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. 
2. 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

2. 
3. 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}/_4θ
   • 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}/_4θ