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TRIGONOMETRY A 1. The Sine Rule a b c sin A sin B sin C b c B C a We can prove the sine rule like this. Let x and y be the perpendiculars from vertices A and B to the opposite sides a and b. Considering the two right-angled triangles with x as one of the A sides, x c x sin C b sin B b c x y B a C Comparing expressions for x, b sin C c sin B b c sin B sin C Similarly, considering the two right-angled triangles with y as one of the sides, y c y sin C a sin A Comparing expressions for y, a sin C c sin A a c sin A sin C …and so we have the sine rule. We now move onto using the sine rule. In order to use it we must have a combination of two sides and one angle, or one side and two angles. Example 1 : Find the missing angles and sides in the triangle below. A 31 B 40 12cm C The angles of the triangle add up to 180°. C 180 31 40 109 To find b, a b sin A sin B 12 b sin 31 sin 40 12sin 40 b sin 31 14.98 cm To find c, we use a rather than b since a is an integer and therefore more convenient. a c sin A sin C 12 c sin 31 sin109 12sin109 c sin 31 22.03 cm Example 2 : solve the triangle with A 30 , b = 5cm, a = 6cm. Construction : draw the base b and a line leaving A at an angle of 30°. Set compasses to 6cm and draw an arc from C. The point of intersection of the arc and the line leaving A is the vertex B. B 6cm A 30 5cm C Solving using the sine rule (we put the sines in the numerators to make the algebra easier), sin A sin B b 1a– 0.5 – 0.5 0.5 0.5 1 90 180 sin 30 sin B 6 5 5sin 30 sin B 6 0.416 A look at the graph of the sine function tells us that there are two possible solutions to this equation in between 0° and 180°. Solving the equation we have B 24.6 and B 180 24.6 155.4 However B cannot be 155.4°, since A is already 30°, so B = 24·6° and C 180 30 24.6 125.4 . 1 0.5 90 – 0.5 24.6 180 155.4 Finally we calculate c. a c sin A sin C 6 c sin 30 sin125.4 6sin125.4 c sin 30 9.78 cm C2 p20 Ex 2A, p22 Ex 2B Example 3 : Now for an ambiguous case - solve the triangle with A 30 , b = 6cm, a = 4cm. Construction : very similar to the previous example, although note that now the arc crosses the line leaving A at two different points. This means that there are two solutions to the triangle. 1– 0.5 – 0.5 0.5 1 0.5 90 180 B 4cm B A 4cm 30 6cm C sin A sin B a b sin 30 sin B 4 6 6sin 30 sin B 4 0.75 B 48.6, B 131.4 1 0.5 90 – 0.5 48.6 180 131.4 Both of these angles are possible, as they are less than 180 30 150 , so we consider each case. • If B = 48.6°, then C = 180° − 30° − 48.6° = 100.4°. To find c, a c sin A sin C 4 c sin 30 sin100.4 4sin100.4 c sin 30 7.84 cm • If B = 131.4°, then C = 180° − 30° − 131.4° = 18.6°. To find c, a c sin A sin C 4 c sin 30 sin18.6 4sin18.6 c sin 30 2.55 cm C2 p23 Ex 2C 2. The Cosine Rule In a right angled triangle, c 2 a 2 b 2 . However, for the acute-angled triangle shown below, it is clear that c 2 a 2 b 2 . And for the obtuse-angled triangle, c 2 a 2 b 2 . Therefore, we need a modified version of Pythagoras’ theorem for non-right angled triangles. c c b c b a a a acute-angled right-angled b C C obtuse-angled In fact the formula is the cosine rule c 2 a 2 b 2 2bc cos C Notice how if C is acute, then cosC is positive, and c 2 a 2 b 2 . Similarly, if C is obtuse, then cosC is negative, and c 2 a 2 b 2 . In fact the formula appears in three forms… a 2 b 2 c 2 2bc cos A b 2 a 2 c 2 2ac cos B c 2 a 2 b 2 2ab cos C To prove the cosine rule, drop a perpendicular of length h from C to the line AB. Let the base of the unshaded triangle be x. Notice that x b cos A . Using Pythagoras’ rule on the unshaded triangle, b x h b 2 2 a h 2 h2 b2 x 2 Using Pythagoras’ rule on the shaded triangle, x A (c x) h a 2 C 2 2 Substituting h 2 from the first equation, (c x ) 2 b 2 x 2 a 2 c 2 2cx x 2 b 2 x 2 a 2 a 2 b 2 c 2 2cx a 2 b 2 c 2 2bc cos A c B Because the formula contains three sides and one angle, we can use it if we know three sides, or two sides and the angle they enclose. Example 1 : Solve the triangle with a = 8cm, b = 6cm and c = 4cm. To find A, A a 2 b 2 c 2 2bc cos A 1– 1 0.5 0.5 –1 0.5 82 62 42 2 6 10.5 904 cos A 180 64 52 48cos A 6cm 4cm 12 48 A 104.5 cos A B We see from the cosine function that there is only one solution between 0° and 180°, and so we do not have to worry about any ambiguous cases. This is consistent with our experience that three sides uniquely define a triangle. It also means that the cosine rule may be less prone to human error than the sine rule! C 8cm 1 104.5 0.5 90 180 – 0.5 –1 To find B, we can use either the cosine rule or the sine rule. The cosine rule is marginally preferable as with the sine rule we would have to use the calculated value of A, rounding it off in the process. Choosing the cosine rule, b 2 a 2 c 2 2ac cos B 62 82 42 2 8 4 cos B 36 80 64 cos B 44 64 B 46.6 cos B Finally we calculate C, C 180 104.5 46.6 29.0 If this answer looks like it should be 28.9, then remember we must use the full calculator values for A and B, not the rounded off ones. Example 2 : Solve the triangle with a = 7cm, b = 5cm and C = 119°. To find c, C c 2 a 2 b 2 2bc cos C 7 2 52 2 7 5cos119 74 70 cos119 c 10.39 cm 5cm A 119 7cm B We can now use the sine or the cosine rule to calculate A or B. Choosing the sine rule for A, sin A sin C a c sin A sin119 7 10.39 7 sin119 sin A 10.39 A 36.1, A 143.9 We reject the solution A = 143.9° as there are not enough degrees in the triangle. Finally, B 180 119 36.1 24.9 C2 p31 Ex 2D, p29 Ex 2E 3. Applications of the Sine and Cosine Rules Example 1 : Find the angle between the hour hand and the minute hand of a clock at 12.15. If the hour hand is 8cm long and the minute hand is 10cm long, how far apart are the tips of the hands at this time? The hour hand covers 360° ÷ 12 = 30° in one hour, and therefore 30° ÷ 4 = 7.5° in quarter of an hour. So the angle between the hands is 90° – 7.5° = 82.5°. C By the cosine rule, a 2 b 2 c 2 2bc cos A 10 8 2 10 8cos82.5 164 160 cos82.5 a 11.96 cm 2 8cm 2 A 82.5 B 10cm Example 2 : A boat is sailing directly towards the foot of a cliff. The angle of elevation of a point on top of the cliff, and directly ahead of the boat, increases from 8° to 12° as the boat sails 100m. Find a) the original distance from the boat to the point on top of the cliff b) the height of the cliff. C a) We have A = 168°, B = 8°, c = 100m. C 180 168 8 4 cliff 12 168 A 8 100m B To find a, a c sin A sin C a 100 sin 68 sin 4 100sin168 a sin 4 298.05 m The original distance from the boat to the top of the cliff is 298.05m. b) To find the height of the cliff, height of cliff a sin B 298.05sin 8 41.48 m Example 3 : A ship leaves a port P and sails 5 nautical miles on a bearing of 052°, followed by 6 nautical miles on a bearing of 205° to arrive at the next port Q. Find the distance and bearing of P from Q. To find the angle X of the triangle, we use the property of ‘interior angles’. X X 380 205 (180 52) 27 Using the cosine rule, x p q 2 pq cos X 2 2 5 miles 52 P 2 62 52 2 6 5cos 27 61 60 cos 27 x 2.75 miles 205 6 miles Q Using the sine rule, sin Q sin X q x sin Q sin 27 5 2.75 5sin 27 sin Q 2.75 Q 55.8, Q 124.2 We reject the solution Q 124.2 as it does not fit with our diagram. The bearing of P from Q is given by 360 (55.8 25) 329.2 . C2 p31 Ex 2F 4. The Area of a Triangle Suppose we label the angles of a triangle A, B and C. Then we label the side opposite angle A as a, the side opposite angle B as b, and the side opposite angle C as c. We can find a formula for the area of a triangle using this notation. A area of triangle 12 base height b c 12 a b sin C 12 ab sin C h B C a Clearly we could also have these alternative expressions for the area : 12 bc sin A or 12 ac sin B . Example 1 : Find the area of this triangle. C 7cm 100 A B 5cm area of triangle 12 ac sin B 12 7 5 sin100 17.23 cm2 Example 2 : The area of a triangle is 36 cm². Two sides of the triangle are 8cm and 10cm. Find the two possible angles between these two sides. area of triangle 12 bc sin A 1– 0.5 – 0.5 0.5 1 0.5 90 180 C 36 8 10 sin 1 2 10cm 36 40 64.2, 115.8, sin A 8cm B A small sketch of the sine curve between 0° and 180° illustrates the last step in the calculation. 1 0.5 90 – 0.5 64.2 180 115.8 C2 p34 Ex 2G 5. Radian Measure One radian is defined as the angle subtended at the centre of a circle, by an arc equal in length to its radius. In general, arc length angle in radians radius Therefore, for one revolution, 2 r angle in radians 2 . r So 2π radians = 360°, and one radian is approximately 57°. r 1 rad r r It is important to be able to convert between degrees and radians fluently. The table below shows some common angles in both units. degrees 0 30 45 60 90 120 135 150 180 240 270 300 360 2 3 5 4 3 5 2 radians 0 3 4 6 3 2 3 6 4 3 2 The symbol for radians is a small superscript c, and so 2 radians is written 2c or 2 rad. If units are omitted, then an angle is in radians, not degrees! Example 1 : Convert the following angles into degrees. a) 79 rad b) 2 rad a) 7 9 180 140 b) 2 180 114.6 (1 d.p.) Example 2 : Convert the following angles into radians. a) 24° b) 7° 7 152 180 or 0.122 a) 24 b) 7 180 180 When angles are measured in radians, we have several convenient formulas concerning circles… C2 p81 Ex 6A 6. Arc Length Consider the arc PQ, of length s, subtending angle θ radians at the P centre of the circle. Now r arc length angle in radians radius O s s r r Q And so s r So a length of arc is found by multiplying the radius by the angle (measured in radians) subtended by the arc. In fact, two points P and Q on a circle define two arcs, a shorter one (known as the minor arc) and a longer one (known as the major arc). Example 1 : Find the length of the minor arc PQ, We first write 135° as 34 radians. P s r 8 34 O 135 6 cm 8cm Q Example 2 : Find the perimeter of the shape opposite. The perimeter consists of two arcs and two straight lines. We first write 60° as 13 radians. 2cm perimeter 5 7 2 2 1 3 1 3 4 4 cm 60 Example 3 : A road tunnel 60m long has the cross-section shown opposite, consisting of an arc centre O and a straight line. It is required to paint the inside of the tunnel (not including the road itself). What area is to be painted? We first find the angle θ. Splitting the isosceles triangle into two right-angled triangles gives us sin 1.25 1.5 1 2 1 2 0.985 1.970 radians The arc subtends an angle of 2 1.970 radians. arc length 1.5 (2 1.970) 2.5 6.469 m The internal area is therefore area 6.469 60 388.2 m2 C2 p83 Ex 6B 5cm O 1.5m 2.5m 1.5m 7. Area of a Sector Similarly, we also have minor sectors and major sectors. Consider the sector OPQ, with area A. Now area of sector fraction of circle P area of circle r A 2 O A 2 r And so r A 12 r 2 Q P Example 1 : Find the area of the major sector OPQ, The major sector subtends 2 34 54 . A 12 r 2 8 1 2 2 5 4 O 135 8cm 40 cm 2 Q Example 2 : A shelf has the shape of a sector of a circle of radius 8cm, with a concentric sector of radius 5cm removed, as shown in the diagram below. If the shelf has an area of 50cm2, find the angle subtended by the sectors. 1 2 82 12 52 50 19.5 50 3cm 5cm 2.56 radians Example 3 : Find a formula for the area A of the segment below. Hence find the area of the segment if r = 6cm and θ = 1.5 rad. P area of segment area of sector area of triangle r 12 r 2 12 ab sin C 12 r 2 12 r 2 sin 12 r 2 ( sin ) Substituting, area of segment 12 2 62 (1.5 sin1.5) 9.045 cm 2 O A r Q Example 4 : This shape is made up of arcs centered on the corners of an equilateral triangle of side length r. Show that the area of the shape is given by 12 r 2 3 . area 3 sector 2 triangle 3 12 r 2 12 r 2 2 1 3 1 2 r 2 sin r 1 3 3 2 r 2 12 r 2 3 Example 5 : The diagram shows a triangle, and an arc whose centre is the point P. a) Show that 1.330 rad 5cm b) Find the perimeter of the region R. 4cm c) Find the area of the region R. P R 10cm 4cm 3cm a) Using the cosine rule, a 2 b 2 c 2 2bc cos A 102 92 7 2 2 9 7 cos 100 81 49 126 cos 30 126 1.330... rad We only get the answer in radians if our calculator is in radian mode! cos b) The perimeter is an arc length added to three straight edges. perimeter 4 1.330... 3 10 5 23.32 cm (2 d.p.) c) The area is the triangle subtract the sector. area 12 9 7 sin1.330... 12 42 1.330... 30.594... 10.643... 19.95 cm 2 (2 d.p.) Here it is essential to have our calculator is in radian mode. C2 p87 Ex 6C 8. Sine, Cosine and Tangent of Common Angles Sines, cosines and tangents are commonly used with right-angled triangles. opp adj opp sin cos tan hyp hyp adj hyp opp adj Activity 1 : Do the Sine, Cosine and Tangent of Common Angles worksheet. Encourage students to learn these values. sin cos tan 0° 0 1 0 1 1 3 30° 2 3 2 45° 60° 90° 1 2 1 2 3 2 1 1 2 0 1 3 – y 9. The Unit Circle However sines, cosines and tangents also have a close relationship with the unit circle. The diagram shows a circle whose centre O is at (0, 0), and whose radius is 1 unit. Let P be a point on the unit circle. If we want to tell someone exactly where P is, we can give – 1 the angle between the positive x-axis and the line OP. We shall call this angle θ, and we shall always measure it anticlockwise, starting from the positive x-axis. 1 P 1 O x –1 Another way to specify the position of P is to give its coordinates. We can find them by considering the right-angled triangle in the diagram below. The x-coordinate is the distance OA. Now adj cos hyp OA 1 OA The y-coordinate is the distance AP. Now opp hyp AP 1 AP y 1 P –1 O sin –1 1 A x So the coordinates of P are (cos , sin ) . So we define sine and cosine as the y and x opp adj coordinates of the point P, rather than sin and cos . hyp hyp The unit circle is divided by the x and y axes into four quadrants, which are numbered anticlockwise as shown here. If the angle θ is greater than 90°, then P will be in the 2nd, 3rd or 4th quadrants. We can now give a meaning to sines and cosines of angles greater than 90°, by simply saying that –1 cos x coordinate of P sin y coordinate of P y 1 1st 2nd 1 O 4th 3rd For example, when θ = 130°, we find from a calculator that cos θ = –0.643, and sin θ = 0.766. We can see why by looking at the unit circle : when θ = 130°, the coordinates of P are (–0.643, 0.766). x –1 Activity 2 : Use the Unit Circle worksheet to find the coordinates of points on the unit circle at angles 50°, 120°, 255° and 305°. Use calculators to check that the coordinates are indeed (cos θ, sin θ). Alternatively, let students choose their own points within each quadrant. As an extension, investigate the sine and cosine of 430° and –65°. 0.5 1 – 0.5 0.5 1 –1 0.5 1 90 180 270 360 10. The Sine Function The sine of θ is the y-coordinate of the point P on the unit circle. Because of this, it is clear that... y x sin 0° = 0, sin 90° = 1, sin 180° = 0, sin 270° = –1, sin 360° = 0. If we plot a graph of sin θ against θ, we find it looks like this. sin y 1 0.5 90 x – 0.5 – 1 180 270 360 180 360 540 720 – 360 180 But after 360°, P is back to the start of the circle (1, 0) again. So the graph repeats itself every 360°. We say it is periodic, with period 360°. sin 1 0.5 – 360 – 180 180 360 540 720 – 0.5 –1 This graph is called the sine function. 0.5 1 – 0.5 0.5 1 –1 0.5 1 90 180 270 360 y 11. The Cosine Function The cosine of θ is the x-coordinate of the point P on the unit circle. Because of this, it is clear that... x cos 0° = 1, cos 90° = 0, cos 180° = –1, cos 270° = 0, cos 360° = 1. If we plot a graph of cos θ against θ, we find it looks like this. cos x 1 0.5 90 y 1– 1 0.5 0.5 180 360 540 720 – 360 180 180 270 360 – 0.5 –1 Clearly the cosine function is periodic, with period 360°, in exactly the same way as the sine function. cos 1 0.5 – 360 – 180 180 – 0.5 –1 This graph is called the cosine function. 360 540 720 0.5 2 1.5 1 – 0.5 1 1.5 0.5 1 1.5 2 –2 0.5 1 1.5 2 90 180 270 360 12. The Tangent Function There are two ways of thinking about the tangent function. Firstly, we can draw the tangent with equation x 1 to the unit circle. Now consider the line OP extrapolated until it cuts the tangent at B. Let the point A be (1, 0). y 1 P opp –1 adj AB 1 AB So the tangent of is the distance up the tangent between (1, 0) and the place where the extrapolated radius cuts the tangent. O tan –1 Activity 3 : Use the Tangent Function worksheet. This has extrapolated radii every 15° up to 360°. This will make the fact that we have asymptotes at 90° and 270°, and that the period of the tangent graph is 180°, very obvious! tan y 4– 4 2 2 180 360 – 360 180 2 1.5 1 0.5 x 90 – 0.5 –1 – 1.5 –2 180 Notice that the period of the tangent function is 180°. tan 4 2 – 360 – 180 180 –2 –4 360 270 360 B A 1 x Alternatively, we can define tangent in terms of sine and cosine. For a right-angled triangle, we have opp opp hyp sin tan adj adj cos hyp hyp opp adj So the tangent of θ is found by dividing the sine of θ by the cosine of θ. 1– 1 0.5 0.5 90 180 – 180 90 Activity 4 : Now distribute the Handy Graphs of Trigonometric Functions worksheet for future reference. Look carefully at the sine and cosine graphs and then the tangent graph. sin Discuss how the graphs confirm that tan . cos 13. Sine, Cosine and Tangent of any Angle From the unit circle we can see that cos(–50°) is the same as cos 50°. This is also clear from the graph. cos y 1 0.5 50 50 – 180 x – 90 90 180 – 0.5 –1 1– 1 0.5 0.5 90 180 – 180 90 In general, cos( ) cos . This means the cosine function is symmetrical about the y-axis. (Functions with this property are known as even functions). From the unit circle, we can see that sin(–50°) is the exact negative of sin50°. This is also clear from the graph. sin y 1 0.5 50 50 x – 180 – 90 90 180 – 0.5 –1 In general, sin( ) sin . This means the sine function has rotational symmetry about the origin. (Functions with this property are known as odd functions). 1– 1 0.5 0.5 90 180 – 180 90 We also have sin( ) cos( ) sin cos tan So the tangent function is also an odd function, with rotational symmetry about the origin. This is confirmed by the graph of the tangent function. tan( ) Example 1 : Write the following as trigonometric functions of acute angles a) cos(−110°) b) sin 220° c) tan(−110°) a) cos(−110°) = cos 110° = −cos 70° cos y 1 0.5 1– 0.5 –1 0.5 1 1 0.5 0.5 90 180 270 360 70 x – 180 – 90 90 180 – 0.5 –1 b) sin 220° = −sin 140° = −sin 40° sin y 1 0.5 40 90 x 180 270 360 – 0.5 –1 c) tan(−110°) = tan 70°, since tan has a period of 180°. Example 2 : Write down the values of these in surd form. a) sin 120° b) cos 300° c) tan 300° sin120 sin 60 tan 300 tan(60) a) c) b) tan 60 3 2 d) d) sin 330° e) cos 225° cos 225 cos135 e) cos 45 3 sin 330 sin( 30) sin 30 12 C2 p113 Ex 8B, p117 Ex 8C, p118 Ex 8D, p120 Ex 8E 1 2 14. The Pythagorean Relationship Consider the right-angled triangle with hypotenuse 1. The opposite has length sin , and the adjacent has length cos . Using Pythagoras’ rule we have sin 2 cos 2 1 1 sin cos With this Pythagorean relationship, we can prove other identities (and solve certain trigonometric equations - see next section). Example 1 : Prove the following identities. a) (sin cos )2 (sin cos )2 2 . 1 . cos 2 x c) cos 4 sin 4 1 2 cos 2 . It is usual to take the more ‘complicated’ side (often the LHS), and reduce it to the form of the ‘simpler’ side. a) (sin cos ) 2 (sin cos ) 2 sin 2 2sin cos cos 2 sin 2 2sin cos cos 2 b) tan 2 x 1 2 sin 2 cos 2 2 1 2 b) c) 2 sin x 1 cos 2 x sin 2 x cos 2 x cos 2 x 1 cos 2 x tan 2 x 1 cos 4 sin 4 1 cos 2 sin 2 cos 2 sin 2 1 cos 2 sin 2 1 cos 2 cos 2 2 cos 2 Example 2 : Given that x cos and y tan , express y in terms of x. y tan sin cos 1 cos 2 cos 1 x2 x Example 3 : Given that cos 23 , and that θ is an acute angle, find sin and tan . We can use this right-angled triangle. A simple application of Pythagoras shows that the opposite is 5 . We now have 3 5 3 5 tan 2 sin 5 2 Example 4 : Given that cos 23 , and that θ is a reflex angle, find sin and tan . Since θ is reflex, we cannot use a right-angled triangle, and so we use the identity… sin 2 cos 2 1 4 1 9 5 sin 2 9 sin 2 5 3 Notice we take the negative root here. This is because the sine of a reflex angle is negative. sin tan cos sin 5 3 2 3 5 2 Example 5 : Given that tan 247 , and that θ is an acute angle, find sin and cos . How would your answers change if θ is reflex? We can use this right-angled triangle. sin 7 25 cos 24 25 25 24 If θ is reflex, then because tan is positive, θ must be in the third quadrant, and so both sin and cos are negative. We therefore have sin 257 cos 24 25 C2 p145 Ex 10A 7 1– 1 0.5 0.5 90 180 270 360 15. Solving Simple Trigonometric Equations Always do a sketch graph, unless the equation involves tan. Example 1 : Solve the equation sin x 0.2, 0 x 360 . sinx 1 From the calculator we have x 11.5 , and by the symmetry of the sine function we also have 1 0.5 – 0.5 0.5 1 –1 0.5 1 x 180 11.5 168.5 90 180 270 360 0.5 90 180 270 360 x 90 180 270 360 – 0.5 –1 Example 2 : Solve the equation cos x 0.6, 0 x 360 . From the calculator we have x 126.9 , and by the symmetry of the cosine function we also have x 360 126.9 233.1 cosx 1 0.5 x – 0.5 –1 Example 3 : Solve the equation tan x 2.6, 0 x 360 1– 0.5 –but 0.5 this solution is not in the required range. By the 1 0.5 1 From the calculator we have x 69.0 , 10.5 90 180 270 360 periodicity of the tangent function we have the solutions x 69.0 180 111.0 and x 111.0 180 291.0 . Notice that we do not need to sketch the tangent function, as it is very simple to ‘hop’ from one solution to another (simply add or subtract 180°). Note that simple sine, cosine and tangent equations like the ones above always have two solutions in a 360° range (unless we are solving something like sin x 1 ). Example 4 : Solve sin x 90 34 , 0 x 360 . Although the equation involves the sine of x 90 , we still sketch the ordinary sine curve. The values on the x-axis are 0.5 1 –1 0.5 the relevant values of x 90 . 90 180 – 180 90 x 90 48.6, 131.4 x 138.6, 221.4 sinx 1 0.5 90 180 270 360 x 90 180 x – 0.5 –1 cosx Example 5 : Solve cos x 30 , 180 x 180 . 1 2 1 x 30 120, 120 x 150, 90 0.5 – 180 – 90 – 0.5 –1 1– 1 0.5 0.5 90 180 – 180 90 Example 6 : Solve sin x 60 12 , 180 x 180 . sinx 1 x 60 30, 150 , 210 1– 1 0.5 0.5 0.5 1 –1 0.5 x 90, 210 , 150 90 180 270 360 0.5 We notice that adding 60° will take one of the solutions out of range, so we go back a period from 150°. – 180 – 90 90 180 x – 0.5 – 1 cosx Example 7 : Solve cos x 50 0.8, 0 x 360 . 0.5 1 – 0.5 0.5 1 –1 0.5 1 90 180 270 360 x 50 36.9 , 323.1,396.9 1 0.5 x 13.1 , 273.1,346.9 We notice that subtracting 50° will take one of the solutions out of range, so we go forward a period from 36·9°. Example 8 : Solve sin 2 x 3 , 0 x 360 . 2 2 x 60, 120, 420, 480 180 270 360 90 180 270 360 x 90 180 x – 0.5 –1 sinx 1 0.5 x 30, 60, 210, 240 Notice that since we divide by two, we have twice as many solutions as normal. This is because the graph of y sin 2 x is a horizontally ‘squashed’ version of the y sin x graph. x 90 – 0.5 –1 Example 9 : Solve tan 3x 15 1, 0 x 360 This time, we need to find six solutions. 3x 15 45, 225, 405, 585, 765, 945 0.5 1 –1 0.5 90 90 , 600, 780, 960 180 3x 60, 240180 ,– 420 x 20, 80, 140, 200, 260, 320 C2 p148 Ex 10B, p150 Ex 10C Example 10 : Solve 3cos2 x cos x 0, 180 x 180 . cosx 1 0.5 – 180 – 90 – 0.5 –1 3cos 2 x cos x 0 cos x(3cos x 1) 0 cos x 0 x 90 cos x 13 x 109.5 Notice that we factorise in the working above. Never divide by cos x , as this loses the solutions of cos x 0 . Example 11 : Solve 2 tan 2 x tan x 3, 180 x 180 2 tan 2 x tan x 3 2 tan 2 x tan x 3 0 (2 tan x 3)(tan x 1) 0 1– 0.5 –1 0.5 1 1 0.5 0.5 90 180 270 360 tan x 32 56.3, 123.7 tan x 1 x 135, 45 Example 12 : Solve tan x 2sin x, 0 x 360 tan x 2sin x cosx sin x 2sin x cos x sin x 2sin x cos x 2sin x cos x sin x 0 1 0.5 sin x(2 cos x 1) 0 sin x 0 x 0, x 180, x 360 90 180 270 360 x – 0.5 cos x 12 x 60, x 300 –1 Example 13 : Solve tan 2 x 3, 0 x 2 . Notice that as there is no ° symbol, the angles are measured in radians. The 2 is another big clue. tan 2 x 3 tan x 3 1– 0.5 –1 0.5 1 1 0.5 0.5 tan x 23 3 x 13 , 22 tan x 3 x 23 , 4 3 5 3 To get from one solution to the next we add π radians, as this is the radian equivalent of 180°, the period of the tangent function. Example 14 : Solve sin 2 x 16 12 , 0 x 2 sinx 1 0.5 – 0.5 –1 2 3 2 2 x sin x sin 2 x 16 1 2 1 2 1 1 x 6 4 , 43 , 1 6 x 121 , 7 12 , 5 4 , 13 12 7 4 , 19 12 22 Example 15 : Solve 4 sin 4cos2 , 0 2 Using the Pythagorean identity, sin 4 sin 4 1 sin 2 1 4 sin 4 4sin 2 0.5 4sin 2 sin 0 sin (4sin 1) 0 sin 0 sin 14 1– 0.5 –1 0.5 1 1 0.5 0, , 0.5 3 2 2 22 0.253, 2.889 – 0.5 (3 d.p.) 2 3 2 2 –1 The last two solutions are best obtained by having your calculator in radian mode. Example 16 : Solve 8 tan 3cos , 0 2 . 8 tan 3cos sin 8sin 3cos cos 8sin 3cos 2 1 0.5 8sin 3 1 sin 2 3sin 8sin 3 0 – 0.5 3 2 3 2 2 2 2 (3sin 1)(sin 3) 0 1– 0.5 –1 0.5 1 1 0.5 0.5 3 2 22 –1 sin 13 0.340, 2.802 sin 3 no solutions Example 17 : Solve sin tan 3, 0 2 sin tan 3 sin cos sin 3 cos sin 2 3cos 1 0.5 1 cos 2 3cos – 0.5 cos 2 3cos 1 0 3 32 4 1 1 2 1 3 13 2 3 13 cos 1.263, 5.020 2 3 13 cos no solutions 2 cos C2 p153 Ex 10D –1 2 2 16. Transformations of Trigonometric Graphs recommended that Autograph and a projector are used to deliver this section of the notes. Students can discuss the effect of a change in the function before it is plotted. The animation feature is also very useful. a) Translation • If we translate the graph of y sin x by a units vertically upwards, we get the graph of y sin x a . 3– 2 1 2 1 90 180 270 360 –It360 90is 180 270 y 3 y = sinx + 2 2 1 y = sinx – 360 – 270 – 180 – 90 90 180 270 360 x y = sinx – 1 –1 –2 1– 1 0.5 0.5 90 180 270 360 – 360 90 180 270 • If we translate the graph of y sin x by a units to the right, we get the graph of y sin( x a) . To understand this, consider the graph of y sin( x 30) . We can think of this as clock running 30 minutes slow. When x is 30°, we are taking the sine of only 30 30 0 . So our graph is doing at 30° what the ordinary sine graph does at 0°, and so the whole graph is shifted 30° to the right. Similar reasoning means that if we translate the graph of y sin x by a units to the left, we get the graph of y sin( x a) . y 1 y = sin(x + 30) y = sinx 0.5 y = sin(x – 30) – 360 – 270 – 180 – 90 – 0.5 2– 2 1 1 90 180 270 360 – 360 90 180 270 90 180 270 360 x – 1 b) Stretches • If we stretch the graph of y sin x vertically by scale factor a, we get the graph of y a sin x . y 2 y = 2sinx 1 y = sinx – 360 – 270 – 180 – 90 90 180 –1 –2 1 y = sinx 2 270 360 x 1– 1 0.5 0.5 90 180 270 360 – 360 90• 180 270 If we stretch the graph of y sin x horizontally by scale factor a, we get the graph of x y sin .Consider the situation when a 2 . We can think of this as a clock running at a half its normal speed. When x is 180°, we are taking the sine of only 90°. So our graph is doing at 180° what the ordinary sine graph does at 90°, and so the whole graph is twice as wide. y y = sin 1 x 2 y = sinx 0.5 – 360 – 270 – 180 – 90 – 0.5 1– 1 0.5 0.5 90 180 270 360 – 360 90 180 270 90 180 270 360 x –1 • In a similar way, the graph of y sin 2 x is a horizontally ‘squashed’ version of y sin x . We can think of this as a clock running at twice its normal speed. When x is 90°, we are taking the sine of 180°. So our graph is doing at 90° what the ordinary sine graph does at 180°, and so the whole graph is half as wide. y y = sin 2x 1 y = sinx 0.5 – 360 – 270 – 180 – 90 – 0.5 1– 1 0.5 0.5 90 180 270 360 – 360 90 180 270 90 180 270 360 x – 1 c) Reflection • Reflection in the x-axis changes the sign of the y-coordinate. So y sin x becomes y sin x on reflection in the x-axis. y = – sinx y 1 y = sinx 0.5 – 360 – 270 – 180 – 90 – 0.5 –1 90 180 270 360 x 1 3 2 1 2 1– 2 3 –3 1 2 3 90 180 270 360 When sketching transformed trig curves, always indicate maxima, minima, and points of intersection with the axes. Example 1 : Sketch the graph of y 3sin 2 x for 0 x 360 . 1 vertical stretch S.F. 3 2 y sin x y sin 2 x y 3sin 2 x horizontal stretch S.F. y (225, 3) (45, 3) 3 2 1 90 180 270 360 x –1 –2 2 1 3 1 1– 2 3 –1 90 180 270 360 –3 (315, -3) (135, -3) Example 2 : Sketch the graph of y 1 2 cos x for 0 x 360 . vertical stretch S.F. 2 translation 1 unit up y cos x y 2cos x y 1 2cos x The graph crosses the x-axis when y 3 1 2 cos x 0 cos x 12 (360, 3) (0, 3) 2 x 120, x 240 (120, 0) 1 90 1 2 1 2 3 1– 2 –4 1 2 3 4 90 180 270 360 180 360 x 270 –1 (240, 0) (180, -1) Example 3 : Sketch the graph of y 3sin x 1 for 0 x 360 . vertical stretch S.F. 3 translation 1 unit down y sin x y 3sin x y 3sin x 1 The graph crosses the x-axis when 3sin x 1 0 sin x 1 3 x 19.5, x 160.5 y (90, 2) 2 (160.5, 0) 1 (0, -1) x 90 180 270 –1 –2 (19.5, 0) –3 –4 (270, -4) 360 4– 2 –4 2 4 4 2 2 90 180 270 360 Example 4 : Sketch the graph of y tan x 1 for 0 x 360 . translation 1 unit down y tan x y tan x 1 The graph crosses the x-axis when y 4 tan x 1 0 tan x 1 2 x 45, x 225 (225, 0) 90 (0, -1) –2 1 3 2 1– 2 3 –1 1 90 180 270 360 180 270 360 x (45, 0) –4 Example 5 : Sketch the graph of y 1 2 cos( x 90) for 0 x 360 . translation 90 left vertical stretch S.F. 2 y cos x y cos( x 90) y 2cos( x 90) translation 1 unit up y 1 2cos( x 90) The graph crosses the x-axis when 1 2 cos( x 90) 0 cos( x 90) 12 x 90 120, 240 x 30, x 150 y (270, 3) 3 2 (0, 1) 1 (30, 0) 90 –1 (90, -1) 180 270 360 x (150, 0) Example 6 : Sketch the graph of y tan(45 x) for 0 x 360 . Whenever the function has a negative x, we must make it positive by using one of the following relationships. Remember that cos is different! 4– 2 –4 2 4 4 2 2 90 180 270 360 sin( x) sin x cos( x) cos x tan( x) tan x So we rewrite our function as y tan( x 45) . translation 45 right reflection in x -axis y tan x y tan( x 45) y tan( x 45) The graph crosses the x-axis when tan( x 45) 0 x 45 90, 270 x 45, x 225 y 4 2 (0, 1) (225, 0) (45, 0) –2 –4 90 180 270 360 x 1– –3 1 2 3 1 1 2 90 180 270 360 Example 7 : Sketch the graph of y 2 cos(30 x) 1 for 0 x 360 . We first rewrite the function as y 2 cos( x 30) 1 . translation 30 right vertical stretch S.F. 2 y cos x y cos( x 30) y 2cos( x 30) translation 1 unit down y 2cos( x 30) 1 The graph crosses the x-axis when 2 cos( x 30) 1 0 cos( x 30) 1 2 x 30 60, 300 x 90, x 330 The graph crosses the y-axis at y 0, (30, 1) 1 3 1 (90, 0) 90 –1 (330, 0) 180 (0, 1) –2 2cos30 1 3 1 –3 (210, -3) C2 p124 Ex 8F Topic Review : Trigonometry 270 360 x