Radians, arc length, sector area, exact values, the Pythagorean identity, compound-angle and double-angle formulas, proving identities, and solving trigonometric equations in [0, 2π].
Trigonometry is the heaviest Paper 01 topic (9 of 45 items) and a recurring element of Paper 02. The section builds from exact values and graphs through to identity proofs and equation solving, all using radians as the default angle unit.
Radians
One radian is the angle subtended at the centre of a circle when the arc length equals the radius.
180°=π radians
Degrees to radians: multiply by 180π
Radians to degrees: multiply by π180
Degrees
0°
30°
45°
60°
90°
120°
135°
150°
180°
270°
360°
Radians
0
6π
4π
3π
2π
32π
43π
65π
π
23π
2π
Arc Length and Sector Area
For a circle of radius r and sector angle θ in radians:
Arc length:s=rθSector area:A=21r2θ
Exam Tip
Both formulas require θ in radians. Convert degrees before substituting. A common error is substituting degrees directly into s=rθ.
Exact Trigonometric Values
These must be memorised. They derive from the equilateral triangle (for 30°,60°) and the isosceles right triangle (for 45°).
θ
sinθ
cosθ
tanθ
0
0
1
0
6π(30°)
21
23
31
4π(45°)
22
22
1
3π(60°)
23
21
3
2π(90°)
1
0
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For related angles in other quadrants, use the CAST diagram.
The CAST Diagram
The sign of each trig function depends on the quadrant. CAST tells you which functions are positive in each quadrant (all others are negative):
Quadrant
Degrees
Positive functions
I (first)
0° to 90°
All: sin, cos, tan
II (second)
90° to 180°
Sin only
III (third)
180° to 270°
Tan only
IV (fourth)
270° to 360°
Cos only
Related angle method: For any angle θ outside [0°,90°], find the reference angle (acute angle to the nearest x-axis), evaluate the trig function for that acute angle, then apply the sign from CAST.
Example
Find sin150°.
150° is in Quadrant II (sin positive). Reference angle: 180°−150°=30°.
sin150°=+sin30°=21
Example
Find cos(65π).
65π=150°, Quadrant II (cos negative). Reference angle: 30°.
cos(65π)=−cos30°=−23
Graphs of Trigonometric Functions
For y=sinkx, y=coskx, and y=tankx:
Property
sinkx / coskx
tankx
Amplitude
1
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Period
k2π
kπ
Range
[−1,1]
R
Increasing k compresses the graph horizontally (shorter period). The sin and cos graphs are identical in shape; cos is sin shifted 2π to the left.
For k=2 specifically: y=sin2x and y=cos2x each complete two full cycles in [0,2π], so their period is π. This is the most commonly tested value of k beyond k=1.
Trigonometric Identities
The fundamental Pythagorean identity derives from the unit circle:
sin2θ+cos2θ≡1
Dividing through by cos2θ:
tan2θ+1≡sec2θ
The identity tanθ≡cosθsinθ is equally essential.
Compound-Angle Formulas
sin(A±B)=sinAcosB±cosAsinB
cos(A±B)=cosAcosB∓sinAsinB
tan(A±B)=1∓tanAtanBtanA±tanB
Note the sign pattern: in cos(A+B) the right-hand side uses minus, and in cos(A−B) it uses plus (opposite to the left-hand side). Students are not expected to prove these formulas, but must apply them accurately.
Example
Find the exact value of sin75°.
sin75°=sin(45°+30°)=sin45°cos30°+cos45°sin30°
=22⋅23+22⋅21=46+42=46+2
Double-Angle Formulas
Setting B=A in the compound-angle formulas gives:
sin2A=2sinAcosA
cos2A=cos2A−sin2A=2cos2A−1=1−2sin2A
tan2A=1−tan2A2tanA
The three forms of cos2A are all equivalent (each comes from substituting sin2A+cos2A=1 into the first). Which to use depends on what the question asks for: if you need cos2A in terms of sinA only, use 1−2sin2A; in terms of cosA only, use 2cos2A−1; if both appear, the first form cos2A−sin2A is most flexible.
Proving Trigonometric Identities
An identity is true for all valid values of the variable. To prove it, work on one side only (usually the more complex side) and transform it step by step until it matches the other side. The reason you cannot move terms across the equals sign is that doing so would assume the identity is already true before you have proved it, which is circular reasoning. The goal is to show both sides are equal independently.
Identity proofs almost always require substituting double-angle formulas and then simplifying. Look for opportunities to cancel or factor. If you are stuck, try expressing everything in terms of sin and cos.
Solving Trigonometric Equations
The syllabus requires solutions in 0≤θ≤2π (radians). General solutions are not required.
Standard method:
Isolate the trig function (e.g. sinθ=21).
Find the reference angle using exact values or a calculator.
Use CAST to identify which quadrants give solutions in the required range.
List all solutions.
Example
Solve cosθ=−23 for 0≤θ≤2π.
Reference angle: cos−1(23)=6π.
Cosine is negative in Quadrants II and III.
Quadrant II: θ=π−6π=65π
Quadrant III: θ=π+6π=67π
Equations Using Identities
Many trig equations require substituting an identity to reduce to a standard form.
Example
Solve 2cos2θ−1=0 for 0≤θ≤2π.
Recognise 2cos2θ−1=cos2θ, but it is simpler to solve directly:
cos2θ=21⇒cosθ=±22
Reference angle: 4π. Cosine is positive in I, IV and negative in II, III:
θ=4π,43π,45π,47π
Example
Solve sin2θ=sinθ for 0≤θ≤2π.
2sinθcosθ=sinθ⇒sinθ(2cosθ−1)=0
Either sinθ=0: θ=0,π,2π.
Or cosθ=21: θ=3π,35π.
Solutions: θ=0,3π,π,35π,2π.
Remember
In the equation above, sinθ=0 gives θ=0, π, and 2π within [0,2π]. Including the endpoints of a closed interval is required. Never automatically exclude 0 or 2π.