The identities and reduction rules that crack Paper 2 trig — simplifying, proving identities, using the CAST diagram, reducing angles and finding exact values — each worked in full, with a worksheet.
Also known as: trigonometric identities · reduction formulae · trig identities
In Grade 11, trigonometry grows beyond right-angled triangles into identities (always true) and reduction formulae (rules for angles in different quadrants). This small toolkit lets you simplify expressions, prove identities and find exact values.
1The two identities you must know
The square identity is the star: rearrange it to sin2θ = 1 − cos2θ or cos2θ = 1 − sin2θ — that's how most 'prove the identity' questions are cracked.
2Type 1: Simplify an expression
- 1Cancel one sinθ: (sinθ·cosθ)/sin2θ = cosθ/sinθ.sin2θ = sinθ × sinθ.
- 2Recognise cosθ/sinθ = 1/tanθ.It's the reciprocal of tanθ = sinθ/cosθ.
3Type 2: Prove an identity
Work on the more complicated side only, using the identities, until it equals the other side.
- 1Left side: 1 − cos2θ = sin2θ.Rearranged square identity.
- 2So LHS = sin2θ/cosθ = sinθ · sinθ/cosθ.Split the fraction.
- 3= sinθ · tanθ = RHS. ✓Since tanθ = sinθ/cosθ.
4Type 3: CAST: sign of a ratio by quadrant
CAST tells you which ratios are positive in each quadrant: 1st All positive, 2nd only Sine, 3rd only Tangent, 4th only Cosine.
- 1In the 4th quadrant only cosine is positive.CAST — the 'C' quadrant.
- 2So sin θ is negative.sin2θ = 1 − 0,36 = 0,64 → sinθ = −0,8 (negative in Q4).
5Type 4: Reduction formulae
Reduction formulae rewrite the ratio of a large or negative angle in terms of an acute angle. Key ones: sin(180° − θ) = sinθ, cos(180° − θ) = −cosθ, sin(360° − θ) = −sinθ, cos(−θ) = cosθ.
- 1180° − θ lands in the 2nd quadrant, where sine is positive.CAST: Q2 is 'S'.
- 2So sin(180° − θ) = sin θ.Same acute reference angle, positive sign.
6Type 5: Exact values using special angles
- 1120° = 180° − 60°, in the 2nd quadrant where cosine is negative.Reduce to the acute reference angle 60°.
- 2cos 120° = −cos 60° = −1/2.cos 60° = 1/2; the sign is negative in Q2.
Keep a single summary page of the identities, CAST and reduction rules and revise it weekly. In the exam, recognising which rule to use is 90% of the battle.
sin2θ means (sinθ)2, not sin(θ2). The square is on the whole ratio, not the angle.
This toolkit reappears throughout Grade 12 (identities, general solutions, 2D/3D problems), so the effort here pays off directly in matric. Practise every type below.
7For which values is the identity valid?
The ATP asks you to determine for which values of a variable an identity holds, and papers ask it as a two-mark add-on to a proof. The rule is simple: an identity fails wherever it would require division by zero.
cosθ = 0 at 90° and 270°, and at every angle a full revolution away from those. So the restriction is written:
- 1The expression is undefined when the denominator is zero: sin θ = 0.Look at the denominator, not the whole expression.
- 2sin θ = 0 at θ = 0°, 180° and 360°.Within the given interval.
- 3So the identity holds for all θ except 0°, 180° and 360°.
When a question says “prove the identity and state for which values of θ it is undefined”, the second part is about the denominator of the original expression — not the simplified one. Simplifying can cancel away the very factor that caused the restriction.
Giving the restriction from the answer instead of the question. If (sinθ cosθ)/cosθ simplifies to sinθ, the simplified form is defined everywhere — but the original still needed cosθ ≠ 0, so θ ≠ 90° + k·180° is the answer.
Practice exercises
Work each one out, then click to reveal the answer.
- 1Simplify sinθ · tanθ · cosθ, and state the restriction on θ. (simplify)
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sinθ·sinθ/cosθ·cosθ = sin2θ ✓✓
The original expression contains tanθ, so cosθ ≠ 0, i.e. θ ≠ 90° + k·180° ✓ - 2Simplify (1 − sin2θ). (identity)
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cos2θ - 3Simplify cos2θ/(1 − sin2θ). (identity)
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cos2θ/cos2θ = 1 - 4Prove tanθ · cosθ = sinθ, and say where the identity does not hold. (prove)
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tanθ = sinθ/cosθ ✓, so sinθ/cosθ·cosθ = sinθ ✓
tanθ is undefined where cosθ = 0, so the identity does not hold at θ = 90° + k·180° ✓ - 5In which quadrant are only sine positive? (CAST)
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2nd quadrant - 6sinθ = 0,5, θ in Q2. Is cosθ + or −? (CAST)
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Q2 → cosine negative - 7Simplify cos(180° − θ). (reduction)
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−cosθ - 8Simplify sin(360° − θ). (reduction)
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−sinθ - 9Exact value of sin 150°. (exact)
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150° = 180° − 30°, Q2 sine + → sin 30° = 1/2 - 10Exact value of cos 210°. (exact)
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210° = 180° + 30°, Q3 cosine − → −cos 30° = (−√3)/2 - 11For which values of θ is tanθ undefined? Give the general answer.
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θ = 90° + k · 180°, k ∈ ℤ ✓✓ (cosθ = 0 there, and tanθ = sinθ/cosθ) - 12For θ ∈ [0° ; 360°], state the values for which (sinθ × tanθ)/cosθ is undefined.
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cosθ = 0, so θ = 90° and θ = 270° ✓✓ - 13Explain why the restriction must be read off the ORIGINAL expression and not the simplified one.
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Simplifying can cancel the factor that caused the division by zero ✓; the original expression is still undefined there, so the restriction stands ✓.
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