LHS
Replace \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\):
Numerator: \(\dfrac{\sin\theta}{\cos\theta}+\sin\theta = \sin\theta\left(\dfrac{1}{\cos\theta}+1\right) = \sin\theta\cdot\dfrac{1+\cos\theta}{\cos\theta}\)
Denominator: \(\sin\theta\left(\dfrac{1}{\cos\theta}-1\right) = \sin\theta\cdot\dfrac{1-\cos\theta}{\cos\theta}\)
Numerator: \(\dfrac{\sin\theta}{\cos\theta}+\sin\theta = \sin\theta\left(\dfrac{1}{\cos\theta}+1\right) = \sin\theta\cdot\dfrac{1+\cos\theta}{\cos\theta}\)
Denominator: \(\sin\theta\left(\dfrac{1}{\cos\theta}-1\right) = \sin\theta\cdot\dfrac{1-\cos\theta}{\cos\theta}\)
Simplify
Divide: \(\dfrac{\frac{1+\cos\theta}{\cos\theta}}{\frac{1-\cos\theta}{\cos\theta}} = \dfrac{1+\cos\theta}{1-\cos\theta}\)
Convert to sec
Divide top & bottom by \(\cos\theta\):
\(= \dfrac{\frac{1}{\cos\theta}+1}{\frac{1}{\cos\theta}-1} = \dfrac{\sec\theta+1}{\sec\theta-1}\) = RHS ✓
\(= \dfrac{\frac{1}{\cos\theta}+1}{\frac{1}{\cos\theta}-1} = \dfrac{\sec\theta+1}{\sec\theta-1}\) = RHS ✓
Past Year Paper Questions
Cambridge IGCSE 0606 style — Trigonometry
0606 Style
Transformations & Equation
[11]
(a) Sketch the graph of \(y = 3\sin(2x) - 1\) for \(0° \leq x \leq 360°\). State the amplitude, period, and the coordinates of all turning points. [5]
(b) Hence solve \(3\sin(2x) - 1 = 0\) for \(0° \leq x \leq 360°\), giving answers to 1 decimal place. [4]
(c) State the number of solutions of \(3\sin(2x) = 1 + k\) in the interval \(0° \leq x \leq 360°\) when (i) \(k = 3\), (ii) \(k = -4\). [2]
(b) Hence solve \(3\sin(2x) - 1 = 0\) for \(0° \leq x \leq 360°\), giving answers to 1 decimal place. [4]
(c) State the number of solutions of \(3\sin(2x) = 1 + k\) in the interval \(0° \leq x \leq 360°\) when (i) \(k = 3\), (ii) \(k = -4\). [2]
FULL WORKED SOLUTION
(a) Parameters
Amplitude = 3, Period = \(\frac{360°}{2}=180°\), vertical shift = \(-1\)
Max: \(3-1=2\), Min: \(-3-1=-4\)
Turning points at \(x=45°,135°,225°,315°\)
Max: \(3-1=2\), Min: \(-3-1=-4\)
Turning points at \(x=45°,135°,225°,315°\)
(b) Setup
\(3\sin(2x)=1 \Rightarrow \sin(2x)=\frac{1}{3}\)
Range for \(2x\): \(0°\leq2x\leq720°\)
Principal: \(2x = \sin^{-1}(\frac13)\approx19.5°\)
Range for \(2x\): \(0°\leq2x\leq720°\)
Principal: \(2x = \sin^{-1}(\frac13)\approx19.5°\)
(b) Solutions
Q1&Q2 in each 360° cycle:
\(2x = 19.5°, 160.5°, 379.5°, 520.5°\)
\(\boldsymbol{x \approx 9.7°, 80.3°, 189.7°, 260.3°}\)
\(2x = 19.5°, 160.5°, 379.5°, 520.5°\)
\(\boldsymbol{x \approx 9.7°, 80.3°, 189.7°, 260.3°}\)
(c)(i) \(k=3\)
Equation becomes \(3\sin(2x)=4\), so \(\sin(2x)=\frac43 > 1\). No solution. 0 solutions.
(c)(ii) \(k=-4\)
Equation becomes \(3\sin(2x) = -3\), so \(\sin(2x)=-1\).
\(2x=270°, 630°\) ⇒ \(x=135°, 315°\). 2 solutions.
\(2x=270°, 630°\) ⇒ \(x=135°, 315°\). 2 solutions.
0606 Style
Identities & Equation
[9]
(a) Prove: \(\dfrac{\cos\theta}{1-\sin\theta} - \dfrac{\cos\theta}{1+\sin\theta} \equiv 2\tan\theta\). [3]
(b) Hence solve \(\dfrac{\cos\theta}{1-\sin\theta} - \dfrac{\cos\theta}{1+\sin\theta} = 1\) for \(0°<\theta<360°\). [3]
(c) Solve \(2\sec^2\theta - 3\tan\theta - 4 = 0\) for \(0°\leq\theta\leq360°\). [3]
(b) Hence solve \(\dfrac{\cos\theta}{1-\sin\theta} - \dfrac{\cos\theta}{1+\sin\theta} = 1\) for \(0°<\theta<360°\). [3]
(c) Solve \(2\sec^2\theta - 3\tan\theta - 4 = 0\) for \(0°\leq\theta\leq360°\). [3]
FULL WORKED SOLUTION
(a) LHS
Common denominator \((1-\sin\theta)(1+\sin\theta)=1-\sin^2\theta=\cos^2\theta\):
\(\dfrac{\cos\theta(1+\sin\theta)-\cos\theta(1-\sin\theta)}{\cos^2\theta} = \dfrac{2\cos\theta\sin\theta}{\cos^2\theta} = \dfrac{2\sin\theta}{\cos\theta} = 2\tan\theta\) ✓
\(\dfrac{\cos\theta(1+\sin\theta)-\cos\theta(1-\sin\theta)}{\cos^2\theta} = \dfrac{2\cos\theta\sin\theta}{\cos^2\theta} = \dfrac{2\sin\theta}{\cos\theta} = 2\tan\theta\) ✓
(b) Solve
From (a): \(2\tan\theta=1 \Rightarrow \tan\theta=\frac12\)
Principal: \(\theta=26.6°\). Positive in Q1 & Q3:
\(\boldsymbol{\theta\approx26.6°}\) and \(\boldsymbol{\theta\approx206.6°}\)
Principal: \(\theta=26.6°\). Positive in Q1 & Q3:
\(\boldsymbol{\theta\approx26.6°}\) and \(\boldsymbol{\theta\approx206.6°}\)
(c) Substitute
Use \(\sec^2\theta = 1+\tan^2\theta\):
\(2(1+\tan^2\theta)-3\tan\theta-4=0\)
\(2\tan^2\theta-3\tan\theta-2=0\)
\((2\tan\theta+1)(\tan\theta-2)=0\)
\(2(1+\tan^2\theta)-3\tan\theta-4=0\)
\(2\tan^2\theta-3\tan\theta-2=0\)
\((2\tan\theta+1)(\tan\theta-2)=0\)
(c) Answer
\(\tan\theta=-\frac12\): Q2&Q4: \(\theta\approx153.4°, 333.4°\)
\(\tan\theta=2\): Q1&Q3: \(\theta\approx63.4°, 243.4°\)
\(\boldsymbol{\theta\approx63.4°, 153.4°, 243.4°, 333.4°}\)
\(\tan\theta=2\): Q1&Q3: \(\theta\approx63.4°, 243.4°\)
\(\boldsymbol{\theta\approx63.4°, 153.4°, 243.4°, 333.4°}\)