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HIGHER MATHEMATICS Unit 2 – Topic 3.2 Compound Angle Formula REMINDERS y Let OP = r & POX = A This gives the following P’(y ,x) A O r A x sinA = P(x,y) y x Now reflect OP in the line y = x x sin(90 - A) = r y cos(90 - A) = r = cosA = sinA y r cosA = x r y tanA = x So OP = OP’ = r & P’OY = A P’OX = 90 - A y P(x,y) r O A -A Let OP = r & POX = A Now reflect OP in the x-axis So OP’ = r & P’OX = -A x r P’(x,-y) -y sin(-A) = = -sinA r x cos(-A) = = cosA r Let OP = r & POX = A y P’(-x,y) A P(x,y) A O x Now reflect OP in the y-axis So OP’ = r & P’OX = 180 - A -y sin(180-A) = = sinA r -x cos(180-A) = = -cosA r y P(x,y) r O A x Consider: Sin A Cos A = = = = y r = y ÷ x r r x r y r x r x yr rx y = Tan A x Sin A So Tan A = Cos A y P(x,y) So Sin2A + Cos2A r O A Consider: x = = = = y2 + 2 r y2 + x2 r2 r2 r2 x2 r2 1 So Sin2A + Cos2A = 1 SUMMARY sin(90 - A) = cosA cos(90 - A) = sinA sin(-A) = -sinA cos(-A) = cosA sin(180 - A) = sinA cos(180 - A)= -cosA sinA cosA = tanA sin2A + cos2A = 1 Exercises from MIA book: Page 153 Ex 1 All Qu. Exercises from Heinemann book: Page ? Ex ? Qu ?? Formulae for cos(A + B) and cos(A – B) Let the circle’s radius be 1 and Q be the point (x , y) Q x cosA = r x = cosA rcosA A O -B y sinA = r y = rsinA sinA So Q is (cosA , sinA) P Consider the point P(x , y) y x cos(-B) = r sin(-B) = r rsin(-B) x = cos(-B) rcos(-B) y = sin(-B) x = cosB x = -sinB So P is (cosB , -sinB) P(cosB , -sinB) & Q(cosA , sinA) Use the distance formula to find PQ PQ2 = (cosA – cosB)2 + (sinA + sinB)2 PQ2 PQ2 PQ2 PQ2 = = = = Q’ cos2A – 2cosAcosB + cos2B + sin2A + 2sinAsinB + sin2B cos2A + sin2A + sin2B + cos2B – 2cosAcosB + 2sinAsinB 1 + 1 – 2cosAcosB + 2sinAsinB 2 – 2(cosAcosB - sinAsinB) Q Now rotate ΔPOQ anti-clockwise P’ through angle B A+B A O -B P Q’ is (x , y) and P’ is (1 , 0) x cos(A+B) = r x = cos(A+B) rcos(A+B) y sin(A+B) = r y = sin(A+B) rsin(A+B) P’(1 , 0) & Q’(cos(A+B) , sin(A+B)) Use the distance formula to find P’Q’ P’Q’2 = (1 – cos(A+B))2 + (0 – sin(A+B))2 P’Q’2 = 1 – 2cos(A+B) + cos2(A+B) + sin2(A+B) P’Q’2 = 1 – 2cos(A+B) + 1 P’Q’2 = 2 – 2cos(A+B) Remember that PQ = P’Q’ so PQ2 = P’Q’2 2 – 2cos(A+B) = 2 – 2(cosAcosB - sinAsinB) Giving: Cos(A + B) = CosACosB - SinASinB We get another expansion by replacing B with -B Cos(A + (-B)) = CosACos(-B) – SinASin(-B) Giving: Cos(A - B) = CosACosB + SinASinB EXAMPLE 1 Acute angles P and Q are such that and cosQ = 3 5 sinP = 12 13 Show that cos(P - Q) = 63 65 13 12 P 5 4 5 3 Q sinP = 12 13 cosP = 5 13 sinQ = 4 5 cosQ = 3 5 Cos(P - Q) = CosPCosQ + SinPSinQ = 5 x 3 + 12 x 4 5 13 5 13 = 15 + 48 65 65 = 63 as required 65 EXAMPLE 2 Acute angles X and Y are such that and tanY = 3 4 sinX = 8 17 Show that cos(X + Y) = 36 85 8 17 15 3 X 5 4 Y sinX = 8 17 cosX = 15 17 sinY = 3 5 cosY = 4 5 Cos(X + Y) = CosXCosY - SinXSinY = 15 x 4 - 8 x 3 5 17 5 17 = 60 - 24 85 85 = 36 as required 85 EXAMPLE 3 Use the formula for cos(A + B) to simplify cos(360 + y) cos(A + B) = cosAcosB - sinAsinB cos(360 + y) = cos360ocosy - sin360osiny cos(360 + y) = (1)cosy - (0)siny cos(360 + y) = cosy EXAMPLE 4 Using the fact that 105 = 60 + 45 Show that the EXACT VALUE of cos105o is 2 - 6 4 cos(A + B) = cosAcosB - sinAsinB cos(105) = cos(60 + 45) = cos60cos45 - sin60sin45 1 1 = x - √3 x 1 2 √2 2 √2 = 1 - √3 2√2 2√2 = 1 - √3 x 2 2√2 2 (1 - √3) √2 = 2√2√2 √2 - √6 As required = 4 1 2 60º √3 45º √2 45º 30º 1 1 Trying to get 2 - 6 4 Exercises from MIA book: Page 154 Ex 2 All Qu. Exercises from Heinemann book: Page ? Ex ? Qu ?? Formulae for sin(A + B) and sin(A – B) Remember that sinX = cos(90 – X) So sin(A + B) = cos(90 – (A + B)) = cos(90 – A - B) = cos((90 – A) - B) = cos(90–A)cosB + sin(90-A)sinB = sinAcosB + cosAsinB Giving: Sin(A + B) = SinACosB + CosASinB Now what happens when we replace B with -B Sin(A + (-B)) = SinACos(-B) + CosASin(-B) Giving: Sin(A - B) = SinACosB - CosASinB EXAMPLE 5 Acute angles X and Y are such that sinX = 1 √5 and tanY = 3 Find the exact value of sin(X - Y) √5 1 X 2 3 √10 1 Y sinX = 1 √5 cosX = sinY = 3 √10 cosY = 2 √5 1 √10 sin(X - Y) = sinXcosY - cosXsinY = 1 x 1 - 2 x 3 √5 √10 √10 √5 = 1 - 6 √50 √50 = -5 = -1 = -5 5√2 √2 √50 EXAMPLE 6 For the diagram, express sin(ABC) as a fraction. C sina = 4 5 cosa = 3 5 sinb = 12 13 cosb = 5 13 sinABC = sin(a + b) sin(a + b) = sinacosb + cosasinb 12 13 D 4 A 5 3 ao bo B = 4 x 5 + 3 x 12 13 5 13 5 = 20 + 36 65 65 = 56 65 Exercises from MIA book: Page 156 Ex 3 All Qu. Exercises from Heinemann book: Page ? Ex ? Qu ?? A Few More Formulae. sin(A + B) = sinAcosB + cosAsinB Replace B with A sin(A + A) = sinAcosA + cosAsinA sin2A = 2sinAcosA So sin2A = 2sinAcosA Cos(A + B) = cosAcosB - sinAsinB cos(A + A) = cosAcosA - sinAsinA cos2A = cos2A – sin2A Remember that cos2A + sin2A = 1 So cos2A = 1 - sin2A & sin2A = 1 - cos2A Replace B with A Hence: cos2A = cos2A – sin2A = cos2A – (1 - cos2A) = 2cos2A – 1 So cos2A = 2cos2A – 1 And: cos2A = cos2A – sin2A = 1 - sin2A – sin2A = 1 – 2sin2A So cos2A = 1 – 2sin2A These also give: So cos2A = ½(1 + cos2A) So sin2A = ½(1 - cos2A) SUMMARY Cos(A + B) = CosACosB - SinASinB Cos(A - B) = CosACosB + SinASinB Sin(A + B) = SinACosB + CosASinB Sin(A - B) = SinACosB - CosASinB sin2A = 2sinAcosA cos2A = cos2A – sin2A cos2A = 2cos2A – 1 cos2A = 1 – 2sin2A cos2A = ½(1 + cos2A) sin2A = ½(1 - cos2A) EXAMPLE 7 8 17 15 15 cosA = 17 Given that 0 < A < 90 and that sinA = 8 , obtain exact values for 17 sin2A and cos2A A When finding cos2A you can choose any of the 3 formulae that you want sin2A = 2sinAcosA cos2A = 2cos2A - 1 = 2 x 8 x 15 17 17 = = 240 289 = = = 2 15 2 x - 1 2 17 2 x 225 - 1 289 450 - 289 289 289 161 289 EXAMPLE 8 1 4 √15 sinX = cosX = 1 4 √15 4 Given that 0 < A < 90 and that tanX = 1 √15 , obtain exact values for sin2X, cos2X and hence sin4X X When finding cos2A you can choose any of the 3 formulae that you want sin2X = 2sinXcosX = 2 x = 2√15 16 = √15 8 1 x √15 4 4 cos2X = 2cos2X - 1 = 2 x = 2 x = 30 16 = 14 16 152 - 1 42 15 - 1 16 - 16 16 = 7 8 sin4X = 2sin2Xcos2X = 2 x √15 x 8 = 14√15 64 = 7√15 32 7 8 Exercises from MIA book: Page 158 Ex 4A/B All Qu. Exercises from Heinemann book: Page ? Ex ? Qu ?? Solving Trig Equations EXAMPLE 9 Solve the equation 2sin2A = 1, 0<A<2 2sin2A = 1 sin2A = ½ sinA = √½ sinA = ± 1 2 A = 45, A = 45, A = π 4 S T 180 – 45, 135, 225, 3π 5π 4 4 A C √2 45 1 1 180 + 45, 360 – 45 315 7π 4 EXAMPLE 10 Solve the equation 4cos2A = 3, 0 < A < 2 4cos2A = 3 cos2A = ¾ cosA = √¾ cosA = ±3 2 A = 30, A = 30, A = π 6 S A T C 180 – 30, 150, 210, 5π 7π 6 6 1 2 60º √3 180 + 30, 360 – 30 330 11π 6 30º EXAMPLE 11 Solve the equation sin2x - cosx = 0, 0 ≤ x ≤ 2 sin2x - cosx = 0 2sinxcosx - cosx = 0 cosx(2sinx – 1) = 0 cosx = 0 2sinx – 1 = 0 S A T C sinx = ½ x = 90, 270, x = 30, 180 – 30 x = 30, 90, 150, 270 5π 3π x = π π 6 2 6 2 1 2 60º √3 30º EXAMPLE 12 Solve the equation cos2x + 7cosx + 4 = 0, 0 ≤ x ≤ 2 cos2x + 7cosx + 4 = 0 2cos2x - 1 + 7cosx + 4 = 0 2cos2x + 7cosx + 3 = 0 (2cosx + 1)(cosx + 3) = 0 cosx = -½ S A T C cosx = -3 Not possible x = 180 - 60, 180 + 60 x = 120, 240 x = 2π 4π 3 3 1 2 60º √3 30º EXAMPLE 13 Solve the equation 3cos2x + sinx - 1 = 0, 0 ≤ x ≤ 360 3cos2x + sinx - 1 = 0 3(1 - 2sin2x) + sinx - 1 = 0 -6sin2x + sinx + 2 = 0 -(6sin2x - sinx - 2) = 0 (3sinx + 2)(2sinx - 1) = 0 sinx = - 2 3 x = 180 + 41.81, 360 – 41.81 x = 30, 150, 221.81, 318.59 S T A C sinx = ½ x = 30, 180 – 30 1 2 60º √3 30º EXAMPLE 14 Solve the equation 3sin(2x + 10) = 2, 0 ≤ x ≤ 180 3sin(2x + 10) = 2 2 sin(2x + 10) = 3 2x + 10 = 41.81, 180 – 41.81 S T 2x + 10 = 41.81, 138.19, 401.81, 498.19 2x = 31.81, 128.19, 391.81, 488.19 x = 15.9, 64.1, 195.9, 244.1 x = 15.9, 64.1 A C EXAMPLE 15 Solve the equation cosxcos50 –sinxsin50 = 0.45, 0 ≤ x ≤ 180 cosxcos50 –sinxsin50 = 0.45 cos(x + 50) = 0.45 x + 50 = 63.26, 360 – 63.26 x + 50 = 63.26, 296.74 x 13.26, 246.74 = S A T C Exercises from MIA book: Page 161 Ex 5 All Qu. Exercises from Heinemann book: Page ? Ex ? Qu ??