6.2 Trigonometric Identities & Equations
Key Takeaways
- Primary identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, 1 + cot²θ = csc²θ—use them to rewrite before expanding.
- Angle-addition and double-angle formulas convert products/sums; sin 2θ = 2 sin θ cos θ and cos 2θ = cos²θ − sin²θ are highest-yield.
- Solve trig equations by isolating a single function, using general solutions (e.g. sin θ = k ⇒ θ = (−1)ⁿα + nπ), then filter the asked interval.
- Casting rule / ASTC (All–Sin–Tan–Cos) fixes signs by quadrant; reference angle magnitude stays positive.
- On timed CAE maths MCQs, reduce with an identity first—algebraic expansion without a plan burns the ~30 s budget.
Trigonometry as Algebra with Angles
On the PAF CAE academic maths paper, trigonometry is less about drawing triangles and more about manipulating identities and solving equations at FSc Pre-Engineering depth. With maths commonly reported as ~50 MCQs in ~25 minutes (verify your induction notice), you need automatic recall of core identities and a clean method for general solutions—not lengthy derivations.
Fundamental (Pythagorean) Identities
Starting from the unit circle definition, x = cos θ, y = sin θ, and x² + y² = 1:
| Identity | Equivalent forms |
|---|---|
| sin²θ + cos²θ = 1 | sin²θ = 1 − cos²θ; cos²θ = 1 − sin²θ |
| 1 + tan²θ = sec²θ | sec²θ − tan²θ = 1 |
| 1 + cot²θ = csc²θ | csc²θ − cot²θ = 1 |
Worked — simplify. Express (1 − cos²θ)/cos²θ in terms of tan θ.
(1 − cos²θ)/cos²θ = sin²θ / cos²θ = tan²θ.
Worked — prove-style MCQ. If sec θ − tan θ = 2, find sec θ + tan θ.
Use (sec θ − tan θ)(sec θ + tan θ) = sec²θ − tan²θ = 1.
So 2(sec θ + tan θ) = 1 ⇒ sec θ + tan θ = 1/2.
(Then you can solve the linear pair if asked for sec or tan individually.)
Reciprocal and Quotient Relations
| Function | Reciprocal | Quotient |
|---|---|---|
| sin θ | csc θ = 1/sin θ | tan θ = sin θ / cos θ |
| cos θ | sec θ = 1/cos θ | cot θ = cos θ / sin θ |
| tan θ | cot θ = 1/tan θ | — |
Domain reminders that appear as traps: tan and sec undefined when cos θ = 0; cot and csc undefined when sin θ = 0.
Compound-Angle (Addition) Formulas
| Formula |
|---|
| sin(A ± B) = sin A cos B ± cos A sin B |
| cos(A ± B) = cos A cos B ∓ sin A sin B |
| tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B) |
Worked — exact value. Find sin 75°.
sin 75° = sin(45° + 30°) = sin 45 cos 30 + cos 45 sin 30
= (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4.
Worked — cos difference. cos 15° = cos(45° − 30°) = cos 45 cos 30 + sin 45 sin 30 = (√6 + √2)/4 (same radical pair; sin 75 and cos 15 match as cofunctions of complementary angles).
Double-Angle and Related Forms
From A = B = θ:
| Form | Expression |
|---|---|
| sin 2θ | 2 sin θ cos θ |
| cos 2θ | cos²θ − sin²θ = 2 cos²θ − 1 = 1 − 2 sin²θ |
| tan 2θ | 2 tan θ / (1 − tan²θ) |
Half-angle style rearrangements used in MCQs:
sin²θ = (1 − cos 2θ)/2, cos²θ = (1 + cos 2θ)/2.
Worked — double angle. If sin θ = 3/5 and θ is acute, find sin 2θ and cos 2θ.
cos θ = 4/5 (Pythagorean, first quadrant).
sin 2θ = 2·(3/5)·(4/5) = 24/25.
cos 2θ = (4/5)² − (3/5)² = (16 − 9)/25 = 7/25.
Signs by Quadrant (ASTC)
| Quadrant | Positive functions | Mnemonic |
|---|---|---|
| I (0°–90°) | All | All |
| II (90°–180°) | sin, csc | Sin |
| III (180°–270°) | tan, cot | Tan |
| IV (270°–360°) | cos, sec | Cos |
Worked — sign. cos 240°: 240° is in QIII, reference angle 60°, cos negative ⇒ cos 240° = −1/2.
Solving Trigonometric Equations
Standard FSc pattern:
- Rewrite using identities so one trig function remains (or a known double angle).
- Solve for the principal/reference value.
- Write the general solution.
- Restrict to the interval asked (often 0 ≤ θ < 2π or 0° ≤ θ < 360°).
Common general solutions (θ real):
| Equation | General solution |
|---|---|
| sin θ = sin α | θ = nπ + (−1)ⁿ α, n ∈ ℤ |
| cos θ = cos α | θ = 2nπ ± α, n ∈ ℤ |
| tan θ = tan α | θ = nπ + α, n ∈ ℤ |
Worked — equation. Solve 2 cos²θ − cos θ − 1 = 0 for 0 ≤ θ < 2π.
Let u = cos θ: 2u² − u − 1 = 0 ⇒ (2u + 1)(u − 1) = 0 ⇒ u = 1 or u = −1/2.
cos θ = 1 ⇒ θ = 0 (in the interval; 2π excluded if upper bound is exclusive).
cos θ = −1/2 ⇒ θ = 2π/3, 4π/3.
Solutions: 0, 2π/3, 4π/3.
Worked — double-angle equation. Solve sin 2θ = √3 / 2 for 0° ≤ θ < 180°.
Because 0° ≤ θ < 180°, the double angle satisfies 0° ≤ 2θ < 360°. In that interval, sin φ = √3/2 at φ = 60° and φ = 120°.
So 2θ = 60° or 120° ⇒ θ = 30° or 60°.
Product-to-Sum (Recognition Level)
Occasionally useful:
2 sin A cos B = sin(A+B) + sin(A−B),
2 cos A cos B = cos(A+B) + cos(A−B),
2 sin A sin B = cos(A−B) − cos(A+B).
On a timed paper, use these when a product is staring at you; otherwise stay with Pythagorean and double-angle tools.
Exam Strategy
- Simplify before expanding—replace 1 − sin² with cos², etc.
- Memorize sin/cos of 0°, 30°, 45°, 60°, 90° and build 15°/75° via addition.
- For equations, factor (as with 2u² − u − 1) rather than jumping to the quadratic formula when integers work.
- Always apply ASTC after finding the reference angle.
- Reject candidates that make tan/sec undefined even if they satisfy a squared equation.
These identities are the same toolkit you will reuse in heights-and-distances and bearing problems in the next section—and in physics (SHM, projections) elsewhere on the academic papers.
If tan θ = 3/4 and θ is acute, what is sec²θ?
sin 2θ equals which of the following?
In which quadrant is cos θ negative and sin θ positive?
The solutions of cos θ = −1/2 in 0 ≤ θ < 2π include which complete set?