Key Concepts
- 1sin²θ + cos²θ = ?
- 21 + tan²θ = ?
- 31 + cot²θ = ?
- 4sin values: 0°, 30°, 45°, 60°, 90°
- 5cos values: 0°, 30°, 45°, 60°, 90°
- 6tan values: 0°, 30°, 45°, 60°, 90°
- 7Complementary angles
- 8sinθ = 1/cosecθ
- 9tanθ = ?
- 10cotθ = ?
Important Formulas & Facts
1 (Fundamental Pythagorean Identity)
sec²θ
cosec²θ
0, 1/2, 1/√2, √3/2, 1
1, √3/2, 1/√2, 1/2, 0
0, 1/√3, 1, √3, undefined
sin(90°-θ) = cosθ, cos(90°-θ) = sinθ, tan(90°-θ) = cotθ
Reciprocal identities: sinθ·cosecθ = 1, cosθ·secθ = 1, tanθ·cotθ = 1
sinθ/cosθ
cosθ/sinθ = 1/tanθ
Must-Know Questions
Q1If tanθ = 4/3, then find the value of (4sinθ − 3cosθ)/(3sinθ + 2cosθ).
tanθ = 4/3, so sinθ = 4/5, cosθ = 3/5 (using Pythagorean triplet 3-4-5). Numerator: 4(4/5) − 3(3/5) = 16/5 − 9/5 = 7/5. Denominator: 3(4/5) + 2(3/5) = 12/5 + 6/5 = 18/5. Value = (7/5)/(18/5) = 7/18.
Q2If sin2A = (√3/2)tan²45° where A is acute, find A.
tan²45° = 1. sin2A = √3/2. 2A = 60° → A = 30°.
Q3(sec²θ − 1)(cosec²θ − 1) is equal to:
sec²θ − 1 = tan²θ. cosec²θ − 1 = cot²θ. tan²θ × cot²θ = (sinθ/cosθ)² × (cosθ/sinθ)² = 1.
Q4If cotθ = 7/8, find sinθ and cosθ.
cotθ = 7/8 → adj/opp = 7/8. Hypotenuse = √(49 + 64) = √113. sinθ = 8/√113, cosθ = 7/√113.
Q5Prove that 2(sin⁶θ + cos⁶θ) − 3(sin⁴θ + cos⁴θ) = −1.
Let s = sin²θ, c = cos²θ. s + c = 1. sin⁶θ + cos⁶θ = s³ + c³ = (s + c)(s² − sc + c²) = s² − sc + c² = (s + c)² − 3sc = 1 − 3sc. sin⁴θ + cos⁴θ = s² + c² = (s + c)² − 2sc = 1 − 2sc. LHS = 2(1 − 3sc) − 3(1 − 2sc) = 2 − 6sc − 3 + 6sc = −1 = RHS.
Practice Introduction to Trigonometry
Reinforce what you just revised with practice questions