Grade 12 Revision – Trigonometry (Compound Angles) Prepared by: Mr. C. Hull “Compound” angles are simply the sum of, or the difference between, two other angles This angle is an example of a compound angle Easy to calculate trig ratios of angles α and β … but what about trig ratios of (α – β) or (α + β)? P (x ; 0)

Prepared by: Mr. C. Hull Essential Compound Angle Formulae – YOU MUST KNOW THESE BY HEART! * Also known as the Addition Formulae cos (α – β) = cos α cos β + sin α sin β …… think “cos cos sin sin” Note the opposite signs cos (α + β) = cos α cos β – sin α sin β sin (α – β) = sin α cos β – cos α sin β …… think “sin cos cos sin” Note the same signs sin (α + β) = sin α cos β + cos α sin β Grade 12 Revision – Trigonometry (Compound Angles)

Prepared by: Mr. C. Hull Worth knowing, easy to prove …. But NOT examinable  Grade 12 Revision – Trigonometry (Compound Angles)

Prepared by: Mr. C. Hull Double Angle Formulae (just special cases of the previous Compound Angle Formulae) sin 2α = sin (α + α) sin 2α = sin α cos α + cos α sin α sin 2α = 2 sin α cos α = 2 (sin α)(cos α) … ( note that sin α and cos α are numbers, or factors in this case) cos 2α = cos (α + α) cos 2α = cos α cos α – sin α sin α cos 2α = cos 2 α – sin 2 α cos 2α = (1 – sin 2 α) – sin 2 α OR cos 2α = cos 2 α – (1 – cos 2 α) cos 2α = 1 – 2 sin 2 α cos 2α = 2 cos 2 α – 1 Grade 12 Revision – Trigonometry (Compound Angles)

Prepared by: Mr. C. Hull Applications of Compound and Double Angle Formulae 1.Miscellaneous manipulations and calculations 2.Simplifying expressions / algebraic manipulations 3.Proving identities 4.Solving trig. equations 5.Graphs 6.Anything else of a miscellaneous nature e.g. combinations of the above! ALWAYS BEAR I N MIND THE “REVERSIBILITY” OF THE FORMULAE e.g. cos (α – β) = cos α cos β + sin α sin β Grade 12 Revision – Trigonometry (Compound Angles)

Prepared by: Mr. C. Hull EXAMPLE 1: Evaluate cos 2 15° - sin 2 15° SOLUTION: cos 2 15° - sin 2 15° = = = ✓ 1. Miscellaneous manipulations and calculations EXAMPLE 2: Evaluate sin 105° SOLUTION: sin 105° = sin 105° = …………… ( Reduce to an angle LESS THAN 90°) sin 105° = ……… ( Special angles) sin 105° = sin 105° = ✓ ✓ * rationalise the denominator if required Grade 12 Revision – Trigonometry (Compound Angles)

Prepared by: Mr. C. Hull EXAMPLE 3: If sin 21° = a, express cos 42° and sin 81° in terms of a. SOLUTION: (i) cos 42° = = = (ii) sin 81° = = =. +. = = ✓ 1. Miscellaneous manipulations and calculations (cont.) EXAMPLE 4: If sin 22° cos 12° = a and sin 12° cos 22° = b, express sin 34° in terms of a and b. SOLUTION: sin 34° = = = ✓ Grade 12 Revision – Trigonometry (Compound Angles)

Prepared by: Mr. C. Hull 2. Simplifying expressions / algebraic manipulations EXAMPLE 1: Simplify to a single trig ratio of β. SOLUTION: = = … ( 2 sinα cosα = sin 2α) = = ✓ Grade 12 Revision – Trigonometry (Compound Angles)

Prepared by: Mr. C. Hull 2. Simplifying expressions / algebraic manipulations (cont.) EXAMPLE 2: Simplify without the use of a calculator: SOLUTION: = = = ✓ Grade 12 Revision – Trigonometry (Compound Angles)

Prepared by: Mr. C. Hull EXAMPLE 1: Prove that: SOLUTION: LHS = = …… ( cos 2α = 1 – 2 sin 2 α) = = ✓ 3. Proving identities Grade 12 Revision – Trigonometry (Compound Angles)

Prepared by: Mr. C. Hull 3. Proving identities (cont.) EXAMPLE 1 (cont.): Hence show that. SOLUTION: From the previous question … x = 30°: ✓ Grade 12 Revision – Trigonometry (Compound Angles)

Prepared by: Mr. C. Hull 4. Solving trig. equations EXAMPLE 1: Find the general solution of sin 2 β + sin 2β = 1, where cos β ≠ 0 SOLUTION: sin 2 β + sin 2β = 1 ✓ Grade 12 Revision – Trigonometry (Compound Angles)

Prepared by: Mr. C. Hull 5. Graphs EXAMPLE : The diagram below shows the graphs of f(x) = cos x + 1 and g(x) = sin 2x, for 0° ≤ x ≤ 360°. Use it to find the approximate general solution to 2sin x cos x = cos x + 1. SOLUTION : Note that 2 sin x cos x = sin 2x, hence read from points A and B on x-axis … A(180°; 0) B(≈ 250°; 0) ✓ ✓ A B Grade 12 Revision – Trigonometry (Compound Angles)

Prepared by: Mr. C. Hull 5. Graphs (cont.) EXAMPLE : Draw the graph of h(x) = sin 2 x – cos 2 x for -90° ≤ x ≤ 180°. SOLUTION : First note that sin 2 x – cos 2 x = – cos 2x …. ( cos 2α = cos 2 α – sin 2 α) Thus the question becomes one of drawing the graph of y = – cos 2x as follows: (90°; 1)(-90°; 1) (180°; -1) Period = 180° Amplitude = 1 Range: - 1 ≤ y ≤ 1 Domain: -90° ≤ x ≤ 180° Grade 12 Revision – Trigonometry (Compound Angles)