HIGHER MTHEMTICS Unit 2 Topic 3.2 Compound ngle Formula
REMINDERS y P (y,) Let OP = r & POX = This gives the following sin = y r cos = r P(,y) r y O Now reflect OP in the line y = sin(90 - ) = r = cos tan = y So OP = OP = r & P OY = P OX = 90 - cos(90 - ) = y r = sin
y P(,y) Let OP = r & POX = Now reflect OP in the -ais O r - r P (,-y) So OP = r & P OX = - sin(-) = cos(-) = -y r = -sin r = cos
P (-,y) O y P(,y) Let OP = r & POX = Now reflect OP in the y-ais So OP = r & P OX = 180 - sin(180-) = cos(180-) = -y r = sin - r = -cos
y O r P(,y) Consider: Sin Cos = y r r y = r = yr r y = r r r = y = Tan So Tan = Sin Cos
y P(,y) Consider: O r So Sin 2 + Cos 2 = y 2 2 + r 2 r 2 = y 2 + 2 = r 2 r 2 r 2 = 1 So Sin 2 + Cos 2 = 1
SUMMRY sin(90 - ) = cos sin(-) = -sin sin(180 - ) = sin cos(90 - ) = sin cos(-) = cos cos(180 - )= -cos sin cos = tan sin 2 + cos 2 = 1
Eercises from MI book: Page 153 E 1 ll Qu. Eercises from Heinemann book: Page? E? Qu??
Formulae for cos( + B) and cos( B) O -B Q Let the circle s radius be 1 and Q be the point (, y) cos = r = rcos cos sin = y r y = rsin So Q is (cos, sin) P Consider the point P(, y) cos(-b) = r = rcos(-b) = cosb sin(-b) = y r y = rsin(-b) = -sinb So P is (cosb, -sinb)
P(cosB, -sinb) & Q(cos, sin) Use the distance formula to find PQ PQ 2 = (cos cosb) 2 + (sin + sinb) 2 PQ 2 = cos 2 2coscosB + cos 2 B + sin 2 + 2sinsinB + sin 2 B PQ 2 = cos 2 + sin 2 + sin 2 B + cos 2 B 2coscosB + 2sinsinB PQ 2 = 1 + 1 2coscosB + 2sinsinB PQ 2 = 2 2(coscosB - sinsinb) Q Q +B O -B P P Now rotate ΔPOQ anti-clockwise through angle B Q is (, y) and P is (1, 0) cos(+b) = r = rcos(+b) sin(+b) = y r y = rsin(+b)
P (1, 0) & Q (cos(+b), sin(+b)) Use the distance formula to find P Q P Q 2 = (1 cos(+b)) 2 + (0 sin(+b)) 2 P Q 2 = 1 2cos(+B) + cos 2 (+B) + sin 2 (+B) P Q 2 = 1 2cos(+B) + 1 P Q 2 = 2 2cos(+B) Remember that PQ = P Q so PQ 2 = P Q 2 2 2cos(+B) = 2 2(coscosB - sinsinb) Giving: Cos( + B) = CosCosB - SinSinB We get another epansion by replacing B with -B Cos( + (-B)) = CosCos(-B) SinSin(-B) Giving: Cos( - B) = CosCosB + SinSinB
EXMPLE 1 12 13 cute angles P and Q are such that sinp = 12 13 and cosq = 3 5 Show that cos(p - Q) = 63 65 sinp = 12 13 cosp = 5 13 4 5 3 5 P Q sinq = 4 5 cosq = 3 5 Cos(P - Q) = CosPCosQ + SinPSinQ = 5 3 + 12 4 13 5 13 5 = 15 + 48 65 65 = 63 65 as required
EXMPLE 2 8 17 cute angles X and Y are such that sinx = 8 17 and tany = 3 4 Show that cos(x + Y) = 36 85 sinx = 8 17 cosx = 15 17 3 15 4 5 X Y siny = 3 5 cosy = 4 5 Cos(X + Y) = CosXCosY - SinXSinY = 15 4-8 3 17 5 17 5 = 60-24 85 85 = 36 85 as required
EXMPLE 3 Use the formula for cos( + B) to simplify cos(360 + y) cos( + B) = coscosb - sinsinb cos(360 + y) = cos360 o cosy - sin360 o siny cos(360 + y) = (1)cosy - (0)siny cos(360 + y) = cosy EXMPLE 4 Using the fact that 105 = 60 + 45 Show that the EXCT VLUE of cos105 o is 2-6 4
cos( + B) = coscosb - sinsinb cos(105) = cos(60 + 45) = cos60cos45 - sin60sin45 1 1 = - 3 1 2 2 2 2 = 1-2 2 = 1-3 2 2 = = 3 2 2 2 2 (1-3) 2 2 2 2 2-6 4 s required 1 60º 2 2 3 45º 1 30º 45º 1 Trying to get 2-6 4
Eercises from MI book: Page 154 E 2 ll Qu. Eercises from Heinemann book: Page? E? Qu??
Formulae for sin( + B) and sin( B) Remember that sinx = cos(90 X) So sin( + B) = cos(90 ( + B)) = cos(90 - B) = cos((90 ) - B) = cos(90 )cosb + sin(90-)sinb = sincosb + cossinb Giving: Sin( + B) = SinCosB + CosSinB Now what happens when we replace B with -B Sin( + (-B)) = SinCos(-B) + CosSin(-B) Giving: Sin( - B) = SinCosB - CosSinB
EXMPLE 5 cute angles X and Y are such that sinx = 1 5 and tany = 3 Find the eact value of sin(x - Y) 1 3 5 X 2 10 Y 1 sinx = 1 5 cosx = 2 5 siny = 3 10 cosy = 1 10 sin(x - Y) = sinxcosy - cosxsiny = 1 1-2 3 5 10 5 10 = 1-6 50 50 = -5 = -5 = -1 50 5 2 2
EXMPLE 6 For the diagram, epress sin(bc) as a fraction. C sina = 4 5 sinb = 12 13 cosa = 3 5 cosb = 5 13 sinbc = sin(a + b) D 4 12 3 5 a obo 13 B sin(a + b) = sinacosb + cosasinb = 4 5 + 5 13 = 20 + 65 = 56 65 36 65 3 5 12 13
Eercises from MI book: Page 156 E 3 ll Qu. Eercises from Heinemann book: Page? E? Qu??
Few More Formulae. sin( + B) = sincosb + cossinb sin( + ) = sincos + cossin sin2 = 2sincos Replace B with So sin2 = 2sincos Cos( + B) = coscosb - sinsinb cos( + ) = coscos - sinsin cos2 = cos 2 sin 2 Replace B with Remember that cos 2 + sin 2 = 1 So & cos 2 = 1 - sin 2 sin 2 = 1 - cos 2
Hence: cos2 = cos 2 sin 2 = cos 2 (1 - cos 2 ) = 2cos 2 1 nd: So cos2 = 2cos 2 1 cos2 = cos 2 sin 2 = 1 - sin 2 sin 2 = 1 2sin 2 So cos2 = 1 2sin 2 These also give: So cos 2 = ½(1 + cos2) So sin 2 = ½(1 - cos2)
SUMMRY Cos( + B) = CosCosB - SinSinB Cos( - B) = CosCosB + SinSinB Sin( + B) = SinCosB + CosSinB Sin( - B) = SinCosB - CosSinB sin2 = 2sincos cos2 = cos 2 sin 2 cos2 = 2cos 2 1 cos2 = 1 2sin 2 cos 2 = ½(1 + cos2) sin 2 = ½(1 - cos2)
EXMPLE 7 17 8 15 Given that 0 < < 90 and that sin = sin2 and cos2 8 17, obtain eact values for When finding cos2 you can choose any of the 3 formulae that you want cos = 15 17 sin2 = 2sincos = 2 8 17 = 240 289 15 17 cos2 = 2cos 2-1 = 2 15 2-1 17 2 = 2 225-1 289 = 450-289 = 161 289 289 289
EXMPLE 8 4 1 15 Given that 0 < < 90 and that tanx =, obtain eact values for sin2x, cos2x and hence sin4x X 1 15 When finding cos2 you can choose any of the 3 formulae that you want sinx = 1 4 cosx = 15 4 sin2x = 2sinXcosX = 2 1 4 = 2 15 16 = 15 8 15 4 cos2x = 2cos 2 X - 1 = 2 15 2-1 4 2 = 2 15-1 16 = 30-16 16 16 = 14 = 7 16 8
sin4x = 2sin2Xcos2X = 2 15 7 8 8 = 14 15 64 = 7 15 32
Eercises from MI book: Page 158 E 4/B ll Qu. Eercises from Heinemann book: Page? E? Qu??
Solving Trig Equations EXMPLE 9 2sin 2 = 1 sin 2 = ½ sin = ½ sin = ± 1 2 = 45, Solve the equation 2sin 2 = 1, 0<<2 S T C 2 45 180 45, 180 + 45, 360 45 = 45, 135, 225, 315 = π 3π 5π 7π 4 4 4 4 1 1
EXMPLE 10 Solve the equation 4cos 2 = 3, 0 < < 2 4cos 2 = 3 cos 2 = ¾ cos = ¾ cos = ±3 2 = 30, S T C 180 30, 180 + 30, 360 30 = 30, 150, 210, 330 = π 5π 7π 11π 6 6 6 6 1 60º 2 3 30º
EXMPLE 11 Solve the equation sin2 - cos = 0, 0 2 sin2 - cos = 0 2sincos - cos = 0 cos(2sin 1) = 0 cos = 0 2sin 1 = 0 S T C 1 60º 2 3 30º sin = ½ = 90, 270, = 30, 180 30 = 30, 90, 150, 270 = π π 5π 3π 6 2 6 2
EXMPLE 12 Solve the equation cos2 + 7cos + 4 = 0, 0 2 cos2 + 7cos + 4 = 0 2cos 2-1 + 7cos + 4 = 0 2cos 2 + 7cos + 3 = 0 (2cos + 1)(cos + 3) = 0 cos = -½ cos = -3 Not possible S T C = 180-60, = 120, 240 = 2π 4π 3 3 180 + 60 1 60º 2 3 30º
EXMPLE 13 3cos2 + sin - 1 = 0 3(1-2sin 2 ) + sin - 1 = 0-6sin 2 + sin + 2 = 0 -(6sin 2 - sin - 2) = 0 (3sin + 2)(2sin - 1) = 0 sin = - 2 3 = 180 + 41.81, 360 41.81 Solve the equation 3cos2 + sin - 1 = 0, 0 360 S T C sin = ½ = 30, 180 30 = 30, 150, 221.81, 318.59 1 60º 2 3 30º
EXMPLE 14 3sin(2 + 10) = 2 2 sin(2 + 10) = 3 2 + 10 = 41.81, Solve the equation 3sin(2 + 10) = 2, 0 180 180 41.81 2 + 10 = 41.81, 138.19, 401.81, 498.19 2 = 31.81, 128.19, 391.81, 488.19 = 15.9, 64.1, 195.9, 244.1 = 15.9, 64.1 S T C
EXMPLE 15 Solve the equation coscos50 sinsin50 = 0.45, 0 180 coscos50 sinsin50 = 0.45 cos( + 50) = 0.45 + 50 = 63.26, 360 63.26 + 50 = 63.26, 296.74 = 13.26, 246.74 S T C
Eercises from MI book: Page 161 E 5 ll Qu. Eercises from Heinemann book: Page? E? Qu??