Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
2008 SLSS LEARNING OUTCOMES Outline and emphasise best practice in teaching trigonometry Teaching for understanding as basic principle in teaching trigonometry Using ICT in the classroom to enhance understanding Using the calculator as a teaching resource Provide a summary of trigonometry which can be used in the classroom for revision of topic Assessment: Current strengths and weaknesses in trigonometry Why Trig ? Connections between Trigonometry and the other parts of the course. Coordinate Geometry of the Circle. Q1. Paper 2 Parametric form of a circle x 4 3cos , y 2 3sin Vectors. Q2. Paper 2 Dot product a . b | a | | b | cos ( where is the abgle between a and b ) Coordinate Geometry of the Line. Q3. Paper 2 Slope of a line: Angle between two lines. Slope = tan ratio : m1 m2 tan = 1 m1m2 Complex Numbers. Q3. Paper 1 Polar Form of a complex number x yi r cos i sin . Differentiation. Q6 and Q7. Paper 1 The differential of trigonometrical functions Integration. Q8. Paper 1 Integral of trigonometrical functions: Using trigonometrical identities. 2cos4 cos2 d = (cos6 + cos2 ) d Further Calculus Option. Q8. Paper 1 Maclaurin Series for sin, cos and tan-1 : Max and Min Modelling: Integration by parts. Syllabus Revise Meaning of Main Facts and Theorems from Junior Cert. Geometry LEAVING CERT TRIGONOMETRY SUMMARY OF THE SECTIONS IN L.C. TRIG. The Unit Circle. Simple identities with proofs. Radian Measure ( Length of arc and Area of sector) Angles on page 9 of tables. Solving Right Angled Triangles TRIG Sin Formula Cos Formula Solving Non-Right Angled Triangles Area of triangle More Identities from page 9 of tables Solving Trig equations. Limits of Trig functions. Proofs required. Unit Circle: Page 9 UNIT CIRCLE (x,y) 1 A x y y SinA x CosA NOW THE TRIANGLE IS 1 sinA A cosA 1 A CosA SinA (cosA,sinA) 1 SinA (CosA,SinA) (x,y) A CosA x CosA y SinA SinA 1 (0,0) A CosA Using Pythagoras Sin2 A Cos 2 A 1 opp Sin A Tan A adj Cos A Tan A Sin A Cos A These are our first two identities on page 9 in tables. Sin2 A Cos 2 A 1 Sin2 A 1 Cos 2 A Cos 2 A 1 Sin2 A Sin A Tan A Cos A Sin A CosA Tan A SinA TanA .CosA 2 Sin A Tan 2 A Cos 2 A 1 Sec A 1 Tan A Cos 2 A 2 2 1 Ex .1 Show that 1 Tan A Cos 2 A 2 LHS = 1 tan 2 A 2 Sin A = 1 2 Cos A Cos 2 A Sin 2 A = Cos 2 A 1 RHS 2 Cos A Use the CD on this Ex .2 Show that Cos 4 A Sin4 A Cos 2 A Sin2 A LHS = Cos 4 A Sin4 A = (Cos2 A Sin2 A)(Cos2 A Sin2 A) = (Cos2 A Sin2 A)(1) = Cos A Sin A = RHS 2 2 BASIC GUIDES IN PROVING IDENTITIES. 1. Start with one side (usually the more complicated). 2. Convert unfamiliar terms to sin and cos, if possible. 3. Keep an eye on the target. 4. At each step do what seems sensible. 5. Try another approach if things become too complicated. Note: Emphasise algebraic procedures. Ex .3 Show that Sec - Tan .Sin Cos LHS = Sec -Tan .Sin 1 Sin .Sin Cos Cos 1 Sin2 Cos Cos 2 Cos Cos = RHS Radian Measure o Radius = r a Radius = r o 2 rads 1 rad b a b Length of arc = r Measure of the angle in radians = Length of arc ab =2r Length of arc Length of radius Semi-Circle Radius = r Length of semi-circle = r = (r) = r units Measure of the angle in radians = Length of Arc(semi-circle) Radius r radians r 180 = radians ANGLES ON PAGE 9 IN TABLES NOTE : radians 180 0 180 90 60 45 30 Link to Java Script. ANGLES > 900 90o (-270o ) SIN + ALL + 180o (-180o ) 180o - 180o + 360o - 0o 360o TAN + COS + 270o (-90o ) Link to Java Script UNIT CIRCLE (0,1) (1,0) (-1,0) (0,-1) Find the Sin 120 Step 1. Draw a circle with the four quadrants Step 2. Mark in the 120 line Step 3. Take 120 from the BASE LINE angle 180 60 Sin120 Sin60 As it is in the 2 nd 3 2 quadrant it is a answer. 120 90o 0o 360o 180o 270o Other examples : (i) Sin 123 (ii) Cos 300 (iii) Tan 290 Answer s (i) 0.8387 (ii) 0.5 (iii) - 2.7475 Use your calculator to check answers making sure you understand why you get a positive or negative answer. 1 Ex .1 If 0 A 360 , find the values of A for which Sin A 2 S wT A Cw Forget about the sign at first. 1 If Sin A 2 A 30 As Sin is Negative the angle can only be in the 3rd or 4 th quadrant. Add 30 to 180 and subtract 30 from 360 So A {210 and 330 } 3 Ex .2 If Cos A , 0 A 360 , find two values of A. 2 wS A wT C Forget about the sign at first. If Cos A 3 2 A 30 As Cos is Negative the angle can only be in the 2 nd or 3rd quadrant. So A {150 and 210 } SOLVING RIGHT ANGLED TRIANGLES a Ex .1 Find ab in the following triangle 3 c ab 3 2 4 2 2 ab 9 16 25 2 b 4 ab 25 5 Some common triplets used 5 3 4 3 ,4 , 5 10 6 13 5 8 12 6, 8, 10 5, 12, 13 USE OF SIN , COS, TAN FORMULAE. A ADJACENT 900 A 900 ADJACENT Opp Sin Hyp Adj Cos Hyp Opp Tan Adj 8 Ex .1 Find the length of the side x in the diagram. 30 x x cos 30 8 3 x 2 8 3 x 8 4 3 2 Ex .2 Find the size of the angle x in the diagram. 5 tan x 4 tan x 1.25 x tan1 1.25 x 4 5 5 Ex.3 If Cos , find Sin and Tan2 , 0 90 13 Note: If given ratio always draw right angled triangle x Adj = 5 Adj 5 Cos Hyp 13 By Pythagoras 13 2 x 2 5 2 x 12 (Note triplet) Opp 12 Sin Hyp 13 2 144 Opp 12 2 2 Tan (Tan ) 25 Adj 5 2 Ex3. If SinA 1 , for 0 A 90 Find (i) 2 SinA 2 (i ) 1 2SinA 2( ) 2 Multiply above and below by 2 2 2 . 2 2 ( ii ) SinA (ii) Sin2A 2 2 2 2 2 1 2 From the handy angles (page 9) A 45 Sin 2 A Sin90 0 2A 90 2 SOLVING NON RIGHT ANGLED TRIANGLES We use three things here. 1. Area of triangle. 2. Sine Rule 1 a .b .SinC 2 a b c SinA SinB SinC 3. Cosine Rule a 2 b 2 c 2 2b.c.CosA B a c A A B C 180 C b 1. Area of triangle. 1 a .b .SinC 2 It is important to emphasise that in order to use the area of triangle formula that you need two sides and the included angle. Example 1. Find the area of the triangle , correct to one decimal place 8cm 37o 10cm 1 Area 8.10.Sin 37 2 24.1cm 2 Example 2. The area of the triangle shown is 16.2 cm 3 Find the measure of the angle A. 5 A 8 1 Area of ( one side)(other side)(Sin of Included Angle) 16.2 2 1 Area of ( 8 )( 5 )SinA 16.2 2 20 SinA 16.2 16.2 SinA 20 SinA 0.81 A 54 06' USE OF SIN AND COS RULE. MOST IMPORTANT RULE. You use the COS RULE if given (a) Three sides (b) Two sides and the included angle. You use the SIN RULE in all other cases. Ex .1 Find the size of the largest angle in the triangle 7 9 8 Given three sides so you have to use the Cos Rule The largest side is always across from the largest angle. You always start with the side across from the angle you are dealing with. 7 X0 8 9 9 2 7 2 8 2 2( 7 )( 8 )Cos x 81 47 64 112Cos x 81 111 112Cos x 81 111 112Cos x 30 112Cos x Cos x .2679 x 7428' 30 112 x Cos 1 .2679 Cos x Ex .2 Find the length of the side x in the following diagram. 9 x 320 10 Given two sides and the included angle Cos Rule. You always start with the side across from the angle you are dealing with. x 2 9 2 10 2 2(9)(10)Cos 32 x 2 81 100 180(.8480 ) x 2 181 152.64 x 2 28.36 x 28.36 x 5.3254 a Ex 3. Given triangle abc find bc . 120o 5cm. 30o b bc 5 Sin120 Sin 30 bc 5 Sin120 Sin30 Using page 9 bc 3 2 5 1 2 5 3 2 21 5 3 2 . 21 5 3 c a Ex 4. Given triangle abc find abc . 120o b 8 5 Sin120 Sinabc 8 5 3 Sinabc 2 5 3 Sinabc 16 Sinabc .54127 abc Sin 1.54127 32.77 8cm 5cm. c MORE DIFFICULT PROBLEMS ON SIN AND COS RULES. Ex .1 A and B are the goalposts at one end of a football pitch and F is the corner flag. (as shown in the diagram), A, Band F lie in a straight line with AB 7.4m and BF 28m. A player is at position P, such that PBF 85 and PFB 25 A 7.4 m B 28 m F P Calculate (i) PF (ii) PA (iii) APB 7.4 m A B 950 28 m 850 F 250 700 P This is the Master Diagram. Fill in all the angles you know. Now split up the diagram into the individual triangles. A 7.4 m B B 28 m 850 950 F A 35.4 m 250 250 700 P P F P B 28 m 850 F 250 700 P To find PF use the Sin rule PF 28 Sin85 Sin70 28 Sin85 PF Sin70 PF 28(.9962 ) .9397 PF 29.68 m 35.4 m A F 250 29.69 m P To find AP use the Cos rule. AP 35.4 2 29.69 2 2( 35.4 )( 29.69 )Cos 25 2 AP 1253.16 881.5 2102.1Cos 25 2 AP 2134.5 1905.15 2 AP 229.35 15.14 A 7.4 m B 950 15.14 m P To find APB use the Sin rule. 15.14 7.4 Sin95 SinAPB 7.4 Sin95 SinAPB 15.14 SinAPB .4896 APB Sin1 .4896 29.31 Ex .2. In a triangle abc, ab 7cm , bc 8 cm and the area of the triangle abc is 14cm 2 . Find two values of the angle abc. Make a sketch of the resulting triangles c 8 cm a 7 cm b Area of abc 21 ( 7 )( 8 )Sinabc 14 cm 2 28 Sinabc 14 Sinabc A wS w T C 14 28 Sinabc The angle is in the first and second quadrent. abc 150 abc 30 1 2 Sketch of the two triangles c 8 cm 300 a a 7 cm c b 1500 7 cm b 8 cm THREE D PROBLEMS Ex .1 pqr are points on level ground. rw is a TV mast which is held in place by two cables pw and qw. If rw is 5m and the wpr wpr is 45 and wqr 120 , show that pq 5 3 w 5m r 1200 450 p 450 q Isolate the various triangles NOTE the right angles. w 5m w 5m r 5m r 450 1200 5m r 450 p Using the Cos rule we q p 5 Tan45 pr 5 1 pr 5 Tan45 qr 5 1 qr pr 5 m qr 5 m q can now find pq pq 5 2 5 2 2( 5 )( 5 )Cos120 2 pq 25 25 50( .5 ) 2 pq 75 2 pq 75 pq 5 3 2007 Paper 2, Q5 (c). s s h h r r 60 p s c r 60 30 q h r 30 p p q c q s s h 3 p h r h 3 q c h To find |prq|, use the cosine formula r r 2 60 30 p q tan 60 h pr h pr h | pr | 3 3 tan 30 h rq 1 h 3 rq 2 h h c h 3 2 h 3 cos prq 3 3 h2 2 c 3h2 2h2 cos prq 3 h2 2 c 3h2 2h2 cos prq 3 3c 2 h2 9h2 6h2 cos prq 2 But given 3c2 13h2 13h2 h2 9h2 6h2 cos prq 3h2 6h2 cos prq 1 cos prq 2 prq 120 LENGTH OF ARC OF CIRCLE Length of arc r ( is in radians) Ex .1 Find the length of the arc ab in the diagram below. o 0 7 30 b a METHOD 1 Length of arc r ( is in radians) Length of arc 7 6 3.142 Length of arc 7 6 Length of arc 3.665 cm METHOD 2 Length of circle 2R We have 30 parts of a full circle 30 Length of arc of a circle 360 1 Length of arc of 2R 12 1 Length of arc 2 7 3.665 12 AREA OF SECTOR OF CIRCLE 1 Area of sector r 2 ( is in radians) 2 Ex .1 Find the area of the sector oab in the diagram below. o 0 7 45 METHOD 1 1 2 Area r ( is in radians) 2 1 2 Area (7) 2 4 Area 19.24 sq .units b a METHOD 2 Area of circle R 2 We have 45 parts of a full circle 45 Area of sector of a circle 360 1 2 Area of sector R 8 1 Area of sector ( 7 )2 19.24 8 Ex .2 2000 Q.4 (a) Paper 2. The area of a sector of a circle is 27cm 2 . The radius is 6cm. Find in radians the measure of the angle in the sector. 1 2 Area r ( is in radians) 27 2 1 Area ( 6 )2 27 2 18 27 6cm 27 3 radians 18 2 NOTE You can convert radians to degrees as follows. radians 180 1 radian 180 3 180 3 radians . 85.94 2 2 COMPOUND ANGLE FORMULAE Sin( A B ) SinACosB CosASinB Sin( A B ) SinACosB CosASinB NOTE Sign changes Cos( A B ) CosACosB SinASinB Cos( A B ) CosACosB SinASinB NOTE Sign changes TanA TanB Tan( A B ) 1 TanATanB TanA TanB Tan( A B ) 1 TanATanB NOTE Sign changes COFUNCTION FORMULAE Sin(90 A) CosA Ex 1. Ex 2. 1 Sin(90 45) Cos 45 2 3 Sin(90 30) Cos60 2 Cos(90 A) SinA Ex 1. 1 Cos(90 45) Sin45 2 Ex 1. 3 Cos(90 30) Sin60 2 EXAMPLES ON COMPOUND ANGLE FORMULAE Ex .1 Find an expression in surd form for Sin75 Sin75 Sin( 45 30 ) Sin45Cos 30 Cos 45 Sin30 1 3 1 1 . . 2 2 2 2 31 2 2 31 2 . 2 2 2 6 2 4 Ex .2 If Tan(x 45 ) a find Tanx in terms of a. Tanx Tan45 Tan( x 45 ) a 1 TanxTan45 Tanx 1 a 1 Tanx .1 Tanx 1 a aTanx Tanx aTanx a 1 Tanx( 1 a ) a 1 Tanx a1 1 a Sin(A B) Ex .3 Prove that TanA TanB CosACosB Sin(A B) SinACosB CosASinB CosACosB CosACosB Divide each term by CosACosB SinACosB CosASinB CosACosB CosACosB CosACosB CosACosB SinA SinB CosA CosB 1 TanA TanB DOUBLE ANGLE FORMULAE Cos 2 A Cos A Sin A 2 2 This formula is easy to derive. Cos 2 A Cos( A A ) Tan2 A Tan( A A ) CosACosA SinASinA Cos A Sin A 2 2 Sin2 A 2 SinACosA This formula is easy to derive. Sin2 A Sin( A A ) SinACosA CosASinA 2 SinACosA 2TanA 1 Tan2 A This formula is easy to derive. Tan2 A TanA TanA 1 TanATanA 2TanA 1 Tan2 A EXAMPLES ON DOUBLE ANGLE FORMULAE Ex .1 Express Sin3A in terms of SinA Sin3 A Sin( 2 A A ) Sin2 ACosA Cos 2 A .SinA 2 SinACosACosA ( 1 Sin2 A )( SinA ) 2 SinA( Cos 2 A ) SinA 2 Sin3 A 2 SinA( 1 Sin2 A ) SinA 2 Sin2 A 2 SinA 2 Sin3 A SinA 2 Sin3 A 3 SinA 4 Sin3 A 3 , wher 0 A 90 , find the value of 5 (i) Sin2A (ii) Cos2A Ex 2 If SinA From the right angled triangle. 5 3 A 4 CosA 5 4 ( i ) Sin2A 2SinACosA 3 4 2. . 5 5 24 25 ( ii ) Cos2A Cos 2 A Sin2 A 2 4 3 5 5 16 9 25 25 7 25 2 DOUBLE ANGLE FORMULAE IN TAN 1 Tan2 A Cos 2 A 1 Tan2 A Pr oof of this Cos 2 A Cos 2 A Sin2 A Cos 2 A Cos 2 ATan2 A Cos 2 A( 1 Tan2 A ) 1 2 ( 1 Tan A) 2 Sec A 1 Tan2 A 1 Tan2 A 2TanA Sin2 A 1 Tan2 A Pr oof of this Sin2 A 2 SinACosA 2 SinA .Cos 2 A CosA 2TanA . 1 Sec 2 A 2TanA 1 Tan2 A HALF ANGLE FORMULAE 1 Cos 2 A ( 1 Cos 2 A ) 2 This formula is easy to derive. 1 Sin A ( 1 Cos 2 A ) 2 2 This formula is easy to derive. Cos 2 A Cos 2 A Sin2 A Cos 2 A Cos 2 A Sin2 A Cos 2 A Cos 2 A ( 1 Cos 2 A ) Cos 2 A ( 1 Sin2 A ) Sin2 A Cos 2 A 2Cos 2 A 1 Cos 2 A 1 2 Sin2 A 2Cos 2 A Cos 2 A 1 2 Sin2 A 1 Cos 2 A 1 Cos A ( 1 Cos 2 A ) 2 1 Sin A ( 1 Cos 2 A ) 2 2 2 CHANGING PRODUCTS INTO SUMS AND DIFFERENCE S THE PROOFS OF THESE ARE NOT REQUIRED 2CosACosB Cos( A B ) Cos( A B ) 2 SinACosB Sin( A B ) Sin( A B ) 2 SinASinB Cos( A B ) Cos( A B ) 2CosASinB Sin( A B ) Sin( A B ) SUMS AND DIFFERENCE S INTO PRODUCTS THE PROOFS OF THESE ARE NOT REQUIRED A B A B CosA CosB 2Cos Cos 2 2 CosA CosB 2 Sin SinA SinB 2 Sin A B A B Sin 2 2 A B A B Cos 2 2 SinA SinB 2Cos A B A B Sin 2 2 These last two sets of identities are used mainly for solving Trig Equations TRIG EQUATIONS Type 1. Quadratic Ex .1 Find all the solutions of Cos 2 x Cosx Sin2 x 0 in the domain 0 x 360 Cos 2 x Cosx Sin2 x 0 Change Sin2 x into 1 - Cos 2 x Cos 2 x Cosx ( 1 Cos 2 x ) 0 Cos 2 x Cosx 1 Cos 2 x 0 2Cos 2 x Cosx 1 0 Factorise ( 2Cosx 1 )( Cosx 1 ) 0 2Cosx 1 0 Cosx 1 0 1 Cosx 2 Cosx 1 We now have two simple trig equations. The reference angle is 60 It is a negative answer so it is in the quadrants shown. wS A wT C x 120 x 240 x 0 x 360 SOLUTION ON AUTOGRAPH 4 y 3 2 1 x –30 30 –1 –2 –3 –4 60 90 120 150 180 210 240 270 300 330 360 390 Ex .2 Find all the solutions of 2Sin2 2 x Cos 2 x 1 in the domain 0 x 360 2 Sin2 2 x Cos 2 x 1 Change Sin2 2 x into 1 - Cos 2 2 x 2( 1 Cos 2 2 x ) Cos 2 x 1 2 2Cos 2 2 x Cos 2 x 1 2Cos 2 2 x Cos 2 x 1 0 Factorise ( 2Cos 2 x 1 )( Cos 2 x 1 ) 0 ( 2Cos 2 x 1 )( Cos 2 x 1 ) 0 2Cos 2 x 1 0 1 Cos 2 x 2 The reference angle is 60 It is a negative answer so it is in the quadrents shown. wS A wT C 2x 120 + n 360 2x 240 + n 360 n = 0 2x 120° x = 60° n = 0 2x 240° x = 120° n = 1 2x 480° x = 240° n = 1 2x 600° x = 300° n = 2 2x 840° x = 420° n = 2 2x 960° x = 480° Outside the domain 0 x 360 Cos 2 x 1 0 Cos 2 x 1 2x 0 + n 360 n = 0 2x 0° x = 0° n = 1 2x 360° x = 180° n = 2 2x 720° x = 360° n = 3 2x 1080° x = 540° Outside the domain 0 x 360 All the values of x are 60, 120, 240,300, 0, 180 and 360 Ex .3 x 0 and x 60 are two solutions of aSin2 2 x Cos 2 x b Find the value of a and the value of b, where a , b N. Using these values of a and b, find all the solutions of the equation where 0 x 360. aSin2 2 x Cos 2 x b 0 At x 0 aSin2 0 Cos0 b 0 a( 0 ) 1 b 0 b1 At x 60 aSin2 60 Cos60 b 0 2 3 1 a b0 2 2 3 1 a b0 4 2 3a 3 a2 0 4 2 Now the equation is 2 Sin2 2 x Cos 2 x 1 0 which can be solved in the same way as the previous example. Type 2. Double Angles Ex .1 Write Cos2x in terms of Sinx. Hence, find all the solutions of the equation Cos2x - Sinx 1, in the domain 0 x 360. Cos 2 x Cos 2 x Sin2 x Cos 2 x ( 1 Sin2 x ) Sin2 x Cos 2 x 1 2 Sin2 x Now we have to solve the equation Cos2x - Sinx 1 Replacing Cos2x with 1 - Sin2 x we have. 1 2 Sin2 x Sinx 1 2 Sin2 x Sinx 0 Factorise Sinx( 2 Sinx 1 ) 0 Sinx 0 x 0 x 180 x 360 2 Sinx 1 0 1 Sinx 2 x 210 x 330 Type 3. Sums and difference into products Ex .1 (i) Express Sin5x Sin3x as a product of Sin and Cos. (ii) Find all the solutions to the equation Sin5x Sin3x 0 in the domain 0 x 180. (i) 5x 3x 5x 3x Sin5 x Sin3 x 2 Sin Cos 2 2 2 Sin4 xCosx ( ii ) Sin5x Sin3x 0 2 Sin4 xCosx 0 Factorise NOTE : This is already factorised (2Sin4 x)( xCosx) 0 2 Sin4 x 0 Sin4 x 0 4x 0 + n 360 4x 180 + n 360 n = 0 4x 0° x = 0° n = 0 4x 180° x = 45° n = 1 4x 360° x = 90° n = 1 4x 540° x = 135° n = 2 4x 720° x = 180° n = 2 4x 900° x = 225° n = 3 4x 1080° x = 270° Outside the domain 0 x 180 Outside the domain 0 x 180 All the values of x are 0, 90, 180, 45 and 135 Cosx 0 x 90 SOLUTION ON AUTOGRAPH Limits 1 1. The first thing normally tried in evaluating any limit is to evaluate f(a) 1 2 3 4 f (a) = Lim f ( x) = k R k xa k R, k 0 k kR k kR 0 k 0 Limits 2 If f (a) does not give one of the previous answers, we need to try something else. 0 i.e. if we got , , or , - (indeterminate forms) 0 2. Find a common factor and cancel sin 3. Lim = 1 Special Limit x 0 Examples: Evaluate the following limits 2 2 x x-6 x 4 x 3 4. Lim 1. Lim x2 x2 x-2 x3 sin 5 sin 4 5. Lim 2. Lim 0 0 2 4 3. Lim cos 4 6. Lim 4 cos 2 9 0 0 7. Lim 4 tan 2 2 x x 0 sin 2 x 8. Lim 0 3x 2 sin 4 x - sin 2 x 9. Lim 0 3x d Area of abc Area of sector abc Area of adc b r a 1 2 r sin 2 sin r c 1 2 r 2 sin sin sin As 0 1 sin Limit Limit 0 0 sin sin sin tan 1 1 2 r tan 2 tan sin 1 cos 1 1 1 Limits Link to Excel Example 1. Find limit 0 Limit 0 sin5θ sin4θ 5 Sin5 4 4 5 Sin4 5 1 1 4 5 4 2007 Q4 (cosA sinA)2 cos 2 A + sin2 A + 2cosAsinA 1 + 2cosAsinA 1 + sin2A 6cos 2 x sinx - 5 0 6(1- sin2 x) sinx - 5 0 6sin 2 x sinx - 1 0 (2sinx 1)(3sinx + 1) 0 2sinx 1 0 1 or sinx 2 3sinx + 1 0 1 sinx 3 x 30 x 199 x 150 x 341 (i) c What if ? 90o a o ac cos ab ab cos ac c b ab diaameter = 2r ac 2rcos a o b (ii) c a 1 2 Area of semicicle = r 2 o b Area of region abc = Area of aoc + Area of sector obc. 1 1 Area of region abc = ao ac sin + (r)(r)(2 ). 2 2 1 Area of region abc = r(2rcos )sin + r 2 . 2 1 2 Area of region abc = r (2cos sin ) + r 2 . 2 1 2 1 2 2 Area of region abc = r sin2α + r α = πr 2 4 1 2 sin2α + 2α = π ( multiplying each term by 2 ) 2 r 2006 Q 4 2006 Q 5 2006 Q 5 2002 Q.4 (a) T he length of an arc of a circle is 10cm. The radius of the circle is 4cm. The measure of the angle at the centre of the circle is . (i) Find in radians. (ii) Find in degrees, correct to the nearest degree. ( b ) (i) Write Cos2x in terns of Sinx. (ii) Hence find all the solution of the equation Cos 2x - Sinx 1 in the domain 0 x 360. ( c ) A triangle has side a, b and c. The angles opposite a, b and c are A, B and C respective ly. (i) Prove that a 2 b 2 c 2 2 bcCosA . (ii) Show that c(bCosA - aCosB) b 2 a 2 c C A a B c (a) (i) Length of arc r ( in Radians) 10 4 r 10 r 2.5 radians 4 ( ii ) radians 180 180 1 radian 2.5 radians 180 2.5 140 (b) Identity ( c ) (i) This is the proof of the Cos rule. ( ii ) To show that c(bCosA - aCosB) b 2 a 2 2 2 2 a b c As a 2 b 2 c 2 2 bcCosA CosA 2 bc b2 a2 c 2 2 2 2 As b a c 2 acCosB CosB 2 ac Sub in for CosA and CosB into c(bCosA - aCosB) b 2 a 2 a2 b2 c 2 b2 a2 c 2 c(b -a ) b2 a2 - 2bc 2 ac Simplify this and we get b 2 a 2 b 2 a 2 2002 Q.5 Sin7 0 Sin 2 (b) xyz is a triangle where xy 8cm. and yz 6cm. (a) Evaluate Limit Given that the area of the triangle xyz is 12cm 2 , find (i) the two possible values of xyz (ii) the two possible values of xz , correct to one decimal place. (c) A is an obtuse angle such that 4 3 Sin A Sin A 6 6 5 (i) Find SinA and TanA. 1 (ii) Given that Tan(A B) , find TanB and express you answer 2 p in the form where p,q Z and q 0 q (a) Limit 0 Sin7 Sin2 Sin7 7 7 = Limit 0 Sin2 2 2 7 7 = 2 2 1 ( b ) Area a .b .SinC 2 1 Area 8.6.Sin 12 2 24 Sin 12 1 Sin 2 30 and 150 z x 6 8 y Case 1 Case 2 z x x 8 30 z 6 8 130 y xz 8 2 6 2 2( 8 )( 6 )Cos 30 2 3 64 36 72 2 37.646 xz 37.646 xz 6.136 xz 6.1 6 y xz 8 2 6 2 2( 8 )( 6 )Cos130 2 3 64 36 72 2 162.353 xz 162.353 xz 12.741 xz 12.7 ( c ) (i) 4 3 Sin A Sin A 6 6 5 Using the formula SinA SinB A 6 A 6 A 6 A 6 Sin A Sin A 2 Sin Cos 6 6 2 2 A 6 A 6 A 6 A 6 2 Sin Cos 2 2 2 6 2 A 2 Sin Cos 2 Sin ACos 2 6 2 3 4 3 4 2 SinA SinA 2 5 5 Drawing a right angled triangle 5 4 A 3 4 TanA 3 ( ii ) 1 Tan(A B) 2 TanA TanB Tan(A B) 1 - TanA.TanB 4 1 3 TanB 4 2 1 - 3 .TanB 8 4 2TanB 1 TanB 3 3 4 8 2TanB TanB 1 3 3 6TanB 4TanB 3 8 3 3 10TanB 5 1 TanB 2 Trigonometric Formulae: Proofs Required. © SLSS 2007