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hsn.uk.net Higher Mathematics Trigonometry Contents Trigonometry 1 2 3 4 5 6 7 8 9 10 11 12 13 Radians Exact Values Solving Trigonometric Equations Trigonometry in Three Dimensions Compound Angles Double-Angle Formulae Further Trigonometric Equations Expressing pcosx + qsinx in the form kcos(x – a) Expressing pcosx + qsinx in other forms Multiple Angles Maximum and Minimum Values Solving Equations Sketching Graphs of y = pcosx + qsinx 1 EF EF RC EF EF EF RC EF EF EF EF RC EF CfE Edition 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/ 1 1 2 5 8 11 12 14 15 16 17 18 20 Higher Mathematics Trigonometry Trigonometry 1 Radians EF Degrees are not the only units used to measure angles. The radian (RAD on the calculator) is a measurement also used. Degrees and radians bear the relationship: π radians = 180°. The other equivalences that you should become familiar with are: 30° = π6 radians 45° = π4 radians 60° = π3 radians 90° = π2 radians 135° = 34π radians 360° =2π radians. Converting between degrees and radians is straightforward. • To convert from degrees to radians, multiply by π × π and divide by 180. 180 Degrees • To convert from radians to degrees, multiply by 180 × 180 π and divide by π . Radians π = 5 π radians . For example, 50°= 50 × 180 18 2 Exact Values EF The following exact values must be known. You can do this by either memorising the two triangles involved, or memorising the table. DEG RAD 1 1 hsn.uk.net 2 1 sin x cos x tan x 0 0 0 1 0 30 π 6 π 4 3 2 1 2 1 3 45 60 π 3 1 2 3 90 π 2 1 2 1 2 3 2 1 0 – Page 1 1 Tip You’ll probably find it easier to remember the triangles. CfE Edition Higher Mathematics 3 Trigonometry Solving Trigonometric Equations RC 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 T C 360° − x ° 180° + x ° Since sin x° Remember The exact value triangle: positive First quadrant solution: 2 30° 3 x = sin −1 ( 12 ) = x 30 x = 30 = 30. or 180 − 30 or 150. is 60° 1 2. Solve cos x° = − 1 for 0 < x < 360 . 5 cos x° = − 1 5 180° − x ° x° S A T C 360° − x ° 180° + x ° negative Since cos x° is ( 5) x = cos −1 1 = 63·435 (to 3 d.p.). 180 − 63·435 or 180 + 63·435 x= x = 116·565 or 243·435. 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 2 CfE Edition Higher Mathematics Trigonometry 4. Solve tan x° = −5 for 0 < x < 360 . 180° − x ° x° tan x° = −5 S A T C 180° + x ° 360° − x ° Since tan x° is negative x = tan −1 ( 5 ) = 78·690 (to 3 d.p.). 180 − 78·690 or 360 − 78·690 x= x = 101·310 or 281·310. 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 3tan 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° 2sin 2 x ° =1 S A sin 2 x ° = 12 T C 360° − 2 x ° 180° + 2 x ° 2 x = sin −1 ( 12 ) = 30. = 2 x 30 or 180 − 30 or 360 + 30 or 360 + 180 − 30 or 360 + 360 + 30 Remember The exact value triangle: 2 30° 3 60° 1 2 x = 30 or 150 or 390 or 510 x = 15 or 75 or 195 or 255. hsn.uk.net 0 ≤ x ≤ 360 0 ≤ 2 x ≤ 720 Page 3 Note There are more solutions every 360°, since sin(30°) = sin(30°+360°) = … So keep adding 360 until 2x > 720. CfE Edition Higher Mathematics 6. Solve Trigonometry 2 cos 2 x = 1 where 0 ≤ x ≤ π . cos 2x = 1 2 2x π − 2x S A π + 2 x T C 2π − 2 x 0≤ x ≤π 0 ≤ 2 x ≤ 2π ( 2) Remember The exact value triangle: 2 x = cos −1 1 = π4 . = 2 x π4 or 2π − π4 2 π 4 π 4 1 1 or 2π + π4 2 x = π4 or 74π x = π8 or 78π . 7. Solve 4cos 2 x = 3 where 0 < x < 2π . ( cos x )2 = 34 cos x = ± 34 cos x = ± 23 x = π6 or x = π6 S A Since cos x can be positive or negative T C Remember The exact value triangle: ( ) x = cos −1 23 = π6 . or π − π6 or π + π6 or 2π − π6 2π + π6 or 56π or 76π or 116π . π 3 1 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 π 6 2 Page 4 CfE Edition 3 Higher Mathematics 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 < 2π . ( ) cos 2 x + π3 = 0·812 S A T C 0 < x < 2π 0 < 2 x < 4π π < 2 x + π < 4π + π 3 3 3 π 1·047 < 2 x + 3 < 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. 4 Trigonometry in Three Dimensions EF 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 5 CfE Edition Higher Mathematics Trigonometry EXAMPLE 1. The triangular prism ABCDEF is shown below. 3 cm F B E 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 = 3 cm A a 10 cm D D Opposite 3 = Adjacent 10 ( ) 3 a = tan −1 10 = 16.699° (or 0.291 radians) (to 3 d.p.). 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 + 62 6 cm = 136 A D 10 cm Therefore: E Opposite tan b = 3 = 3 cm Adjacent 136 C b A hsn.uk.net b = tan −1 ( 1363 ) = 14.426° (or 0.252 radians) (to 3 d.p.). 136 cm Page 6 CfE Edition Higher Mathematics 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 8 cm D 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. (c) Start with a sketch: F tan a = 8 cm A a 12 cm hsn.uk.net a = tan −1 23 = 33.690° (or 0.588 radians) (to 3 d.p.). Let AF and BE intersect at T TBA AB 33.690° = ATB is isosceles, so T= = 180° − ( 33.690° + 33.690° ) ATB Note = 112.620°. T • ( ) Note Angle GDC is the same size as angle FAB. B (d) Again, make a sketch: E F A Opposite 8 = Adjacent 12 • 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 7 CfE Edition Higher Mathematics 5 Trigonometry Compound Angles EF When we add or subtract angles, the result is 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 °. 2 30° 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 = sin ( π2 ) = 1. ) = RHS sin a cos b + cos a sin b = sin π6 cos π3 + cos π6 sin π3 ( = ( 12 × 12 ) + 23 × 23 = 14 + 34 = 1. ) Since LHS = RHS , the claim is true for a = π6 and b = π3 . hsn.uk.net Page 8 Remember The exact value triangle: π 6 2 π 3 1 CfE Edition 3 Higher Mathematics Trigonometry 3. Find the exact value of sin 75° . sin= 75° sin ( 45° + 30° ) = sin 45° cos30° + cos 45° sin30° ( ) ( = 1 × 23 + 1 × 12 2 2 ) 3 + 1 2 2 2 2 = 3 +1 2 2 = 6 +4 2 . = Finding Trigonometric Ratios You should already be familiar with the following formulae (SOH CAH TOA). Hypotenuse a Opposite 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. Opposite p Since sin a = q then: = Hypotenuse q p the side opposite a has length p; a 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 9 CfE Edition Higher Mathematics Trigonometry EXAMPLES 5 . Show that 4. Acute angles p and q are such that sin p = 54 and sinq = 13 63 . sin ( p + q ) = 65 5 sin p = 54 4 13 q 12 cos p = 35 p 3 5 5 sinq = 13 cosq = 12 13 + q ) sin p cos q + cos p sin q sin ( p= ( ) ( 3 5 = 54 × 12 13 + 5 × 13 ) Note Since “Show that” is used in the question, all of this working is required. 48 + 15 = 65 65 63 = 65 . Using compound angle formulae to confirm 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 ) 6. Show that = 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 x sin s sin t tan x . = + cos x cos s cos t = tan s + tan t . hsn.uk.net Page 10 CfE Edition Higher Mathematics 6 Trigonometry Double-Angle Formulae EF Using the compound angle identities with A = B , we obtain expressions for sin 2A and cos 2A . These are called double-angle formulae. sin 2 A = 2sin A cos A cos = 2 A cos 2 A − sin 2 A = 2cos 2 A − 1 = 1 − 2sin 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 θ sin 2θ = 2sin θ cos θ sinθ = 34 4 cos 2θ cos 2 θ − sin 2 θ = = 2 × 54 × 35 cosθ = 35 = 24 25 . 3 ( ) ( ) 3 2− 4 2 5 5 9 − 16 Note = 25 25 Any of the cos2A 7 . = − 25 formulae could have been used here. = 5 , where 0 < x < π , find the exact values of sin x 2. Given that cos 2x = 13 and cos x . Since cos 2 x = 1 − 2 sin 2 x , 5 1 − 2sin 2 x = 13 8 2sin 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 x 2 a= 2 13 − 2 2 = 2 13 is possible. 13 − 4 = 9 = 3. a So cos x = hsn.uk.net 3 . 13 Page 11 CfE Edition Higher Mathematics 7 Trigonometry Further Trigonometric Equations RC We will now consider trigonometric equations where double-angle formulae can be used to find solutions. These equations will involve: sin2x and either sin x or cos x ; Remember The double-angle cos2x and cos x ; formulae are given in cos2x 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 . Replace sin2x using the double angle formula 2sin x ° cos x ° + sin x ° =0 Take all terms to one side, making the sin x ° ( 2cos x ° + 1) =0 equation equal to zero Factorise the expression and solve 2cos x ° + 1 =0 sin x ° =0 S A T C cos x ° = − 12 x = 0 or 180 or 360 −1 1 x= 180 − 60 or 180 + 60 x = cos ( 2 ) 2sin x ° cos x ° = − sin x ° = 60. = 120 or 240. So x = 0 or 120 or 180 or 240 . Solving equations involving cos2x and cosx EXAMPLE 2. Solve cos2 x = cos x for 0 ≤ x ≤ 2π . cos 2 x = cos x 2cos 2 x − 1 = cos x 2cos 2 x − cos x − 1 = 0 0 ( 2cos x + 1)( cos x − 1) = 2cos x + 1 = 0 cos x = − 12 x= π − π3 or π + π3 = 23π or 43π Replace cos2x by 2 cos 2 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 12 CfE Edition Higher Mathematics Trigonometry Solving equations involving cos2x and sinx EXAMPLE 3. Solve cos2 x = sin x for 0 < x < 2π . 1 − 2sin 2 x = sin x 2sin 2 x + sin x − 1 = 0 0 ( 2sin x − 1)( sin x + 1) = 2sin x − 1 = 0 sin x = 12 = x π6 or π − π6 = π6 or 5π 6 Replace cos2x by 1 − 2 sin 2 x Take all terms to one side, making a quadratic equation in cos 2 x = sin x sin x Solve the quadratic equation (using factorisation or the quadratic formula) S A T C x = sin = π6 −1 sin x + 1 = 0 sin x = −1 ( ) 1 2 x = 32π . So x = π6 or 56π or 32π . hsn.uk.net Page 13 CfE Edition Higher Mathematics 8 Trigonometry Expressing pcosx + qsinx in the form kcos(x – a) EF An expression of the form p cos x + q sin x can be written in the form k cos ( x − a ) where: k sin a = k p 2 + q 2 and tan a = . k cos a The following example shows how to achieve this. EXAMPLES 1. Write 5cos x ° + 12sin x ° in the form k cos ( x ° − a ° ) where 0 ≤ a < 360 . Step 1 Expand k cos ( x − a ) using the compound angle formula. Step 2 Rearrange to compare with p cos x + q sin x . Step 3 Compare the coefficients of cos x and sin x with p cos x + q sin x . Step 4 Mark the quadrants on a CAST diagram, according to the signs of k cos a and k sin a . Step 5 Find k and a using the formulae above (a lies in the quadrant marked twice in Step 4). Step 6 State p cos x + q sin x in the form k cos ( x − a ) using these values. hsn.uk.net k cos ( x ° − a ° ) = k cos x ° cos a ° + k sin x ° sin a ° = k cos a ° cos x ° + k sin a ° sin x ° = k cos a ° cos x ° + k sin a ° sin x ° 12 5 k cos a ° =5 k sin a ° =12 180° − a ° a° S A T C 180° + a ° 360° − a ° = k 52 + 122 = 169 = 13 tan a ° = k sin a ° k cos a ° = 12 5 ( ) a = tan −1 12 5 . = 67 4 (to 1 d.p.) 5cos x ° + 12sin = x ° 13cos ( x ° − 67.4° ) Page 14 CfE Edition Higher Mathematics Trigonometry 2. Write 5cos x − 3sin x in the form k cos ( x − a ) where 0 ≤ a < 2π . 5cos x − 3sin x = k cos ( x − a ) = k cos x cos a + k sin x sin a = ( k cos a ) cos x + ( k sin a ) sin x k sin a = −3 π−a tan a = = 34 ( ) a Note tan −1 35 Make sure your = 0.540 (to 3 d.p.). calculator is in radian mode. So = a 2π − 0.540 = 5.743 (to 1 d.p.). S A T C π+a 2π − a Hence a is in the fourth quadrant. 34 cos ( x − 5.743 ) . Hence 5cos x − 3sin= x 9 k sin a = − 35 k cos a First quadrant answer is: 52 + ( −3 )2 = k k cos a = 5 Expressing pcosx + qsinx in other forms EF An expression in the form p cos x + q sin x can also be written in any of the following forms using a similar method: k cos ( x + a ) , k sin ( x + a ) . k sin ( x − a ) , EXAMPLES 1. Write 4cos x ° + 3sin x ° in the form k sin ( x ° + a ° ) where 0 ≤ a < 360 . 4cos x ° + 3sin = x ° k sin ( x ° + a ° ) = k sin x ° cos a ° + k cos x ° sin a ° = ( k cos a ° ) sin x ° + ( k sin a ° ) cos x °. = k k cos a ° =3 k sin a ° =4 180° − a ° 4 2 + 32 = 25 a° S A T C 180° + a ° 360° − a ° =5 tan a ° = So: k sin a ° 4 = k cos a ° 3 ( ) a = tan −1 34 = 53.1 (to 1 d.p.). Hence a is in the first quadrant. Hence 4cos x ° + 3sin = x ° 5sin ( x ° + 53.1° ) . hsn.uk.net Page 15 CfE Edition Higher Mathematics Trigonometry 2. Write cos x − 3 sin x in the form k cos ( x + a ) where 0 ≤ a < 2π . cos x − 3 sin x =k cos ( x + a ) = k cos x cos a − k sin x sin a = ( k cos a ) cos x − ( k sin a ) sin x . k cos a = 1 k= k sin a = 3 12 + ( − 3 ) = 1+ 3 π−a 2 So: a = tan −1 = 4 =2 a S A T C π+a 2π − a k sin a = k cos a = tan a = π3 . 3 ( 3) Hence a is in the first quadrant. ( ) Hence cos x − 3 sin x =2cos x + π3 . 10 Multiple Angles EF We can use the same method with expressions involving the same multiple angle, i.e. p cos (nx ) + q sin (nx ) , where n is a constant. EXAMPLE Write 5cos 2 x ° + 12sin 2 x ° in the form k sin ( 2 x ° + a ° ) where 0 ≤ a < 360 . 5cos 2 x ° + 12sin 2 x ° k sin ( 2 x ° + a ° ) = = k sin 2 x ° cos a ° + k cos 2 x ° sin a ° = = k k cos a ° =12 k sin a ° =5 180° − a ° ( k cos a ° ) sin 2 x ° + ( k sin a ° ) cos 2 x °. a° S A T C 180° + a ° 360° − a ° 122 + 52 = 169 = 13 = tan a ° So: k sin a ° 5 = k cos a ° 12 ( ) 5 a = tan −1 12 = 22.6 (to 1 d.p.). Hence a is in the first quadrant. Hence 5cos 2 x ° + 12sin = 2 x ° 13sin ( 2 x ° + 22.6° ) . hsn.uk.net Page 16 CfE Edition Higher Mathematics Trigonometry 11 Maximum and Minimum Values EF To work out the maximum or minimum values of p cos x + q sin x , we can rewrite it as a single trigonometric function, e.g. k cos ( x − a ) . Recall that the maximum value of the sine and cosine functions is 1, and their minimum is –1. y y y = cos x y = sin x max. = 1 max. = 1 1 1 x min. = –1 O –1 x min. = –1 O –1 EXAMPLE Write 4sin x + cos x in the form k cos ( x − a ) where 0 ≤ a ≤ 2π and state: (i) the maximum value and the value of 0 ≤ x < 2π at which it occurs (ii) the minimum value and the value of 0 ≤ x < 2π at which it occurs. 4sin x + cos x= k cos ( x − a ) = k cos x cos a + k sin x sin a = ( k cos a ) cos x + ( k sin a ) sin x . k = ( −1)2 + 42 k cos a = 1 k sin a = 4 π−a = 17 tan a = k sin a = 4 k cos a So: a = tan −1 ( 4 ) = 1.326 (to 3 d.p.). a S A T C π+a 2π − a Hence a is in the first quadrant. Hence 4sin x + cos x = 17 cos ( x − 1.326 ) . The minimum value of − 17 occurs when: cos ( x − 1.326 ) = −1 The maximum value of 17 occurs when: cos ( x − 1.326 ) = 1 cos −1 (1) x − 1.326 = 0 x − 1.326 = x 1.326 (to 3 d.p.). = hsn.uk.net Page 17 x − 1.326 = cos −1 ( −1) x − 1.326 = π x 4.468 (to 3 d.p.). = CfE Edition Higher Mathematics Trigonometry 12 Solving Equations RC The method of writing two trigonometric terms as one can be used to help solve equations involving both a sin ( nx ) and a cos ( nx ) term. EXAMPLES 1. Solve 5cos x ° + sin x ° = 2 where 0 ≤ x < 360 . First, we write 5cos x ° + sin x ° in the form k cos ( x ° − a ° ) : = x ° k cos ( x ° − a ° ) 5cos x ° + sin = k cos x ° cos a ° + k sin x ° sin a ° = ( k cos a ° ) cos x ° + ( k sin a ° ) sin x °. = k k cos a ° =5 k sin a ° =1 180° − a ° 52 + 12 = tan a ° = 26 a° So: k sin a ° 1 = k cos a ° 5 () a = tan −1 15 = 11.3 (to 1 d.p.). S A T C 180° + a ° 360° − a ° Hence a is in the first quadrant. Hence 5cos x °= + sin x ° 26 cos ( x ° − 11.3° ) . Now we use this to help solve the equation: x° 5cos x ° + sin x ° = 2 180° − x ° S A 26 cos ( x ° − 11.3° ) = 2 180° + x ° T C 360° − x ° cos ( x ° − 11.3° ) = 2 26 x − 11.3 = cos −1 2 ( 26 ) = x − 11.3 66.9 or 360 − 66.9 x − 11.3 = 66.9 or 293.1 x = 78.2 or 304.4. hsn.uk.net Page 18 = 66.9 (to 1d.p.). CfE Edition Higher Mathematics Trigonometry 2. Solve 2cos 2 x + 3sin 2 x = 1 where 0 ≤ x < 2π . First, we write 2cos 2 x + 3sin 2 x in the form k cos ( 2 x − a ) : 2cos 2 x + 3sin 2 x = k cos ( 2 x − a ) = k cos 2 x cos a + k sin 2 x sin a = ( k cos a ) cos 2 x + ( k sin a ) sin 2 x . k cos a = 2 = k 22 + ( −3 )2 = tan a k sin a = 3 = 4+9 So: π−a a S A T C π+a 2π − a = 13 k sin a 3 = k cos a 2 ( ) a = tan −1 32 = 0.983 (to 3 d.p.). Hence a is in the first quadrant. Hence 2cos 2 x + 3sin 2= x 13 cos ( 2 x − 0.983 ) . Now we use this to help solve the equation: 2x 2cos 2 x + 3sin 2 x = 1 π − 2x 0 < x < 2π S A 13 cos ( 2 x − 0.983 ) = 1 0 < 2 x < 4π π + 2 x T C 2π − 2 x 1 . cos ( 2 x − 0 983 ) = 13 2 x − 0.983 = cos −1 1 ( 13 ) = 1.290 (to 3 d.p.). 2 x −= 0.983 1.290 or 2π − 1.290 or 2π + 1.290 or 2π + 2π − 1.290 or 2π + 2π + 1.290 2 x − 0.983 = 1.290 or 4.993 or 7.573 or 11.276 2 x = 2.273 or 5.976 or 8.556 or 12.259 x = 1.137 or 2.988 or 4.278 or 6.130. hsn.uk.net Page 19 CfE Edition Higher Mathematics Trigonometry 13 Sketching Graphs of y = pcosx + qsinx EF Expressing p cos x + q sin x in the form k cos ( x − a ) enables us to sketch the graph= of y p cos x + q sin x . EXAMPLES 1. (a) Write 7 cos x ° + 6sin x ° in the form k cos ( x ° − a ° ) , 0 ≤ a < 360 . (b) Hence sketch the graph = of y 7 cos x ° + 6sin x ° for 0 ≤ x ≤ 360 . (a) First, we write 7 cos x ° + 6sin x ° in the form k cos ( x ° − a ° ) : = x ° k cos ( x ° − a ° ) 7 cos x ° + 6sin = k cos x ° cos a ° + k sin x ° sin a ° = ( k cos a ° ) cos x ° + ( k sin a ° ) sin x °. k cos a ° =7 = k 62 + 7 2 = tan a ° k sin a ° =6 = 36 + 49 So: 180° − a ° = 85 a° S A T C 180° + a ° 360° − a ° k sin a ° 6 = k cos a ° 7 ( ) a = tan −1 76 = 40.6 (to 1 d.p.). Hence a is in the first quadrant. Hence 7 cos x ° += 6sin x ° 85 cos ( x ° − 40.6° ) . (b) Now we can sketch the graph = of y 7 cos x ° + 6sin x ° : y y 7 cos x ° + 6 sin x ° = 85 x O − 85 40.6 hsn.uk.net 360 Page 20 CfE Edition Higher Mathematics Trigonometry 2. Sketch the graph of= y sin x ° + 3 cos x ° for 0 ≤ x ≤ 360 . First, we write sin x ° + 3 cos x ° in the form k cos ( x ° − a ° ) : sin x ° + 3 cos = x ° k cos ( x ° − a ° ) = k cos x ° cos a ° + k sin x ° sin a ° = ( k cos a ° ) cos x ° + ( k sin a ° ) sin x °. = k k cos a ° = 3 k sin a ° =1 180° − a ° 12 + 3 = 2 1+ 3 =2 a° S A T C 180° + a ° 360° − a ° = tan a ° k sin a ° = k cos a ° So: 1 3 ( 3) a = tan −1 1 = 30. Hence a is in the first quadrant. Hence sin x ° + 3 cos = x ° 2cos ( x ° − 30° ) . Now we can sketch the graph of= y sin x ° + 3 cos x ° : y = y sin x ° + 3 cos x ° 2 O −2 x 30 hsn.uk.net 360 Page 21 CfE Edition Higher Mathematics Trigonometry 3. (a) Write 5sin x ° − 11cos x ° in the form k sin ( x ° − a ° ) , 0 ≤ a < 360 . (b) Hence sketch the graph of = y 5sin x ° − 11cos x ° + 2 , 0 ≤ x ≤ 360 . (a) First, we write 5sin x ° − 11cos x ° in the form k sin ( x ° − a ° ) : = x ° k sin ( x ° − a ° ) 5sin x ° − 11cos = k sin x ° cos a ° − k cos x ° sin a ° = ( k cos a ° ) sin x ° − ( k sin a ° ) cos x °. k cos a ° =5 = k 52 + 11 k sin a ° = 11 = 25 + 11 180° − a ° a° S A T C 180° + a ° 360° − a ° = 36 =6 2 = tan a ° So: k sin a ° = k cos a ° 11 5 ( ) a = tan −1 11 5 = 33.6 (to 1 d.p.). Hence a is in the first quadrant. = x ° 6sin ( x ° − 33.6° ) . Hence 5sin x ° − 11cos (b) We can now sketch the graph of = y 5sin x ° − 11cos x °= + 2 6sin ( x ° − 33.6° ) + 2 : y 6 4 y 5sin x ° − 11cos x ° + 2 = 2 x O −4 −6 33.6 hsn.uk.net 360 Page 22 CfE Edition