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
hsn.uk.net Higher Mathematics UNIT 2 OUTCOME 3 Trigonometry Contents Trigonometry 1 2 3 4 5 Solving Trigonometric Equations Trigonometry in Three Dimensions Compound Angles Double-Angle Formulae Further Trigonometric Equations HSN22300 This document was produced specially for the HSN.uk.net website, and we require that any copies or derivative works attribute the work to Higher Still Notes. For more details about the copyright on these notes, please see http://creativecommons.org/licenses/by-nc-sa/2.5/scotland/ 107 107 110 113 116 117 Higher Mathematics Unit 2 – Trigonometry OUTCOME 3 Trigonometry 1 Solving Trigonometric Equations You should already be familiar with solving some trigonometric equations. EXAMPLES 1. Solve sin x ° = 12 for 0 < x < 360 . sin x ° = 12 180° − x ° x° S A Since sin x ° is positive T C 180° + x ° 360° − x ° First quadrant solution: x = sin −1 ( 12 ) = 30. x = 30 or 180 − 30 x = 30 or 150. Remember The exact value triangle: 30° 2 60° 1 2. Solve cos x ° = − 1 for 0 < x < 360 . 5 cos x ° = − 1 180° − x ° x° S A Since cos x ° is negative T C 360° − x ° 180° + x ° 5 ( 5) x = cos −1 1 = 63·435 (to 3 d.p.). x = 180 − 63·435 or 180 + 63·435 x = 116·565 243·435. or 3. Solve sin x ° = 3 for 0 < x < 360 . There are no solutions since −1 ≤ sin x ° ≤ 1 . Note that −1 ≤ cos x ° ≤ 1 , so cos x ° = 3 also has no solutions. hsn.uk.net Page 107 HSN22300 3 Higher Mathematics Unit 2 – Trigonometry 4. Solve tan x ° = −5 for 0 < x < 360 . 180° − x ° x° tan x ° = −5 S A Since tan x ° is negative T C 180° + x ° 360° − x ° x = tan −1 ( 5 ) = 78·690 (to 3 d.p.). x = 180 − 78·690 or x = 101·310 281·310. or 360 − 78·690 Note All trigonometric equations we will meet can be reduced to problems like those above. The only differences are: the solutions could be required in radians – in this case, the question will not have a degree symbol, e.g. “Solve 3 tan x = 1 ” rather than “ 3tan x ° = 1 ”; exact value solutions could be required in the non-calculator paper – you will be expected to know the exact values for 0, 30, 45, 60 and 90 degrees. Questions can be worked through in degrees or radians, but make sure the final answer is given in the units asked for in the question. EXAMPLES 5. Solve 2sin 2 x ° − 1 = 0 where 0 ≤ x ≤ 360 . 180° − 2 x ° 2x° 2sin2 x ° = 1 0 ≤ x ≤ 360 Remember S A 1 sin2 x ° = 2 T C 360° − 2 x ° 0 ≤ 2 x ≤ 720 The exact value triangle: 180° + 2 x ° 2 x = sin −1 ( 12 ) = 30. 2 x = 30 or 180 − 30 30° 2 3 60° 1 or 360 + 30 or 360 + 180 − 30 or 360 + 360 + 30 2 x = 30 or 150 or 390 or 510 x = 15 or 75 or 195 or 255. hsn.uk.net Page 108 Note There are more solutions every 360°, since sin(30°) = sin(30°+360°) = … So keep adding 360 until 2x > 720. HSN22300 Higher Mathematics 6. Solve Unit 2 – Trigonometry 2 cos 2 x = 1 where 0 ≤ x ≤ π . cos 2x = 1 2 2x π − 2x 0≤ x ≤π S A π + 2 x T C 2 π − 2 x 0 ≤ 2 x ≤ 2π 2 x = cos −1 ( ) 1 2 Remember The exact value triangle: = π4 . 2 x = π4 or 2π − π4 π 4 1 2 π 4 1 or 2π + π4 2 x = π4 or 74π x = π8 or 78π . 7. Solve 4cos 2 x = 3 where 0 < x < 2 π . ( cos x )2 = 34 cos x = ± 34 cos x = ± 23 S A Since cos x can be positive or negative T C Remember The exact value triangle: ( ) x = cos −1 23 = π6 . π 6 2 x = π6 or π − π6 or π + π6 or 2π − π6 3 π 3 or 2π + π6 1 x = π6 or 56π or 76π or 116π . 8. Solve 3tan ( 3 x° − 20° ) = 5 where 0 ≤ x ≤ 360 . 3tan ( 3 x ° − 20° ) = 5 tan ( 3 x ° − 20° ) = 53 S A T C 0 ≤ x ≤ 360 0 ≤ 3 x ≤ 1080 −20 ≤ 3 x − 20 ≤ 1060 () 3 x − 20 = tan −1 53 = 59·036 (to 3 d.p.). 3 x − 20 = 59·036 or 180 + 59·036 or 360 + 59·036 or 360 + 180 + 59·036 or 360 + 360 + 59·036 or 360 + 360 + 180 + 59·036 or 360 + 360 + 360 + 59·036 hsn.uk.net Page 109 HSN22300 Higher Mathematics Unit 2 – Trigonometry 3 x − 20 = 59·036 or 239·036 or 419·036 or 599·036 or 779·036 or 959·036 3 x = 79·036 or 259·036 or 439·036 or 619·036 or 799·036 or 979·036 x = 26·35 or 86·35 or 146·35 or 206·35 or 266·35 or 326·35. ( ) 9. Solve cos 2 x + π3 = 0·812 for 0 < x < 2 π . ( ) S A T C cos 2 x + π3 = 0·812 0 < x < 2π 0 < 2 x < 4π π < 2 x + π < 4π + π 3 3 3 1·047 < 2 x + π3 < 13·614 (to 3 d.p.) Remember 2 x + π3 = cos −1 ( 0·812 ) Make sure your = 0·623 (to 3 d.p.). calculator uses radians. 2 x + π3 = 0·623 or 2π − 0·623 or 2π + 0·623 or 2π + 2π − 0·623 or 2π + 2π + 0·623 or 2π + 2π + 2π − 0·623 2 x + π3 = 5.660 or 6.906 or 11.943 or 13.189 2 x = 4.613 or 5.859 or 10.896 or 12.142 x = 2.307 or 2.930 or 5.448 or 6.071. 2 Trigonometry in Three Dimensions It is possible to solve trigonometric problems in three dimensions using techniques we already know from two dimensions. The use of sketches is often helpful. The angle between a line and a plane The angle a between the plane P and the line ST is calculated by adding a line perpendicular to the plane and then using basic trigonometry. T a P S hsn.uk.net Page 110 HSN22300 Higher Mathematics Unit 2 – Trigonometry EXAMPLE 1. The triangular prism ABCDEF is shown below. E 3 cm F B C 6 cm A 10 cm Calculate the acute angle between: (a) The line AF and the plane ABCD. (b) AE and ABCD. (a) Start with a sketch: F tan a = a 10 cm Opposite 3 = Adjacent 10 ( ) 3 a = tan −1 10 3 cm A D = 16.699° (or 0.291 radians) (to 3 d.p.). D Note Since the angle is in a right-angled triangle, it must be acute so there is no need for a CAST diagram. (b) Again, make a sketch: E 3 cm C b angle to be A calculated We need to calculate the length of AC first using Pythagoras’s Theorem: C AC = 102 + 6 2 6 cm = 136. A D 10 cm E Therefore: 3 cm tan b = C b A hsn.uk.net Opposite = 3 Adjacent 136 b = tan −1 ( 1363 ) = 14.426° (or 0.252 radians) (to 3 d.p.). 136 cm Page 111 HSN22300 Higher Mathematics Unit 2 – Trigonometry The angle between two planes The angle a between planes P and Q is calculated by adding a line perpendicular to Q and then using basic trigonometry. P a Q EXAMPLE 2. ABCDEFGH is a cuboid with dimensions 12 × 8 × 8 cm as shown below. H G F E D 8 cm C 8 cm A 12 cm B (a) Calculate the size of the angle between the planes AFGD and ABCD. (b) Calculate the size of the acute angle between the diagonal planes AFGD and BCHE. (a) Start with a sketch: F tan a = 8 cm A Opposite 8 = Adjacent 12 ( ) a = tan −1 23 Note Angle GDC is the same size as angle FAB. = 33.690° (or 0.588 radians) (to 3 d.p.). a 12 cm B (b) Again, make a sketch: Let AF and BE intersect at T. = TBA ɵ = 33.690° . E F ATB is isosceles, so TAB = 180° − ( 33.690° + 33.690° ) ATB Note = 112.620°. T A • hsn.uk.net • The angle could also B So the acute angle is: have been calculated . BTF = ATE = 180° − 112 620° using rectangle DCGH. = 67.380° (or 1.176 radians) (to 3 d.p.). Page 112 HSN22300 Higher Mathematics 3 Unit 2 – Trigonometry Compound Angles When we add or subtract angles, the result is called a compound angle. For example, 45° + 30° is a compound angle. Using a calculator, we find: • sin ( 45° + 30° ) = sin ( 75° ) = 0.966 ; • sin ( 45° ) + sin ( 30° ) = 1.207 (both to 3 d.p.). This shows that sin ( A + B ) is not equal to sin A + sin B . Instead, we can use the following identities: sin ( A + B ) = sin A cos B + cos A sin B sin ( A − B ) = sin A cos B − cos A sin B cos ( A + B ) = cos A cos B − sin A sin B cos ( A − B ) = cos A cos B + sin A sin B . These are given in the exam in a condensed form: sin ( A ± B ) = sin A cos B ± cos A sin B cos ( A ± B ) = cos A cos B ∓ sin A sin B . EXAMPLES 1. Expand and simplify cos ( x ° + 60° ) . cos ( x ° + 60° ) = cos x ° cos 60° − sin x ° sin 60° Remember The exact value triangle: = 12 cos x ° − 23 sin x °. 30° 2 3 60° 1 2. Show that sin ( a + b ) = sin a cos b + cos a sin b for a = π6 and b = π3 . LHS = sin ( a + b ) ( = sin π6 + π3 ) RHS = sin a cos b + cos a sin b = sin π6 cos π3 + cos π6 sin π3 ( = sin ( π2 ) = ( 12 × 12 ) + 23 × 23 = 1. = 14 + 34 = 1. ) Since LHS = RHS , the claim is true for a = π6 and b = π3 . hsn.uk.net Page 113 Remember The exact value triangle: π 6 2 3 π 3 1 HSN22300 Higher Mathematics Unit 2 – Trigonometry 3. Find the exact value of sin75° . sin 75° = sin ( 45° + 30° ) = sin 45° cos30° + cos 45° sin30° ( ) ( = 1 × 23 + 1 × 12 2 2 ) = 3 + 1 2 2 2 2 = 23 +21 = 6 +4 2 . Finding Trigonometric Ratios You should already be familiar with the following formulae (SOH CAH TOA). Hypotenuse Opposite a Adjacent sin a = Opposite Hypotenuse cos a = Adjacent Hypotenuse tan a = Opposite . Adjacent p If we have sin a = q where 0 < a < π2 , then we can form a right-angled triangle to represent this ratio. Since sin a = q a p Opposite p = q then: Hypotenuse the side opposite a has length p; the hypotenuse has length q. The length of the unknown side can be found using Pythagoras’s Theorem. Once the length of each side is known, we can find cos a and tan a using SOH CAH TOA. The method is similar if we know cos a and want to find sin a or tan a . hsn.uk.net Page 114 HSN22300 Higher Mathematics Unit 2 – Trigonometry EXAMPLES 5 . Show that 4. Acute angles p and q are such that sin p = 54 and sin q = 13 sin ( p + q ) = 63 65 . 5 sin p = 54 4 13 q 12 cos p = 35 p 3 5 5 sin q = 13 cos q = 12 13 sin ( p + q ) = sin p cos q + cos p sin q ( ) ( 3 5 = 54 × 12 13 + 5 × 13 ) 48 + 15 = 65 65 = 63 65 . Note Since “Show that” is used in the question, all of this working is required. Confirming Identities EXAMPLES 5. Show that sin ( x − π2 ) = − cos x . sin ( x − π2 ) = sin x cos π2 − cos x sin π2 = sin x × 0 − cos x × 1 = − cos x . sin ( s + t ) = tan s + tan t for cos s ≠ 0 and cos t ≠ 0 . cos s cos t sin ( s + t ) sin s cos t + cos s sin t = cos s cos t cos s cos t sin s cos t cos s sin t = + cos s cos t cos s cos t Remember sin s sin t sin x = tan x . = + cos x cos s cos t = tan s + tan t . 6. Show that hsn.uk.net Page 115 HSN22300 Higher Mathematics 4 Unit 2 – Trigonometry Double-Angle Formulae Using the compound angle identities with A = B , we obtain expressions for sin 2 A and cos2 A . These are called double-angle formulae. sin 2 A = 2 sin A cos A cos 2 A = cos 2 A − sin 2 A = 2 cos 2 A − 1 = 1 − 2 sin 2 A. Note that these are given in the exam. EXAMPLES 1. Given that tanθ = 34 , where 0 < θ < π2 , find the exact value of sin2θ and cos2θ . 5 4 θ sin 2θ = 2sin θ cos θ sin θ = 34 cos θ = 35 3 cos 2θ = cos 2 θ − sin 2 θ 2 2 = 2 × 54 × 35 = 35 = 24 25 . 9 − 16 Note = 25 25 ( ) − ( 54 ) 7 . = − 25 Any of the cos2A formulae could have been used here. 5 , where 2. Given that cos2 x = 13 0 < x < π , find the exact values of sin x and cos x . Since cos2 x = 1 − 2sin 2 x , 5 1 − 2 sin 2 x = 13 8 2 sin 2 x = 13 8 sin 2 x = 26 4 = 13 sin x = ± 2 . 13 We are told that 0 < x < π , so only sin x = 13 2 x a= 2 13 is possible. 2 13 − 2 2 = 13 − 4 = 9 = 3. a So cos x = hsn.uk.net 3 . 13 Page 116 HSN22300 Higher Mathematics 5 Unit 2 – Trigonometry Further Trigonometric Equations We will now consider trigonometric equations where double-angle formulae can be used to find solutions. These equations will involve: sin 2x and either sin x or cos x ; Remember The double-angle cos 2x and cos x ; formulae are given in cos 2x and sin x . the exam. Solving equations involving sin2x and either sinx or cosx EXAMPLE 1. Solve sin 2 x ° = − sin x ° for 0 ≤ x < 360 . 2 sin x ° cos x ° = − sin x ° 2 sin x ° cos x ° + sin x ° = 0 sin x ° ( 2 cos x ° + 1) = 0 sin x ° = 0 Replace sin 2x using the double angle formula Take all terms to one side, making the equation equal to zero Factorise the expression and solve 2cos x ° + 1 = 0 x = 0 or 180 or 360 S A T C cos x ° = − 12 x = 180 − 60 or 180 + 60 x = cos −1 ( 12 ) = 60 = 120 or 240. So x = 0 or 120 or 180 or 240 . Solving equations involving cos2x and cosx EXAMPLE 2. Solve cos 2 x = cos x for 0 ≤ x ≤ 2 π . cos 2 x = cos x 2 cos2 x − 1 = cos x 2 cos2 x − cos x − 1 = 0 ( 2 cos x + 1)( cos x − 1) = 0 2 cos x + 1 = 0 cos x = − 12 x = π − π3 or π + π3 = 23π or 43π Replace cos 2x by 2cos2 x − 1 Take all terms to one side, making a quadratic equation in cos x Solve the quadratic equation (using factorisation or the quadratic formula) S A T C x = cos −1 ( 12 ) cos x − 1 = 0 cos x = 1 x = 0 or 2π . = π3 So x = 0 or 23π or 43π or 2π . hsn.uk.net Page 117 HSN22300 Higher Mathematics Unit 2 – Trigonometry Solving equations involving cos2x and sinx EXAMPLE 3. Solve cos 2 x = sin x for 0 < x < 2 π . cos2 x = sin x 1 − 2 sin 2 x = sin x 2 sin 2 x + sin x − 1 = 0 ( 2 sin x − 1)( sin x + 1) = 0 2 sin x − 1 = 0 sin x = 12 x = π6 or π − π6 = π6 or 5π 6 Replace cos 2x by 1 − 2 sin 2 x Take all terms to one side, making a quadratic equation in sin x Solve the quadratic equation (using factorisation or the quadratic formula) sin x + 1 = 0 sin x = −1 S A T C x = sin −1 ( ) 1 2 x = 32π . = π6 So x = π6 or 56π or 32π . hsn.uk.net Page 118 HSN22300