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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
• 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
1. 2. 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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