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TRIGONOMETRICAL EQUATIONS Trigonometry identities of the form sin(A+B) Double Angle formulae Trigonometric Equations Radians & Trig Basics More Trigonometric Equations Exam Type Questions Trig Identities Supplied on a formula sheet !! The following relationships are always true for two angles A and B. 1a. 1b. sin(A + B) = sinAcosB + cosAsinB sin(A - B) = sinAcosB - cosAsinB 2a. cos(A + B) = cosAcosB – sinAsinB 2b. cos(A - B) = cosAcosB + sinAsinB Quite tricky to prove but some of following examples should show that they do work!! Trig Identities Examples 1 (1) Expand cos(U – V). (use formula 2b ) cos(U – V) = cosUcosV + sinUsinV (2) Simplify sinf°cosg° - cosf°sing° (use formula 1b ) sinf°cosg° - cosf°sing° = sin(f – g)° (3) Simplify cos8 θ sinθ + sin8 θ cos θ (use formula 1a ) cos8 θ sin θ + sin8 θ cos θ = sin(8 θ + θ) = sin9 θ Trig Identities Example 2 By taking A = 60° and B = 30°, prove the identity for cos(A – B). NB: cos(A – B) = cosAcosB + sinAsinB LHS = cos(60 – 30 )° = cos30° = 3/ 2 RHS = cos60°cos30° + sin60°sin30° =(½ Hence LHS = RHS !! X = 3/ 4 = 3/ 2 3/ 2 + ) + (3/2 X ½) 3/ 4 Trig Identities Example 3 Prove that sin15° = ¼(6 - 2) sin15° = sin(45 – 30)° = sin45°cos30° - cos45°sin30° = (1/2 X 3/ 2 = (3/22 - = (3 - 1) 22 = (6 - 2) 4 = ¼(6 - 2) ) - (1/2 X ½) 1/ 22) X 2 2 Trig Identities NAB type Question Example 4 y 41 x 40 Show that 3 4 cos( - ) = 187/ 205 Triangle2 Triangle1 If missing side = x If missing side = y Then x2 = 412 – 402 = 81 Then y2 = 42 + 32 = 25 So So x=9 sin = 9/41 and cos = 40/ 41 y=5 sin = 3/5 and cos = 4/5 Trig Identities sin = 9/41 and cos = 40/ 41 sin = 3/5 and cos = 4/5 cos( - ) = coscos + sinsin = (40/41 X 4/5) + (9/41 = 160/ 205 = 187/ 205 + X 3/ 5 27/ 205 Remember this is a NAB type Question ) Trig Identities NAB type Question Example 5 Solve sinxcos30 + cosxsin30 = -0.966 ALWAYS where 0o < x < 360o work out By1 rule 1a sinxcos30 + cosxsin30 = sin(x + 30) Quad first S A sin(x + 30) = -0.966 180-xo Quad 3 and Quad 4 sin-1 0.966 = 75 Quad 3: angle = 180o + 75o x + 30o = 255o x = 225o xo o 180+xo 360-x T C Quad 4: angle = 360o – 75o x + 30o = 285o x = 255o Trig Identities Example 6 Solve sin5 θ cos3 θ - cos5 θ sin3 θ = By rule 1b. 3/ 2 where 0 < θ < sin5θ cos3θ - cos5θ sin3θ =sin(5θ - 3θ) = sin2θ sin2θ = 3/2 Repeats every Quad 1 and Quad 2 sin-1 3/ 2 = /3 S -θ A θ +θ 2-θ T C Quad 1: angle = /3 Quad 2: angle = - /3 In this example repeats lie out 2 θ = /3 2 θ = 2/3 θ = /6 θ = /3 with limits Trig Identities Example 7 Find the value of x that minimises the expression cosxcos32 + sinxsin32 Using rule 2(b) we get cosxcos32 + sinxsin32 = cos(x – 32) cos graph is roller-coaster min value is -1 when angle = 180 ie x – 32o = 180o ie x = 212o Paper 1 type questions Trig Identities Example 8 Simplify sin(θ - /3) + cos(θ + /6) + cos(/2 - θ) sin(θ - /3) + cos(θ + /6) + cos(/2 - θ) = sin θ cos/3 – cos θ sin/3 + cos θ cos/6 – sin θ sin/6 + cos/2 cos θ + sin/2 sin θ = 1/2 sin θ – 3/ 2cos θ+ 3/ = 2 cos θ – 1/2sin θ + 0 x cos θ + 1 X sin θ sin θ Paper 1 type questions Trig Identities Example 9 Prove that (sinA + cosB)2 + (cosA - sinB)2 = 2(1 + sin(A - B)) LHS = (sinA + cosB)2 + (cosA - sinB)2 = sin2A + 2sinAcosB + cos2B + cos2A – 2cosAsinB + sin2B = (sin2A + cos2A) + (sin2B + cos2B) + 2sinAcosB - 2cosAsinB = 1 + 1 + 2(sinAcosB - cosAsinB) = 2 + 2sin(A – B) = 2(1 + sin(A – B)) = RHS Double Angle Formulae sin 2 A 2sin A cos A cos 2 A cos 2 A sin 2 A 2cos 2 A 1 2 1 2sin A Two further formulae derived from the cos 2 A formulae. cos A 12 (1 cos 2 A) 2 sin A 12 (1 cos 2 A) 2 Double Angle formulae Mixed Examples: 4 Given that A is an acute angle and tan A , calculate sin 2 A and cos 2 A. 3 sin A 4 cos A 3 2 sin 2 A ( 43 sin A)2 1 sin A Similarly: sin 2 A 2sin A cos A Substitute form the tan (sin/cos) equation sin A cos A 1 2 cos A 16 25 3 5 9 16 25 +ve because A is acute 3-4-5 triangle ! 24 25 cos 2 A cos 2 A sin 2 A 4 5 7 25 A is greater than 45 degrees – hence 2A is greater than 90 degrees. Double Angle formulae Find the exact value of sin 75o. sin(75o ) sin(45 30) 2 sin(75 ) sin 45cos30 cos 45sin 30 o o 45 1 1 3 1 1 1 3 2 2 22 2 2 Prove that 1 2 30 o 1 sin( ) tan tan cos cos sin( ) sin cos cos sin cos cos cos cos sin sin cos cos tan tan 3 Double Angle formulae For the diagram opposite show that cos LMN 5 . 5 cos LMN cos( ) Length of LM 18 3 2 Length of MN 10 M 3 2 cos( ) cos cos sin sin 1 3 1 1 2 10 2 10 2 2 20 4 5 5 5 1 5 3 L 3 10 1 N Double Angle formulae Prove that, cos 4 sin 4 cos 2 . cos 4 sin 4 (cos 2 ) 2 (sin 2 ) 2 Using x 2 y 2 ( x y )( x y ) (cos 2 sin 2 )(cos 2 sin 2 ) cos 2 sin 2 1 cos 2 sin 2 cos 2 cos 2 sin 2 cos 2 Trigonometric Equations Double angle formulae (like cos2A or sin2A) often occur in trig equations. We can solve these equations by substituting the expressions derived in the previous sections. Rules for solving equations sin2A = 2sinAcosA when replacing sin2Aequation cos2A = 2cos2A – 1 if cosA is also in the equation cos2A = 1 – 2sin2A if sinA is also in the equation Trigonometric Equations cos 2 x o 4sin x o 5 0 for 0 x 360o. Solve: cos2x and sin x, (1 2sin x) 4sin x 5 0 2 6 4sin x 2sin x 0 2 so substitute 1-2sin2x compare with 6 4 z 2 z 2 0 (6 2sin x)(1 sin x) 0 sin x 1 or sin x 3 x 90 o 0 sin x 1 for all real angles Trigonometric Equations Solve: 5cos 2 x o cos x o 2 for 0 x 360o cos 2x and cos x, so substitute 2cos2 -1 5(2cos 2 x 1) cos x 2 2 10cos 2 x cos x 3 0 (5cos x 3)(2cos x 1) 0 3 1 cos x or cos x 5 2 x 180 60 120 x 53.1o and x 360 53.1 306.9 90o o o x 180 60 240o o 180 and S A T C 270o 3 2 0o Trigonometric Equations Trigonometric Equations The diagram shows the graphs of f ( x) a sin bx o and g ( x) c sin x o for 0 x 360o. 4 y y y f ( x) 2 360 o 0 -2 xx y g ( x) -4 Three problems concerning this graph follow. Trigonometric Equations y i) State the values of a, b and c. y y f ( x) f ( x) a sin bx o g ( x) c sin x o 360o x x The max & min values of sinbx are 1 and -1 resp. The max & min values of asinbx are 3 and -3 resp. y g ( x) a3 f(x) goes through 2 complete cycles from 0 – 360o b 2 g ( x) c sin x o The max & min values of csinx are 2 and -2 resp. c2 Trigonometric Equations ii) Solve the equation f ( x) g ( x) algebraically. From the previous problem we now have: f ( x) 3sin 2 x and g ( x) 2sin x Hence, the equation to solve is: 3(2sin x cos x) 2sin x 6sin x cos x 2sin x 0 3sin 2 x 2sin x Expand sin 2x Divide both sides by 2 3sin x cos x sin x 0 Spot the common factor in the terms? sin x(3cos x 1) 0 Is satisfied by all values of x for which: sin x 0 or 1 cos x 3 Trigonometric Equations iii) find the coordinates of the points of intersection of the graphs for 0 x 360o. From the previous problem we have: sin x 0 and sin x 0 Hence x 0o x x 180o 360o 1 cos x 3 1 cos x 3 x 70.5o x (360 70.5)o 289.5o Radian Measurements Reminders i) Radians radians 180 Converting between degrees and radians: 120o 120. 180 5 5 180 . 6 6 2 radians 3 150o o Degree Measurements Equilateral triangle: ii) Exact Values 45o right-angled triangle: 1 2 o 60 1 2 o 45 1 cos 45o sin 45o tan 45 1 o 30o 1 2 2 3 1 1 2 sin 60o 3 2 sin 30o cos60o 1 2 cos30o 3 2 tan 30o 1 3 tan 60o 3 Radians / Degrees 0o 30o 45o 60o 90o radians 0 0 cos 1 4 1 2 1 2 3 3 2 1 2 2 sin 0 tan 0 6 1 2 3 2 1 3 1 3 degrees Example: 1 What is the exact value of sin 240o ? 240 180 60 sin(180 ) sin sin 240o 3 2 Sine Graph Period = 360o Amplitude = 1 Cosine Graph Period = 360o Amplitude = 1 Tan Graph Period = 180o Amplitude cannot be found for tan function Solving Trigonometric Equations Example: Solve 2cos3x 1 0 (0 x 360o ) Step 2: consider what solutions Step 1: Re-Arrange are expected 2 2cos3x 1 0 2cos3x 1 1 cos3x 2 90o 180 o S A T C 270o 3 2 0o Solving Trigonometric Equations cos 3x is positive so solutions 1 in the first and fourth quadrants cos3x 2 0 x 360o Since x3 Then has 2 solutions x3 0 3x 1080o has 6 solutions Solving Trigonometric Equations Step 3: Solve the equation 1 3 x cos 60o 2 1 cos3x 2 1 o 1st quad 4th quad cos wave repeats every 360 3x = 60o x = 20o 300o 420o 660o 780o 1020o 100o 140o 220o 260o 340o Solving Trigonometric Equations Graphical solution for 1 cos3x 2 Solving Trigonometric Equations Example: Solve 1 2 sin 6t 0 Step 1: Re-Arrange (0 t 180o ) Step 2: consider what solutions are expected 1 sin 6t 2 2 90o sin 6t is negative so solutions in the third and fourth quadrants Since 0 t 180o has 2 solutions x6 x6 Then 0 6t 1080o has 12 solutions 180 o S A T C 270o 3 2 0o Solving Trigonometric Equations Step 3: Solve the equation 1 sin 6t 2 1 6t sin 1 2 3rd quad 4th quad 6t = 225o x = 39.1o 315o 1 o st sin 1 45 always 1 Quad first 2 sin wave repeats every 360o 585o 675o 945o 1035o 52.5o 97.5o 112.5o 157.5o 172.5o Solving Trigonometric Equations 1 Graphical solution for sin 6t 2 The solution is to be in radians – but work in degrees and convertEquations at the Solving Trigonometric end. Example: (0 x 2 ) Solve 2sin(2 x ) 1 3 Step 2: consider what solutions are expected Step 1: Re-Arrange 2 1 sin(2 x 60 ) 2 o 90o (2x – 60o ) = sin-1(1/2) Since 0 x 360 has 2 solutions o x2 Then x2 0 2 x 720 o has 4 solutions 180 o S A T C 270o 3 2 0o Solving Trigonometric Equations Step 3: Solve the equation sin(2 x 60o ) 1 2 1 2 x 60o sin 1 2 1 sin 1 30o 2 (1st quadrant) 2 x 60o 30o and 150o 1st quad 2nd quad 2x = 90o x = 45o 4 210o sin wave repeats every 360o 450o 570o 105o 225o 285o 7 5 19 12 4 12 Solving Trigonometric Equations Graphical solution for sin(2 x 60o ) 1 2 The solution is to be in radians – but work in degrees and convert at the end. Solving Trigonometric Equations Harder Example: Solve tan 2 x 3 (0 x 2 ) Step 2: consider what solutions are expected Step 1: Re-Arrange 2 tan x 3 90o tan x 3 tan x 3 2 solutions 1st and 3rd quads 2 solutions 2nd and 4th quads 180 o S A T C 270o 3 2 0o Solving Trigonometric Equations Step 3: Solve the equation tan x 3 tan x 3 tan 1 3 3 (60o in the 1st quadrant) o 1st quad 2nd quad tan wave repeats every 180 x = 60o 3 120o 240o 300o 2 3 4 3 5 3 Solving Trigonometric Equations Graphical solution for tan x 3 2 Solving Trigonometric Equations Harder Example: Solve Step 1: Re-Arrange 3sin 2 x 4sin x 1 0 (0 x 360o ) Step 2: Consider what solutions (3sin x 1)(sin x 1) 0 1 sin x 3 are expected 2 90o sin x 1 Two solutions One solution 180 o S A T C 270o 3 2 0o Solving Trigonometric Equations Step 3: Solve the equation sin x sin x 1 1 3 Two solutions 1stquad 2nd quad x = 19.5o 160.5o Overall solution One solution 90o x = 19.5o , 90o and 160.5o Solving Trigonometric Equations Graphical solution for 3sin 2 x 4sin x 1 0 The solution is to be in radians – but work in degrees and convert at the end. Solving Trigonometric Equations Harder Example: Solve 5sin 2 x 2 2cos x Step 1: Re-Arrange Step 2: Consider what solutions 5(1 cos x) 2 2cos x 2 are expected 2 Remember othis ! 90 2 2 3 2cos x 5cos 2 x 0 (3 5cos x)(1 cos x) 0 3 cos x 5 (0 x 2 ) cos x 1 sin cos 1 2 cos 2 1 sin A S o 2 180sin 1 cos 2 C T 270o Two solutions One solution 3 2 0o Solving Trigonometric Equations Step 3: Solve the equation cos x 3 5 Two solutions 1stquad 3rd quad x = 53.1o 306.9o Overall solution in radians cos x 1 One solution 180o x = 0.93 , π and 5.35 Solving Trigonometric Equations Graphical solution for 5 sin 2 x 2 2cos x A is the point (8, 4). The line OA is inclined at an angle p radians to the x-axis a) Find the exact values of: i) sin (2p) ii) cos (2p) The line OB is inclined at an angle 2p radians to the x-axis. b) Write down the exact value of the gradient of OB. Draw triangle 80 Pythagoras 4 p 8 8 sin p 80 4 8 2 80 80 cos p Write down values for cos p and sin p Expand sin (2p) sin 2 p 2sin p cos p Expand cos (2p) cos 2 p cos p sin p Use m = tan (2p) tan 2 p Previous 2 2 sin 2 p cos 2 p Quit 8 80 2 4 80 2 4 80 64 4 80 5 64 16 3 80 5 4 3 4 5 5 3 Quit Hint Next In triangle ABC show that the exact value of sin(a b) is Use Pythagoras sin a sin a, cos a, sin b, cos b Substitute values Simplify 10 2 AC 2 CB 10 Write down values for Expand sin (a + b) 2 5 1 2 cos a 1 2 sin b 1 10 cos b 3 10 sin(a b) sin a cos b cos a sin b sin(a b) sin(a b) 3 20 1 20 1 3 2 10 4 20 1 1 2 10 4 4 2 45 2 5 5 Hint Previous Quit Quit Next Using triangle PQR, as shown, find the exact value of cos 2x Use Pythagoras 11 PR 11 Write down values for cos x and sin x 2 cos x 11 7 sin x 11 Expand cos 2x cos 2 x cos 2 x sin 2 x Substitute values cos 2x Simplify Previous cos 2 x 2 11 4 7 11 11 Quit 2 7 11 2 3 11 Quit Hint Next On the co-ordinate diagram shown, A is the point (6, 8) and 10 8 B is the point (12, -5). Angle AOC = p and angle COB = q 6 Find the exact value of sin (p + q). Use Pythagoras OA 10 Write down values for Mark up triangles OB 13 sin p sin p, cos p, sin q, cos q Expand sin (p + q) Substitute values Simplify Previous 12 8 , 10 5 13 cos p 6 , 10 sin q 5 , 13 cos q 12 13 sin ( p q) sin p cos q cos p sin q sin ( p q) sin ( p q) 96 130 30 130 Quit 8 12 10 13 6 5 10 13 126 130 Quit 63 65 Hint Next A and B are acute angles such that tan A Find the exact value of a) sin 2A b) cos 2A Draw triangles c) and tan B sin A sin 2 A 2sin A cos A Expand cos 2A cos 2 A cos A sin A Substitute Previous . 5 13 3 A 5 B 4 12 Hypotenuses are 5 and 13 respectively Expand sin 2A 2 Expand sin (2A + B) 5 12 sin(2 A B) Use Pythagoras Write down sin A, cos A, sin B, cos B 3 4 2 3 , 5 cos A 4 , 5 sin B 3 5 sin 2 A 2 cos 2A 2 4 5 4 3 5 5 5 , 13 2 cos B 12 13 24 25 16 9 25 25 7 25 sin 2 A B sin 2 A cos B cos 2 Asin B sin 2 A B 24 12 7 5 323 25 13 25 13 325 Quit Quit Hint Next If x° is an acute angle such that tan x 4 3 5 4 3 3 sin( x 30) is show that the exact value of 10 4 x 3 Draw triangle Use Pythagoras Write down sin x and cos x Expand sin (x + 30) sin x Hypotenuse is 5 4 , 5 cos x 3 5 sin( x 30) sin x cos 30 cos x sin 30 Substitute sin( x 30) 4 3 3 1 5 2 5 2 Simplify sin( x 30) 4 3 3 10 10 4 3 3 10 Hint Previous Table of exact values Quit Quit Next The diagram shows two right angled triangles 24 ABD and BCD with AB = 7 cm, BC = 4 cm and CD = 3 cm. Angle DBC = x° and angle ABD is y°. cos( x y ) is 20 6 6 35 Show that the exact value of BD 5, AD 24 Use Pythagoras Write down sin x, cos x, sin y, cos y. Expand cos (x + y) Substitute Simplify Previous 5 sin x 3 , 5 cos x 4 , 5 sin y 24 , 7 cos y 5 7 cos( x y ) cos x cos y sin x sin y cos( x y ) 4 5 3 24 5 7 5 7 20 3 4 6 20 6 6 20 3 24 cos( x y ) 35 35 35 35 Quit Quit Hint Next The framework of a child’s swing has dimensions as shown in the diagram. Find the exact value of sin x° Draw triangle Use Pythagoras Draw in perpendicular h 5 3 Use fact that sin x = sin ( ½ x + ½ x) sin sin Write down sin ½ x and cos ½ x Expand sin ( ½ x + ½ x) Substitute Simplify sin x x 2 2 sin x sin x 2 x x 2 2 x x 2 h5 2 3 2 2 3 , cos x 2 x x cos 2 2 2 5 3 x 2 sin cos 3 4 x 2 x 2 2sin cos x 2 5 3 4 5 9 Previous Table of exact values Hint Quit Quit Next Given that tan find the exact value of 11 , 0 3 2 20 sin 2 Draw triangle cos a and sin a a Use Pythagoras Write down values for hypotenuse 3 cos a 20 20 3 11 sin a 20 Expand sin 2a sin 2a 2 sin a cos a Substitute values sin 2a 2 Simplify 6 11 sin 2a 20 Previous 11 11 3 20 20 Quit Quit 3 11 10 Hint Next Find algebraically the exact value of sin q sin q 120 cos(q 150) Expand sin (q +120) sin q 120 sin q cos120 cosq sin120 Expand cos (q +150) cos q 150 cosq cos150 sin q sin150 Use table of exact values Combine and substitute Simplify 1 2 3 2 cos 120 cos 60 cos 150 cos 30 sin 120 sin 150 sin 60 sin q sin q . sin q sin q 1 2 3 cos q 2 sin 30 3 2 1 2 cosq . cosq . sin q . 1 2 3 2 3 cos q 2 3 2 1 2 sin q 1 2 0 Previous Table of exact values Quit Quit Hint Next If cos q a) sin 2q 4 , 0 q 5 2 Find sin 4q Previous 4 3 sin q 5 3 4 24 2 5 5 25 sin 2q 2 sin q cos q Expand sin 4q (4q = 2q + 2q) Expand cos 2q Opposite side = 3 4 cos q 5 cos q and sin q 3 q Use Pythagoras Write down values for Expand sin 2q 5 sin 4q b) Draw triangle find the exact value of sin 4q 2 sin 2q cos 2q cos 2q cos q sin q 2 24 7 sin 4q 2 25 25 Quit 2 16 9 7 25 25 25 336 625 Quit Hint Next For acute angles P and Q sin P 12 and 13 sin Q Show that the exact value of sin ( P Q) Draw triangles Use Pythagoras Write down sin P, cos P, sin Q, cos Q Expand sin (P + Q) 3 5 63 65 5 12 P 3 Q 5 4 Adjacent sides are 5 and 4 respectively sin P 12 , 13 cos P 5 , 13 sin Q 3 , 5 cos Q 4 5 sin P Q sin P cos Q cos P sin Q Substitute sin P Q 12 4 5 3 13 5 13 5 Simplify sin P Q 48 15 65 65 Previous 13 Quit Quit 63 65 Hint Next Solve the equation 3cos(2 x) 10cos( x) 1 0 for 0 ≤ x ≤ correct to 2 decimal places Replace cos 2x with Substitute Simplify cos 2 x 2 cos 2 x 1 3 2 cos x 1 10 cos x 1 0 2 6 cos x 10 cos x 4 0 2 Determine quadrants S A T C 3cos 2 x 5cos x 2 0 Factorise Hence 3cos x 1 cos x 2 0 cos x x 1.23 1 3 cos x 2 Discard Find acute x Previous acute x 1.23 x 5.05 x 1.23 rad Quit or 2 1.23 rads rads rads Hint Quit Next The diagram shows the graph of a cosine function from 0 to . a) State the equation of the graph. b) The line with equation y = -3 intersects this graph at points A and B. Find the co-ordinates of B. Equation y 2 cos 2 x Determine quadrants 2cos 2 x 3 Solve simultaneously cos 2x Rearrange Check range 0 x Find acute 2x Deduce 2x acute 2x 2x Table of exact values T C x 5 7 or 12 12 B 6 Previous A 3 2 0 2 x 2 6 or 6 6 S 6 rads 6 6 Quit Quit is B 7 12 , 3 Next Hint Functions f and g are defined on suitable domains by f(x) = sin (x) and g(x) = 2x a) Find expressions for: i) b) f(g(x)) ii) Determine x g(f(x)) Solve 2 f(g(x)) = g(f(x)) 1st expression f ( g ( x)) f (2 x) sin 2 x 2nd expression g ( f ( x)) g (sin x) 2sin x Form equation 2sin x cos x sin x Rearrange 2sin x cos x sin x 0 Hence cos x 2sin 2x 2sin x sin 2 x sin x Replace sin 2x Common factor sin x 0 x 0, 360 for 0 x 360° 1 2 acute x 60 S A T C Determine quadrants x 60, 300 x 0, 60, 300, 360 sin x 2cos x 1 0 sin x 0 or 2 cos x 1 0 cos x Previous Table of exact values Quit Hint 1 2 Quit Next Functions f ( x) sin x, g ( x) cos x a) Find expressions for b) i) Show that and h( x ) x i) f(h(x)) f (h( x)) 4 are defined on a suitable set of real numbers ii) g(h(x)) 1 1 sin x cos x ii) Find a similar expression for g(h(x)) 2 2 iii) Hence solve the equation f (h( x)) g (h( x)) 1 for 0 x 2 f (h( x)) f x sin x Simplifies to 1st expression 4 4 2nd expression Simplify 1st g (h( x)) g x cos x 4 4 4 4 f (h( x)) sin x cos cos x sin expr. Rearrange: sin x 1 1 sin x 2 2 Use exact values f (h( x)) Similarly for 2nd expr. g (h( x)) cos x cos sin x sin g (h( x)) Form Eqn. 1 2 acute x cos x Determine 4 4 cos x 1 sin x 2 f (h( x)) g (h( x)) 1 Previous Table of exact values Quit acute Quit 2 2 sin x 1 2 2 x 2 1 2 2 2 4 S A T C quadrants x 3 4 , 4 Hint Next a) Solve the equation sin 2x - cos x = 0 in the interval 0 x 180° b) The diagram shows parts of two trigonometric graphs, y = sin 2x and y = cos x. Use your solutions in (a) to write down the co-ordinates of the point P. Replace sin 2x Common factor Hence 2sin x cos x cos x 0 cos x 2sin x 1 0 cos x 0 Determine x or Solutions for where graphs cross x 30, 90, 150 2sin x 1 0 sin x 1 2 cos x 0 x 90, ( 270 out of range) sin x 1 2 acute x 30 S A Determine quadrants T Previous Table of exact values x 150 y cos150 Find y value y Coords, P x 30, 150 for sin x By inspection (P) P 150, 3 2 Hint C Quit Quit 3 2 Next 3cos(2 x) cos( x) 1 Solve the equation for 0 ≤ x ≤ 360° cos 2 x 2 cos 2 x 1 Replace cos 2x with Determine quadrants 3 2 cos x 1 cos x 1 2 Substitute Simplify 6 cos 2 x cos x 2 0 Factorise 3cos x 2 2cos x 1 0 cos x Hence Find acute x acute 2 3 x 48 cos x acute 1 2 x 60 cos x 2 3 cos x acute x 48 acute x 60 S A S A T C x 132 x 228 T C x 60 x 300 Solutions are: x= 60°, 132°, 228° and 300° Previous Table of exact values Quit Quit 1 2 Hint Next Solve the equation 2sin 2 x 6 1 Rearrange sin Find acute x Note range 2x 6 acute 2x 6 for 0 ≤ x ≤ 2 Determine quadrants 2x 6 and for range 6 0 x 2 0 2 x 4 S 0 2 x 2 for range 1 2 2x 6 6 2x 6 5 6 17 6 2 2 x 4 13 6 2x 6 A Solutions are: T x C 6 , 2 , 7 3 , 6 2 Hint Previous Table of exact values Quit Quit Next a) Write the equation cos 2q + 8 cos q + 9 = 0 in terms of cos q and show that for cos q it has equal roots. b) Show that there are no real roots for q Replace cos 2q with Rearrange Divide by 2 Factorise Deduction cos 2q 2 cos 2 q 1 Try to solve: cosq 2 0 2 cos 2 q 8cos q 8 0 cosq 2 cos 2 q 4 cos q 4 0 No solution Hence there are no real solutions for q cosq 2 cosq 2 0 Equal roots for cos q Hint Previous Quit Quit Next Solve algebraically, the equation sin 2x + sin x = 0, 0 x 360 Replace sin 2x 2sin x cos x sin x 0 Determine quadrants sin x 2cos x 1 0 for cos x S A Common factor Hence sin x 0 or Determine x 2 cos x 1 0 cos x 1 2 1 2 acute x 60 x = 0°, Previous Table of exact values C x 120, 240 sin x 0 x 0, 360 cos x T Quit 120°, 240°, 360° Quit Hint Next Find the exact solutions of 4sin2 x = 1, 0 x 2 Rearrange Take square roots Find acute x sin 2 x 1 4 sin x 1 2 x acute Determine quadrants for sin x S 6 + and – from the square root requires all 4 quadrants A T C 5 7 11 x , , , 6 6 6 6 Hint Previous Table of exact values Quit Quit Next Solve the equation Replace cos 2x with cos2x cos x 0 for 0 ≤ x ≤ 360° cos 2 x 2 cos 2 x 1 Determine quadrants cos x 2 cos 2 x 1 cos x 0 Substitute Simplify 2 cos 2 x cos x 1 0 Factorise 2cos x 1 cos x 1 0 cos x Hence Find acute x acute 1 2 x 60 cos x 1 1 2 acute x 60 S A T C x 60 x 300 x 180 Solutions are: x= 60°, 180° and 300° Previous Table of exact values Quit Quit Hint Next cos2x 5cos x 2 0 Solve algebraically, the equation Replace cos 2x with Substitute cos 2 x 2 cos 2 x 1 for 0 ≤ x ≤ 360° Determine quadrants 2 cos x 1 5cos x 2 0 2 cos x acute Simplify 2 cos 2 x 5cos x 3 0 Factorise 2cos x 1 cos x 3 0 Hence Find acute x cos x acute 1 2 x 60 S cos x 3 Discard above 1 2 x 60 A T C x 60 x 300 Solutions are: x= 60° and 300° Previous Table of exact values Quit Quit Hint Next You have completed all 12 questions in this presentation Previous Quit Quit Back to start Table of exact values sin cos tan Return 30° 45° 60° 6 1 2 4 3 1 2 1 2 3 2 3 2 1 3 1 1 2 3