1. Higher Maths Revision Notes The Auxiliary Angle Get Started

2. The Wave Equation find corresponding values of θ Express a c os θ + b s in θ in the form r c os(θ ± α) or r s in(θ ± α) solve equations of the form a c os θ + b s in θ = c f in d maximum and minimum values of expressions of the form a cos θ + b sin θ and the corresponding values of θ

3. r c os(θ ± α) = r cos θ cos α r sinθ sin α If a c os θ + b s in θ = r cos θ cos α r sinθ sin α for all values of θ, then (1)the coefficients of cos θ must be equal: a = r cos α (2)the coefficients of sin θ must be equal: b = r sin α To express a c os θ + b s in θ in the form r c os(θ ± α) Expand the required form Example

4. Square both equations and add them: a 2 + b 2 = r 2 cos 2 α + r 2 sin 2 α =r 2 (cos 2 α + sin 2 α) = r 2 So r = √(a 2 + b 2 ) Example

5. Divide one equation by the other to get Use the equations to decide in which quadrant α lies. Example

6. Express 3 c os θ + 4 s in θ in the form r c os(θ + α) r c os(θ + α) = r cos θ cos α – r sinθ sin α Equate coefficients: (1)3 = r cos α (2)–4 = r sin α Expand the required form Nextsee graph

7. Square both equations and add them: 3 2 + (–4) 2 = r 2 cos 2 α + r 2 sin 2 α =r 2 (cos 2 α + sin 2 α) = r 2 So r = √(3 2 + (–4) 2 ) = √25 = 5 Next

8. Test Yourself? Divide one equation by the other to get Use the equations to decide in which quadrant α lies. So α is in 4 th quadrant. Thus α = 5·356 radians Giving: 3 c os θ + 4 s in = 5 c os(θ + 5·356)

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10. Two swings are set in motion. Their heights above an arbitrary line can be modelled by: Swing 1:h 1 = 5 cos x Swing 2:h 2 = 12 sin x Express the difference in their heights in the form r sin (x – a) reveal

11. h 1 – h 2 = 5 cos x – 12 sin x r sin (x – a) = r sin x cos a – r cos x sin a Equating coefficients: r cos a = –12; –r sin a = 5 Square and add: r 2 (cos 2 a + sin 2 a) = (–12) 2 + (–5) 2 = 169 so r = 13 Divide: tan a = (–5) ÷ (–12) = 0·41666…, So a = 0·395, 3·536, 6·678, … Identify quadrant: sin a = – 5 / 13 … 3 rd or 4 th quadrant. cosa = – 12 / 13 … 2 nd or 3 rd quadrant. Choose the 3 rd quadrant angle: a = 3·536 h 1 – h 2 = 13 sin (x – 3·536)

12. To solve equations of the form a c os θ + b s in θ = c Express the LHS in desired form e.g. r cos (x – a) r cos (x – a) = c: solve for x – a: x – a = cos –1 ( c / r ) or 2π – cos –1 ( c / r ) or 2π + cos –1 ( c / r ) etc Solve for x: x = cos –1 ( c / r ) + a or 2π – cos –1 ( c / r ) + a or 2π + cos –1 ( c / r ) + a etc Test Yourself?

13. When p cos x + q sin x is expressed as r cos(x + a) Max: p cos x + q sin x = r when x + a = 0 Min: p cos x + q sin x = – r when x + a = π When p cos x + q sin x is expressed as r sin(x + a) Max: p cos x + q sin x = r when x + a = π / 2 Min: p cos x + q sin x = – r when x + a = 3π / 2 Note that the graph is unchanged Test Yourself?

14. For what values of x, 0 ≤ x ≤ 2π, is sin x + cos x = 0·5 reveal

15. sin x + cos x = r sin(x + a) = r sin x cos a + r cosx sin a Equating coefficients: r cos a = 1 and r sin a = 1 This gives: r = √2 and a = π / 4 sin x + cos x = 0·5 so √2 sin(x + π / 4 ) = 0·5 x + π / 4 = sin –1 ( 0·353553…), or π – sin –1 ( 0·353553…), or 2π + sin –1 ( 0·353553…), etc x + π / 4 = 0·361, 2·78, 6·64,… x = –0·424, 1·995, 5·859,… In the required domain, x = 1·995, 5·859

16. As the paddle turns, the height of a point on it can be modelled by h = sin x + 2 cos x + 1 where h units is the height above the water. Find the minimum height and the value of x at which it occurs. reveal

17. The minimum of sin (x + 1·107) = –1 when (x + 1·107) = 3π / 2 So the minimum height is 1 – √5 = –1·24 units when x = 3·605 First express sin x + 2 cos x in the form r sin(x + a) sin x + 2 cos x = r sin(x + a) = r sin x cos a + r cosx sin a Equating coefficients: r cos a = 1 and r sin a = 2 This gives: r = √5 and a = 1·107 So h = √5 sin(x + 1·107) + 1.