ZIMSEC A Level · Pure Mathematics · Form 5, Form 6
Radians
π radians = 180°. Arc length s = rθ; sector area = ½r²θ (θ in radians).
Identities
- sin²θ + cos²θ = 1; 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ.
- Compound angles: sin(A ± B) = sinA cosB ± cosA sinB; cos(A ± B) = cosA cosB ∓ sinA sinB.
- Double angles: sin 2A = 2 sinA cosA; cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A; tan 2A = 2tanA/(1 − tan²A).
The R-formula
a sinθ + b cosθ = R sin(θ + α), where R = √(a² + b²) and tan α = b/a.
Worked example: Express 3 sin θ + 4 cos θ in the form R sin(θ + α), and find its maximum value.
R = √(9 + 16) = 5; tan α = 4/3, α ≈ 53.13°
3 sin θ + 4 cos θ = 5 sin(θ + 53.13°); maximum value 5
Solving equations
Use identities to get one function, find the principal value, then all solutions in the interval using the symmetry of the graphs.
Worked example: Solve 2cos²x − sin x − 1 = 0 for 0° ≤ x ≤ 360°.
2(1 − sin²x) − sin x − 1 = 0 → 2sin²x + sin x − 1 = 0 → (2 sin x − 1)(sin x + 1) = 0
sin x = ½: x = 30°, 150°; sin x = −1: x = 270°
Check yourself
- Convert 135° to radians.
- Find the area of a sector with r = 6 cm and θ = 0.5 rad.
- Prove that (1 − cos 2θ)/sin 2θ = tan θ.
Show answers
- 3π/4
- ½ × 36 × 0.5 = 9 cm²
- (1 − (1 − 2sin²θ))/(2 sinθ cosθ) = 2sin²θ/(2 sinθ cosθ) = tan θ
Questions about this topic (0)
No questions about this topic yet.
Sign in or create a free account to join in.