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Trigonometry II

Ratios
  • tan θ = sin θ/cos θ
  • cosec θ = 1/sin θ
  • sec θ = 1/cos θ
  • cot θ = 1/tan θ = cos θ/sin θ
Identities
  • (cos θ)2 + (sin θ)2 ≡ 1
  • 1 + (tan θ)2 ≡ (sec θ)2
  • (cot θ)2 + 1 ≡ (cosec θ)2
Double Angle Identities
  • sin 2A ≡ 2 sin A cos A
  • cos 2A ≡ (cos A)2 – (sin A)2 ≡ 2(cos A)2 – 1 ≡ 1 – 2(sin A)2
  • tan 2A = 2 tan A/1 – (tan A)2
Addition Identities
  • 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
Changing Forms
  • a sin x ± b cos x ⇔ R sin(x ± α)
  • a cos x ± b sin x ⇔ R cos(x ∓ α)
  • where: R = √(a2 + b2) and R cos α = a, R sin α = b with 0 < α < 1/π/2
Examples
  1. The diagram shows curve y = sin2 2x cosx, for 0 ≤ x ≤ π/2, and M is the maximum point. Find the x coordinate of M.
    • Use product rule to differentiate:
      • u = sin2 2x                    v = cos x
      • u´ = 4 sin 2x cos 2x                    v´ = -sin x
      • dy/dx = u´v + uv´
      • dy/dx = (4 sin 2x cos 2x)(cos x) + (sin2 2x)(-sin x)
      • dy/dx = 4 sin 2x cos 2x cos x – sin2 2x sin x
    • Use the following identities:
      • cos 2x = 2 cos2 x – 1
      • sin 2x = 2 sin x cos x
      • sin2 x = 1 – cos2 x
    • Equate dy/dx to zero:
      • dy/dx = 0
      • ∴ 4 sin 2x cos 2x cos x – sin2 2x sin x = 0
      • ∴ 4 sin 2x cos 2x cos x = sin2 2x sin x
    • Cancel sin 2x from both sides: 4 cos 2x cos x = sin 2x sin x
    • Substitute identities: 4 (2 cos2 x – 1) cos x = (2 sin x cos x) sin x
    • Cancel cos x and constant 2 from both sides: 4 cos2 x – 2 = sin2 x
    • Use identities:
      • 4 cos2 x – 2 = 1 – cos2 x
      • 5 cos2 x = 3
      • cos2 x = 3/5
      • cos x = 0.7746
      • x = cos-1 (0.7746)
      • ∴ x = 0.6847 ≈ 0.685
  1. A is a point on the circumference of a circle, center O, radius r. A circular arc, center A, meets the circumference at B and C. Angle OAB is θ radians. the area of the shaded region is equal to half the area of the circle.
    Show that: cos 2θ = 2 sin2θ – π/

    • Express the area of the sector:
      • Sector area = 1/2θr2
      • ∴ OBAC 1/2(2π – 4θ)r2 = (π – 2θ)r2
    • Express the area of sector ABC:
      • ABC = 1/2(2θ)(Length of BA)2
      • Express BA using sine rule: BA = r sin (π – 2θ)/sin θ
      • Use double angles to simplify this expression:
        • ∴ BA = r sin 2θ/sin θ
        • = 2r sin θ cos θ/sin θ
        • ∴ BA = 2r cos θ
      • Substitute BA:
        • ∴ ABC = 1/2(2θ)(2r cos θ)2
        • ∴ ABC = 4θr2 cos2 θ
    • Express the area of kite ABOC:
      • ABOC = 2 × Area of triangle
      • ABOC = 2 × 1/2r2 sin(π – 2θ)
      • ∴ ABOC = r2 sin(π – 2θ)
    • Equate the expression of the shaded region to half the area of the circle:
      •  4θr2 cos2 θ + r2(π – 2θ) – r2 sin(π – 2θ) = 1/2πr2
    • Cancel common terms:
      • 4θ cos2 θ + (π – 2θ) – sin(π – 2θ) = 1/2π
    • Use identities:
      • 4θ cos2 θ + π – 2θ – (sin π cos 2θ + sin 2θ cos π) = 1/2π
      • 4θ cos2 θ + π – 2θ – sin 2θ = 1/2π
      • 4θ (cos 2θ + 1/2) + π – sin 2θ – 2θ = 1/2π
      • 4θ cos 2θ + 4θ + 2π – 2sin 2θ – 4θ = π
      • 4θ cos 2θ + 2π – 2sin 2θ = π
    • Clean up:
      • 4θ cos 2θ = 2sin 2θ – π
      • cos 2θ = 2sin 2θ – π/
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