HIGHER MATHEMATICS. Unit 2 Topic 3.2 Compound Angle Formula
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1 HIGHER MTHEMTICS Unit 2 Topic 3.2 Compound ngle Formula
2 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
3 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
4 P (-,y) O y P(,y) Let OP = r & POX = Now reflect OP in the y-ais So OP = r & P OX = sin(180-) = cos(180-) = -y r = sin - r = -cos
5 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
6 y P(,y) Consider: O r So Sin 2 + Cos 2 = y r 2 r 2 = y = r 2 r 2 r 2 = 1 So Sin 2 + Cos 2 = 1
7 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
8 Eercises from MI book: Page 153 E 1 ll Qu. Eercises from Heinemann book: Page? E? Qu??
9 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)
10 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 = coscosB + 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)
11 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
12 EXMPLE cute angles P and Q are such that sinp = and cosq = 3 5 Show that cos(p - Q) = sinp = cosp = P Q sinq = 4 5 cosq = 3 5 Cos(P - Q) = CosPCosQ + SinPSinQ = = = as required
13 EXMPLE cute angles X and Y are such that sinx = 8 17 and tany = 3 4 Show that cos(x + Y) = sinx = 8 17 cosx = X Y siny = 3 5 cosy = 4 5 Cos(X + Y) = CosXCosY - SinXSinY = = = as required
14 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 = Show that the EXCT VLUE of cos105 o is 2-6 4
15 cos( + B) = coscosb - sinsinb cos(105) = cos( ) = cos60cos45 - sin60sin = = = = = (1-3) s required 1 60º º 1 30º 45º 1 Trying to get 2-6 4
16 Eercises from MI book: Page 154 E 2 ll Qu. Eercises from Heinemann book: Page? E? Qu??
17 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
18 EXMPLE 5 cute angles X and Y are such that sinx = 1 5 and tany = 3 Find the eact value of sin(x - Y) X 2 10 Y 1 sinx = 1 5 cosx = 2 5 siny = 3 10 cosy = 1 10 sin(x - Y) = sinxcosy - cosxsiny = = = -5 = -5 =
19 EXMPLE 6 For the diagram, epress sin(bc) as a fraction. C sina = 4 5 sinb = cosa = 3 5 cosb = 5 13 sinbc = sin(a + b) D a obo 13 B sin(a + b) = sinacosb + cosasinb = = =
20 Eercises from MI book: Page 156 E 3 ll Qu. Eercises from Heinemann book: Page? E? Qu??
21 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
22 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)
23 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)
24 EXMPLE 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 = sin2 = 2sincos = = cos2 = 2cos 2-1 = = = =
25 EXMPLE 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 = = = cos2x = 2cos 2 X - 1 = = = = 14 =
26 sin4x = 2sin2Xcos2X = = =
27 Eercises from MI book: Page 158 E 4/B ll Qu. Eercises from Heinemann book: Page? E? Qu??
28 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 , , = 45, 135, 225, 315 = π 3π 5π 7π
29 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 , , = 30, 150, 210, 330 = π 5π 7π 11π º º
30 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 º sin = ½ = 90, 270, = 30, = 30, 90, 150, 270 = π π 5π 3π
31 EXMPLE 12 Solve the equation cos2 + 7cos + 4 = 0, 0 2 cos2 + 7cos + 4 = 0 2cos cos + 4 = 0 2cos 2 + 7cos + 3 = 0 (2cos + 1)(cos + 3) = 0 cos = -½ cos = -3 Not possible S T C = , = 120, 240 = 2π 4π º º
32 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 = = , Solve the equation 3cos2 + sin - 1 = 0, S T C sin = ½ = 30, = 30, 150, , º º
33 EXMPLE 14 3sin(2 + 10) = 2 2 sin(2 + 10) = = 41.81, Solve the equation 3sin(2 + 10) = 2, = 41.81, , , = 31.81, , , = 15.9, 64.1, 195.9, = 15.9, 64.1 S T C
34 EXMPLE 15 Solve the equation coscos50 sinsin50 = 0.45, coscos50 sinsin50 = 0.45 cos( + 50) = = 63.26, = 63.26, = 13.26, S T C
35 Eercises from MI book: Page 161 E 5 ll Qu. Eercises from Heinemann book: Page? E? Qu??
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