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
www.mathsrevision.com Higher Higher Maths Connection between Radians / degrees & Exact values Solving Basic Trig Equations Solving Harder Trig Equations Solving Trig Equation by Substitution Solving Wave Function www.mathsrevision.com Radians Outcome 3 www.mathsrevision.com Higher Radian measure is an alternative to degrees and is based upon the ratio of arc Length radius L θ θ- theta (angle at the centre) r Circumference 2 r Circumference 2 r So, full circle 360o 2π radians Radians Copy Table 90o π 2 π o 60 3 π o 45 4 30o π 6 360o 2π 270o 3π 2 2π o 120 3 3π o 135 4 240o 5π 6 210o 150o Demo 180o 4π 3 5π o 225 4 7π 6 π 5π 3 7π o 315 4 300o 330o 11π 6 Converting Outcome 3 www.mathsrevision.com Higher For any values ÷180 then X π degrees radians ÷ π then x 180 Converting Outcome 3 www.mathsrevision.com Higher Ex1 72o = Ex2 330o = 330/180 X π = 11 π /6 Ex3 2π /9 =2π /9 ÷ π x 180o = Ex4 72/ 180 X π = 2π /5 2/ 9 23π/18 = 23π /18 ÷ π x 180o = X 180o = 40o 23/ 18 X 180o = 230o Exact Values Outcome 3 www.mathsrevision.com Higher Some special values of Sin, Cos and Tan are useful left as fractions, We call these exact values 2 60º 60º 2 2 3 60º 30º 2 1 This triangle will provide exact values for sin, cos and tan 30º and 60º 60º Exact Values Outcome 3 www.mathsrevision.com Higher x Sin xº Cos xº Tan xº 0º 0 30º 45º 60º ½ 3 3 ½ 2 1 3 90º 2 3 Exact Values Outcome 3 Higher www.mathsrevision.com Consider the square with sides 1 unit 1 1 2 45º 1 45º 1 We are now in a position to calculate exact values for sin, cos and tan of 45o Exact Values Outcome 3 www.mathsrevision.com Higher x Sin xº Cos xº Tan xº 0º 0 30º 45º 60º ½ 1 2 3 3 2 1 2 ½ 1 3 1 3 90º 2 Undefined Solving Trig Equations Outcome 3 www.mathsrevision.com Higher Sin +ve All +ve 180o - xo 1 2 3 4 created by Mr. Lafferty 180o + xo 360o - xo Tan +ve Cos +ve o Solving Trig Equations 90 Graphically what 2 www.mathsrevision.com Higher to Outcome 3are we trying A S o solve 180 Example 1 Type 1: T C Solving the equation sin xo = 0.5 in the range 0o 0o o 270 o to 360 3 2 sin xo = (0.5) xo = sin-1(0.5) 1st Q xo = 30o 2nd Q xo = 150o 1 2 3 4 created by Mr. Lafferty (180o – 30o = 150o) o Solving Trig Equations 90 Graphically what 2 to A Outcome 3are we trying S solve 180o T C Higher www.mathsrevision.com Example 2 : Solving the equation cos xo 0o o 270 o - 0.625 = 0 in the range 0 to 360o 3 2 cos xo = 0.625 xo = cos -1 (0.625) 1st Q xo = 51.3o 2nd Q xo = 308.7o 1 2 3 4 created by Mr. Lafferty (360o - 53.1o = 308.7o) o Solving Trig Equations 90 Graphically what 2 www.mathsrevision.com Higher to Outcome 3are we trying A S o solve 180 Example 6 Type 2 : T C Solving the equation cos2x = 1 in the range 0o to 360o 270o 3 2 cos2 xo = 1 cos xo = ± 1 cos xo = 1 xo = 0o and 360o cos xo = -1 created by Mr. Lafferty xo = 180o 0o Maths4Scotland Higher Find the exact solutions of 4sin2 x = 1, 0 x 2 Rearrange sin 2 x 1 4 Take square roots sin x 1 2 Find acute x acute x + and – from the square root requires all 4 quadrants Determine quadrants for sin x S 6 A T C 5 7 11 x , , , 6 6 6 6 Hint Previous Table of exact values Quit Quit Next o Solving Trig Equations 90 Graphically what 2 www.mathsrevision.com Higher to Outcome 3are we trying A S o solve 180 Example 4 Type 3 : T C Solving the equation sin 2xo + 0.6 = 0 in the range 0o 0o o 270 to 360o 3 2 sin 2xo = (-0.6) 2xo = sin-1(0.6) 2xo = 37o ( always 1st Q First) 2xo = 217o , 323o o , 683o ...... 577 ÷2 created by Mr. Lafferty xo = 108.5o , 161.5o 288.5o , 341.5o Solving Trigonometric Equations Higher www.mathsrevision.com Example: Solve 2cos3x 1 0 Step 1: Re-Arrange (0 x 360o ) Step 2: consider what solutions are expected 2 2cos3x 1 0 2cos3x 1 1 cos3x 2 90o 180 o S A T C 270o 3 2 0o 3x means Solving Trigonometric Equations 3 rows Higher www.mathsrevision.com Step 3: Solve the equation 1 cos3x 2 1st Q 3x = 60o 4th Q 420o 780o 1 3 x cos 60o 2 1 300o 660o 1020o x = 20o 140o 100o 220o 260o 340o Solving Trigonometric Equations Higher www.mathsrevision.com Graphical solution for 1 cos3x 2 Solving Trigonometric Equations Higher www.mathsrevision.com 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 6x means 6 rows Solving Trigonometric but Equations only over 180o so 3 rows Higher www.mathsrevision.com Step 3: Solve the equation 1 sin 6t 2 3rd Q 1 6t sin 1 2 1 o st sin 1 45 always 1 Quad first 2 6t = 225o 4th Q = 315o x = 37.5o 52.5o 585o 675o 97.5o 112.5o 945o 1035o 157.5o 172.5o Solving Trigonometric Equations www.mathsrevision.com Higher Graphical solution for 1 sin 6t 2 90o Solving 2 Higher A S www.mathsrevision.com 180o 0o Example T C5 Type 3 : Trig Equations Graphically what Outcome 3are we trying to solve o 270 Solving the equation 2sin (2xo - 30o) - √3 = 0 in the range 0o to 360o 3 2 2sin (2x - 30o) = √3 sin (2x - 30o) = √3 ÷ 2 (2x - 30o) = sin-1(√3 ÷ 2) (2xo - 30o) = 60o 120o 420o 480o 2xo = 90o 150o 450o 510o ÷2 created by Mr. Lafferty xo = 45o 75o 225o 255o Trig Equations www.mathsrevision.com Higher Example 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 ALWAYS work out Quad 1 first Higher Example 5 Trig Equations www.mathsrevision.com Solve sinxcos30 + cosxsin30 = -0.966 where 0o < x < 360o By rule 1a sinxcos30 + cosxsin30 = sin(x + 30) sin(x + 30) = -0.966 1st Q sin-1 (0.966) = 75 3rd Q S 180-xo A xo o 180+xo 360-x T C (x + 30o) = 255o 4th Q x + 30o = 285o x = 225o x = 255o The solution is to be in radians – but workTrigonometric in degrees and convert at the Solving Equations end. Higher www.mathsrevision.com Example: Solve 2sin(2 x ) 1 3 Step 1: Re-Arrange (0 x 2 ) Step 2: consider what solutions are expected 2 1 sin(2 x 60 ) 2 o (2x – 60o ) = sin-1(1/2) 90o 180 o S A T C 270o 3 2 0o Solving Trigonometric Equations Higher www.mathsrevision.com Step 3: Solve the equation 1 sin(2 x 60 ) 2 o 1 2 x 60 sin 2 1 o 1st Q 30o 2nd Q 150° 2x - 60° = 30o 150o 390o 510o 2x = 90o 210o 450o 570o x = 45o 225o 4 105o 285o 7 12 5 4 19 12 Solving Trigonometric Equations www.mathsrevision.com Higher Graphical solution for 1 sin(2 x 60 ) 2 o 2 o 90 S A 180 0o Example T C7 Type 5 : Higher www.mathsrevision.com Solving Trig Equations o Outcome 3 180 Solving the equation 270o 3sin2x +32sin x - 1 = 0 in the range 0o to 360o 2 Let p = sin x We have 3p2 + 2p - 1 = 0 Factorise (3p – 1)(p + 1) = 0 3p – 1 = 0 p = 1/3 sin x = 1/3 xo = 19.5o and 160.5o o p+1=0 p =-1 sin x = -1 xo = 270o 2 90o S A T C 270o 3 2 0o Solving Trigonometric Equations Higher www.mathsrevision.com Harder Example: Solve Step 1: Re-Arrange 3sin 2 x 4sin x 1 0 (0 x 360o ) Step 2: Consider what solutions are expected (3sin x 1)(sin x 1) 0 1 sin x 3 2 90o sin x 1 Two solutions One solution 180 o S A T C 270o 3 2 0o Solving Trigonometric Equations Higher www.mathsrevision.com Step 3: Solve the equation 1 sin x 3 1st Q x = 19.5o 2nd Q x = 160.5o Overall solution sin x 1 x = 90o x = 19.5o , 90o and 160.5o Solving Trigonometric Equations Higher www.mathsrevision.com Graphical solution for 3sin 2 x 4sin x 1 0 cos(2x – 10o) – 0.5 = 0 Range : 0 ≤ x ≤ 2π tanxo – 0.5 = 0 Range : 0 ≤ x ≤ 180o Name : 2sin2xo + 3sinxo +1 = 0 Higher Trig. 6cos2xo Range : 0 ≤ x ≤ 2π +3=0 Range : 0 ≤ x ≤ 360o Range : 0 ≤ x ≤ 2π sin2xo – 0.5 = 0 cos(2x – 10o) – 0.5 = 0 Range : 0 ≤ x ≤ 2π cos(2xo – 10o) = 0.5 2xo – 10 = cos o -1 S T (0.5) 2x o -10o = 60o ,300o , 420o , 660o A C 2x o = 70o ,310o , 430o , 670o x o = 35o ,155o , 215o , 335o xo = 7 31 43 67 , , , 36 36 36 36 tanxo – 0.5 = 0 Range : 0 ≤ x ≤ 180o tanxo = 0.5 xo = tan xo = -1 26.5o , 206.5o S T Name : A C -1 (0.5) 2π 4π 8π 10π , , , 3 3 3 3 π 2π 4π 5π xo = , , , 3 3 3 3 S A T C (2sinxo + 1 ) = 0 sinxo = -0.5 (sinxo + 1 ) = 0 sinxo = -1 xo = 210o , 330o xo = 270o 6cos2xo + 3 = 0 Range : 0 ≤ x ≤ 2π 2xo = cos Range : 0 ≤ x ≤ 360o Let p = sinxo 2p2 + 3p + 1 = 0 (2p + 1 )(p + 1) = 0 Higher Trig. cos2xo = -0.5 (0.5) 2sin2xo + 3sinxo + 1 = 0 1st Quad xo = sin2xo – 0.5 = 0 sin = 0.5 sinxo 3 sinxo = 1 /√2 2x o = xo = S T xo 2 A C = ±1/√2 π 3π , 4 4 Range : 0 ≤ x ≤ 2π 1st Quad xo = 4 sinxo = - 1/√2 xo = S A T C 5π 7π , 4 4 Trigonometric Equations www.mathsrevision.com Higher 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 Higher cos 2 x o 4sin x o 5 0 for 0 x 360o. www.mathsrevision.com Solve: (1 2sin 2 x) 4sin x 5 0 6 4sin x 2sin x 0 2 cos2x and sin x, 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 Higher www.mathsrevision.com 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 www.mathsrevision.com Higher The solution is to be in radians – but work in degrees and convert at the end. Solving Trigonometric Equations Higher www.mathsrevision.com Harder Example: Solve 5sin 2 x 2 2cos x Step 1: Re-Arrange (0 x 2 ) Step 2: Consider what solutions are expected 5(1 cos 2 x) 2 2cos x 2 Remember othis ! 90 2 2 3 2cos x 5cos 2 x 0 (3 5cos x)(1 cos x) 0 3 cos x 5 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 Higher www.mathsrevision.com Step 3: Solve the equation 3 cos x 5 1st Q 4th Q x = 53.1o x = 306.9o Overall solution in radians cos x 1 x = 180o x = 0.93 , π and 5.35 Solving Trigonometric Equations Higher www.mathsrevision.com Graphical solution for 5 sin 2 x 2 2cos x Maths4Scotland Higher 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 Maths4Scotland Higher Functions f and g are defined on suitable domains by f(x) = sin (x) and g(x) = 2x a) Find expressions for: i) f(g(x)) ii) g(f(x)) Determine x b) Solve 2 f(g(x)) = g(f(x)) for 0 x 360° sin x 0 x 0, 360 1st expression f ( g ( x)) f (2 x) sin 2 x 2nd expression g ( f ( x)) g (sin x) 2sin x Form equation Replace sin 2x 2sin 2x 2sin x sin 2 x sin x Common factor 1 2 acute x 60 S A T C Determine quadrants x 60, 300 2sin x cos x sin x x 0, 60, 300, 360 2sin x cos x sin x 0 Rearrange Hence cos x 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 Maths4Scotland Higher Functions f ( x) sin x, g ( x) cos x a) Find expressions for b) i) 2nd i) f(h(x)) 4 are defined on a suitable set of real numbers ii) g(h(x)) 1 1 f (h( x)) sin x cos x 2 Hence solve the equation2 expression expression Simplify h( x ) x Show that iii) 1st and 1st ii) Find a similar expression for g(h(x)) f (h( x)) g (h( x)) 1 for 0 x 2 sin x Simplifies to 4 g (h( x)) g x cos x f (h( x)) f x 4 4 4 4 4 f (h( x)) sin x cos cos x sin expr. Rearrange: 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 4 4 cos x 1 sin x 2 f (h( x)) g (h( x)) 1 Previous Table of exact values Quit Quit acute Determine quadrants x 3 4 , sin x 2 2 sin x 1 2 2 x 2 1 2 2 2 4 S A T C 4 Hint Next Maths4Scotland a) b) Higher Solve the equation sin 2x - cos x = 0 in the interval 0 x 180° 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 for sin x Previous Table of exact values x 150 y cos150 Find y value y Coords, P x 30, 150 T By inspection (P) P 150, 3 2 Hint C Quit Quit 3 2 Next Maths4Scotland Solve the equation Higher 3cos(2 x) cos( x) 1 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 Maths4Scotland Higher 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 Maths4Scotland Higher 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 cos 2q 2 cos 2 q 1 Rearrange 2 cos 2 q 8cos q 8 0 Divide by 2 cos 2 q 4 cos q 4 0 Factorise Deduction Try to solve: cosq 2 0 cosq 2 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 Maths4Scotland Higher Solve algebraically, the equation sin 2x + sin x = 0, 0 x 360 Replace sin 2x 2sin x cos x sin x 0 Common factor sin x 2cos x 1 0 Hence S A T C sin x 0 or Determine x Determine quadrants for cos x 2 cos x 1 0 cos x 1 2 x 120, 240 sin x 0 x 0, 360 1 2 cos x acute x 60 x = 0°, Previous Table of exact values Quit 120°, 240°, 360° Quit Hint Next Maths4Scotland Higher Solve the equation cos2x cos x 0 Replace cos 2x with cos 2 x 2 cos 2 x 1 for 0 ≤ x ≤ 360° 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 Maths4Scotland Higher 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 Maths4Scotland Higher f ( x) 2 cos x 3sin x a) Express f (x) in the form k cos( x ) b) Hence solve algebraically f ( x) 0.5 Expand kcos(x - a): k cos a 2 Square and add k 2 22 32 Put together: tan a x 56 82 Previous and 0 360 0 x 360 3 2 acute k sin a 3 k 13 a 56 a is in 1st quadrant (sin and cos are both + ) a 56 2cos x 3sin x 13 cos x 56 Solve equation. acute for k 0 k cos( x a) k cos x cos a k sin x sin a Equate coefficients: Dividing: where cos x 56 13 cos x 56 0.5 Cosine +, so 1st & 4th quadrants Quit Quit 0.5 13 x 138 or x 334 Next Hint Maths4Scotland Higher Solve the equation 2sin x 3cos x 2.5 in the interval 0 x 360. Use R cos(x - a): R cos( x a) R cos x cos a R sin x sin a R cos a 3 Equate coefficients: R 2 22 3 Square and add tan a Dividing: Put together: x 146 46 x 192 or Previous acute R 13 a 34 a is in 2nd quadrant a 146 (sin + and cos - ) 2sin x 3cos x 13 cos x 146 Solve equation. acute 2 3 2 R sin a 2 x 460 cos x 146 13 cos x 146 2.5 2.5 13 Cosine +, so 1st & 4th quadrants (out of range, so subtract 360°) Quit x 100 Quit or x 192 Hint Next Maths4Scotland Higher Solve the simultaneous equations k sin x 5 k cos x 2 where k > 0 and 0 x 360 Use tan A = sin A / cos A Divide tan x Find acute angle 5 2 acute Determine quadrant(s) x 68 Sine and cosine are both + in original equations Solution must be in 1st quadrant State solution x 68 Hint Previous Quit Quit Next Solving Trig Higher Example Expand and Equations equate coefficients Outcome 4 www.mathsrevision.com a) Express 2cos 2 x 3sin 2 x in the form k sin(2 x ) b) Hence solve 2cos 2 x 3sin 2 x 1 for 0 x 360o Find tan ratio note:2 o sin(-) and90 cos(-) 3sin 2 x 2cos 2 x k sin(2 x ) k sin 2 x cos k cos2 x sin k sin 2 k cos 3 Square and add k 13 2 sin 2 tan tan 1 33.7o cos 3 180o 33.7o 213.7o 180 o S A T C 270o 3 2 0o Solving Trig Equations Outcome 4 www.mathsrevision.com Higher b) We now have 2cos 2 x 3sin 2 x 13 sin(2 x 213.7)0 We solve 2cos 2 x 3sin 2 x 1 by solving 13 sin(2 x 213.7) 1 sin(2 x 213.7) 1 13 In the 1st quadrant 2x – 213.7 = 16.1o 163.9o 376.1o 523.9o x = 114.9o , 188.8o, 1 0 sin 1 16.1 13 o o 2x = 229.8 377.6 368.8º - 360º 589.8o= 8.8º 737.6o 294.9o, 368.8o Range : 0 ≤ x ≤ π Range : 0 ≤ x ≤ 360o Solving Trig. Equations Range : 0 ≤ x ≤ 2π Name : Are you on Target ! www.mathsrevision.com Higher • Update you log book • Make sure you complete and correct MOST of the Trigonometry questions in the past paper booklet.