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AS-Level Maths: Core 2 for Edexcel C2.4 Trigonometry 1 These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 47 © Boardworks Ltd 2005 The sine, cosine and tangent of any angle Contents The sine, cosine and tangent of any angle The graphs of sin θ, cos θ and tan θ Exact values of trigonometric functions Trigonometric equations Trigonometric identities Examination-style questions 2 of 47 © Boardworks Ltd 2005 The three trigonometric ratios The three trigonometric ratios, sine, cosine and tangent, can be defined using the ratios of the sides of a right-angled triangle as follows: O P P O S I T E H Y P O T E N U S E θ Opposite Sin θ = Hypotenuse SOH Adjacent Cos θ = Hypotenuse CAH Tan θ = Opposite Adjacent TOA ADJACENT Remember: S O H C A H 3 of 47 TOA © Boardworks Ltd 2005 The sine, cosine and tangent of any angle These definitions are limited because a right-angled triangle cannot contain any angles greater than 90°. To extend the three trigonometric ratios to include angles greater than 90° and less than 0° we consider the rotation of a straight line OP of fixed length r about the origin O of a coordinate grid. Angles are then measured anticlockwise from the positive x-axis. y P(x, y) θ r α O 4 of 47 x For any angle θ there is an associated acute angle α between the line OP and the x-axis. © Boardworks Ltd 2005 The sine, cosine and tangent of any angle The three trigonometric ratios are then given by: y sin = r x cos = r y tan = x The x and y coordinates can be positive or negative, while r is always positive. This means that the sign of the required ratio will depend on the sign of the x-coordinate and the y-coordinate of the point P. 5 of 47 © Boardworks Ltd 2005 The sine, cosine and tangent of any angle If we take r to be 1 unit long then these ratios can be written as: y sin = = y 1 x cos = = x 1 sin y tan = tan = x cos The relationship between θ measured from the positive x-axis and the associated acute angle α depends on the quadrant that θ falls into. For example, if θ is between 90° and 180° it will fall into the second quadrant and α will be equal to (180 – θ)°. 6 of 47 © Boardworks Ltd 2005 The sine of any angle If the point P is taken to revolve about a unit circle then sin θ is given by the y-coordinate of the point P. 7 of 47 © Boardworks Ltd 2005 The cosine of any angle If the point P is taken to revolve about a unit circle then cos θ is given by the x-coordinate of the point P. 8 of 47 © Boardworks Ltd 2005 The tangent of any angle Tan θ is given by the y-coordinate of the point P divided by the x-coordinate. 9 of 47 © Boardworks Ltd 2005 The tangent of any angle Tan θ can also be given by the length of tangent from the point P to the x-axis. 10 of 47 © Boardworks Ltd 2005 Remember CAST We can use CAST to remember in which quadrant each of the three ratios are positive. 11 of 47 2nd quadrant 1st quadrant S Sine is positive A All are positive 3rd quadrant 4th quadrant T Tangent is positive C Cosine is positive © Boardworks Ltd 2005 The sine, cosine and tangent of any angle The sin, cos and tan of angles in the first quadrant are positive. In the second quadrant: sin θ = sin α cos θ = –cos α tan θ = –tan α In the third quadrant: sin θ = –sin α cos θ = –cos α tan θ = tan α In the fourth quadrant: sin θ = –sin α cos θ = cos α tan θ = –tan α 12 of 47 where α is the associated acute angle. © Boardworks Ltd 2005 The sine, cosine and tangent of any angle The value of the associated acute angle α can be found using a sketch of the four quadrants. For angles between 0° and 360° it is worth remembering that: when 0° < θ < 90°, α=θ when 90° < θ < 180°, α = 180° – θ when 180° < θ < 270°, α = θ – 180° when 270° < θ < 360°, α = 360° – θ For example, if θ = 230° we have: α = 230° – 180° = 50° 230° is in the third quadrant where only tan is positive and so: sin 230° = –sin 50° cos 230° = –cos 50° tan 230° = tan 50° 13 of 47 © Boardworks Ltd 2005 The graphs of sin θ, cos θ and tan θ Contents The sine, cosine and tangent of any angle The graphs of sin θ, cos θ and tan θ Exact values of trigonometric functions Trigonometric equations Trigonometric identities Examination-style questions 14 of 47 © Boardworks Ltd 2005 The graph of y = sin θ 15 of 47 © Boardworks Ltd 2005 The graph of y = sin θ The graph of sine is said to be periodic since it repeats itself every 360°. We can say that the period of the graph y = sin θ is 360°. Other important features of the graph y = sin θ include the fact that: It passes through the origin, since sin 0° = 0. The maximum value is 1 and the minimum value is –1. Therefore, the amplitude of y = sin θ is 1. It has rotational symmetry about the origin. In other words, it is an odd function and so sin (–θ) = –sin θ. 16 of 47 © Boardworks Ltd 2005 The graph of y = cos θ 17 of 47 © Boardworks Ltd 2005 The graph of y = cos θ Like the graph of y = sin θ the graph of y = cos θ is periodic since it repeats itself every 360°. We can say that the period of the graph y = cos θ is 360°. Other important features of the graph y = cos θ include the fact that: It passes through the point (0, 1), since cos 0° = 1. The maximum value is 1 and the minimum value is –1. Therefore, the amplitude of y = cos θ is 1. It is symmetrical about the y-axis. In other words, it is an even function and so cos (–θ) = cos θ. It the same as the graph of y = sin θ translated left 90°. In other words, cos θ = sin (90° – θ). 18 of 47 © Boardworks Ltd 2005 The graph of y = tan θ 19 of 47 © Boardworks Ltd 2005 The graph of y = tan θ You have seen that the graph of y = tan θ has a different shape to the graphs of y = sin θ and y = cos θ. Important features of the graph y = tan θ include the fact that: It is periodic with a period of 180°. It passes through the point (0, 0), since tan 0° = 0. Its amplitude is not defined, since it ranges from +∞ to –∞. It is symmetrical about the origin. In other words, it is an odd function and so tan (–θ) = –tan θ. tan θ is not defined for θ = ±90°, ±270°, ±450°, … that is, for odd multiples of 90°. The graph y = tan θ therefore contains asymptotes at these points. 20 of 47 © Boardworks Ltd 2005 Exact values of trigonometric functions Contents The sine, cosine and tangent of any angle The graphs of sin θ, cos θ and tan θ Exact values of trigonometric functions Trigonometric equations Trigonometric identities Examination-style questions 21 of 47 © Boardworks Ltd 2005 Sin, cos and tan of 45° A right-angled isosceles triangle has two acute angels of 45°. Suppose the equal sides are of unit length. 45° 2 1 Using Pythagoras’ theorem: 2 2 The hypotenuse 1 1 45° 2 1 We can use this triangle to write exact values for sin, cos and tan 45°: 1 sin 45° = 2 22 of 47 1 cos 45° = 2 tan 45° = 1 © Boardworks Ltd 2005 Sin, cos and tan of 30° Suppose we have an equilateral triangle of side length 2. 60° 30° 2 2 3 60° 1 If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°. Using Pythagoras’ theorem: 2 2 2 1 60° The height of the triangle 3 2 We can use this triangle to write exact values for sin, cos and tan 30°: 1 sin 30° = 2 23 of 47 3 cos 30° = 2 1 tan 30° = 3 © Boardworks Ltd 2005 Sin, cos and tan of 60° Suppose we have an equilateral triangle of side length 2. 60° 30° 2 3 60° 1 2 If we cut the triangle in half then we have a right-angled triangle with acute angles of 30° and 60°. Using Pythagoras’ theorem: 2 2 2 1 60° The height of the triangle 3 2 We can also use this triangle to write exact values for sin, cos and tan 60°: 3 sin 60° = 2 24 of 47 1 cos 60° = 2 tan 60° = 3 © Boardworks Ltd 2005 Sin, cos and tan of 30°, 45° and 60° The exact values of the sine, cosine and tangent of 30°, 45° and 60° can be summarized as follows: sin cos tan 30° 45° 60° 1 2 3 2 1 3 1 2 1 2 3 2 1 2 1 3 Use this table to write the exact value of cos 135° 1 cos 135° = –cos 45° = 2 25 of 47 © Boardworks Ltd 2005 Sin, cos and tan of 30°, 45° and 60° Write the following ratios exactly: 1) cos 300° = 1 2 2) tan 315° = -1 3) tan 240° = 3 4) sin –330° = 1 2 5) cos –30° = 3 2 6) tan –135° = 1 7) sin 210° = 1 2 8) cos 315° = 1 2 26 of 47 © Boardworks Ltd 2005 Trigonometric equations Contents The sine, cosine and tangent of any angle The graphs of sin θ, cos θ and tan θ Exact values of trigonometric functions Trigonometric equations Trigonometric identities Examination-style questions 27 of 47 © Boardworks Ltd 2005 Equations of the form sin θ = k Equations of the form sin θ = k, where –1 ≤ k ≤ 1, have an infinite number of solutions. If we use a calculator to find arcsin k (or sin–1 k) the calculator will give a value for θ between –90° and 90°. There is one and only one solution is this range. This is called the principal solution of sin θ = k. Other solutions in a given range can be found using the graph of y = sin θ or by considering the unit circle. For example: Solve sin θ = 0.7 for –360° < θ < 360° arcsin 0.7 = 44.4° (to 1 d.p.) 28 of 47 © Boardworks Ltd 2005 Equations of the form sin θ = k Using the graph of y = sin θ between –360° and 360° and the line y = 0.7 we can locate the other solutions in this range. y = sin θ y = 0.7 –315.6° –224.4° 44.4° 135.6° So the solutions to sin θ = 0.7 for –360° < θ < 360° are: θ = –315.6°, –224.4°, 44.4°, 135.6° (to 1 d.p) 29 of 47 © Boardworks Ltd 2005 Equations of the form sin θ = k 30 of 47 © Boardworks Ltd 2005 Equations of the form sin θ = k We could also solve sin θ = 0.7 for –360° < θ < 360° by considering angles in the first and second quadrants of a unit circle where the sine ratio is positive. Start by sketching the principal solution 44.4° in the first quadrant. Next, sketch the associated acute angle in the second quadrant. –224.4° –315.6° 135.6° 44.4° Moving anticlockwise from the x-axis gives the second solution: 180° – 44.4° = 135.6° Moving clockwise from the x-axis gives the third and fourth solutions: –(180° + 44.4°) = –224.4° –(360° – 44.4°) = –315.6° 31 of 47 © Boardworks Ltd 2005 Equations of the form cos θ = k and tan θ = k Equations of the form cos θ = k, where –1 ≤ k ≤ 1, and tan θ = k, where k is any real number, also have infinitely many solutions. For example: Solve tan θ = –1.5 for –360° < θ < 360° Using a calculator, the principal solution is θ = –56.3° (to 1 d.p.) Now look at angles in the second and fourth quadrants of a unit circle where the tangent ratio is negative. 123.7° –236.3° –56.3° 303.7° 32 of 47 This gives us four solutions in the range –360° < θ < 360°: θ = –236.3°, –56.3°, 123.7°, 303.7° © Boardworks Ltd 2005 Equations of the form cos θ = k 33 of 47 © Boardworks Ltd 2005 Equations of the form tan θ = k 34 of 47 © Boardworks Ltd 2005 Equations involving multiple or compound angles Multiples angles are angles of the form aθ where a is a given constant. Compound angles are angles of the form (θ + b) where b is a given constant. When solving trigonometric equations involving these types of angles, care should be taken to avoid ‘losing’ solutions. Solve cos 2θ = 0.4 for –180° < θ < 180° Start by changing the range to match the multiple angle: –180° < θ < 180° –360° < 2θ < 360° Next, let x = 2θ and solve the equation cos x = 0.4 in the range –360° ≤ x ≤ 360°. 35 of 47 © Boardworks Ltd 2005 Equations involving multiple or compound angles Now, using a calculator: x = 66.4° (to 1 d.p.) –293.6° 66.4° Using the unit circle to find the values of x in the range –360° ≤ x ≤ 360° gives: x = 66.4°, 293.6°, –66.4°, –293.6° But x = 2θ so: 293.6° –66.4° θ = 33.2°, 146.8°, –33.2°, –146.8° This is the complete solution set in the range –180° < θ < 180°. 36 of 47 © Boardworks Ltd 2005 Equations involving multiple or compound angles Solve tan(θ + 25°)= 0.8 for 0° < θ < 360° Start by changing the range to match the compound angle: 0° < θ < 360° 25° < θ + 25° < 385° Next, let x = θ + 25° and solve the equation tan x = 0.8 in the range 25° < x < 385°. 38.7° Using a calculator: x = 38.7° (to 1 d.p.) Using the unit circle to find the values of x in the range 25° ≤ x ≤ 385° gives: 218.7° x = 38.7°, 218.7° (to 1 d.p.) But x = θ + 25° so: θ = 13.7°, 193.7.7° (to 1 d.p.) 37 of 47 © Boardworks Ltd 2005 Trigonometric equations involving powers Sometimes trigonometric equations involve powers of sin θ, cos θ and tan θ. For example: Solve 4sin2θ – 1= 0 for –180° ≤ θ ≤ 180° Notice that (sin θ)2 is usually written as sin2θ. 4sin2 1= 0 4sin2 = 1 sin2 = 41 sin = 21 When sin θ = 21 , θ = 30°, 150° When sin θ = – 21 , θ = –30°, –150° So the full solution set is θ = –150°, –30°, 30°, 150° 38 of 47 © Boardworks Ltd 2005 Trigonometric equations involving powers Solve 3cos2θ – cos θ = 2 for 0° ≤ θ ≤ 360° Treat this as a quadratic equation in cos θ. 3cos2θ – cos θ = 2 Factorizing: 3cos2θ – cos θ – 2 = 0 (cos θ – 1)(3cos θ + 2) = 0 cos θ = 1 or cos θ = – 32 When cos θ = 1, θ = 0°, 360° When cos θ = – 32 , θ = 131.8°, 22.8.2° So the full solution set is θ = 0°, 131.8°, 228.2°, 360° 39 of 47 © Boardworks Ltd 2005 Trigonometric identities Contents The sine, cosine and tangent of any angle The graphs of sin θ, cos θ and tan θ Exact values of trigonometric functions Trigonometric equations Trigonometric identities Examination-style questions 40 of 47 © Boardworks Ltd 2005 Trigonometric identities Two important identities that must be learnt are: sin tan cos (cos 0) sin2θ + cos2θ ≡ 1 The symbol ≡ means “is identically equal to” although an equals sign can also be used. An identity, unlike an equation, is true for every value of the given variable so, for example: sin24° + cos24° ≡ 1, sin267° + cos267° ≡ 1, sin2π + cos2π ≡ 1, etc. 41 of 47 © Boardworks Ltd 2005 Trigonometric identities We can prove these identities by considering a right-angled triangle: x y and cos = sin = r r r y y sin y = r = = tan as required. x cos θ x r x 2 2 y x sin2 + cos2 = + Also: r r x2 + y 2 = r2 But by Pythagoras’ theorem x2 + y2 = r2 so: 2 r sin2 + cos2 = 2 = 1 as required. r 42 of 47 © Boardworks Ltd 2005 Trigonometric identities One use of these identities is to simplify trigonometric equations. For example: Solve sin θ = 3 cos θ Dividing through by cos θ: for 0° ≤ θ ≤ 360° sin 3cos = cos cos tan = 3 Using a calculator, the principal solution is θ = 71.6° (to 1 d.p.) 71.6° So the solutions in the given range are: θ = 71.6°, 251.6° (to 1 d.p.) 251.6° 43 of 47 © Boardworks Ltd 2005 Trigonometric identities Solve 2cos2θ – sin θ = 1 for 0 ≤ θ ≤ 360° We can use the identity cos2θ + sin2 θ = 1 to rewrite this equation in terms of sin θ. 2(1 – sin2 θ) – sin θ = 1 2 – 2sin2θ – sin θ = 1 2sin2θ + sin θ – 1 = 0 (2sin θ – 1)(sin θ + 1) = 0 So: sin θ = 0.5 or If sin θ = 0.5, θ = 30°, 150° If sin θ = –1, θ = 270° 44 of 47 sin θ = –1 © Boardworks Ltd 2005 Examination-style questions Contents The sine, cosine and tangent of any angle The graphs of sin θ, cos θ and tan θ Exact values of trigonometric functions Trigonometric equations Trigonometric identities Examination-style questions 45 of 47 © Boardworks Ltd 2005 Examination-style question Solve the equation 3 sin θ + tan θ = 0 for θ for 0° ≤ θ ≤ 360° sin : Rewriting the equation using tan cos 3 sin + sin =0 cos 3sin cos + sin = 0 sin (3cos +1) = 0 So sin θ = 0 or 3 cos θ + 1 = 0 3 cos θ = –1 cos θ = – 31 46 of 47 © Boardworks Ltd 2005 Examination-style question In the range 0° ≤ θ ≤ 360°, when sin θ = 0, θ = 0°, 180°, 360°. when cos θ = – 1 3 : cos–1 – 31 = 109.5° (to 1 d.p.) cos θ is negative in the 2nd and 3rd quadrants so the second solution in the range is: 109.5° θ = 250 .5° (to 1 d.p.) So the complete solution set is: 250.5° 47 of 47 θ = 0°, 180°, 360°, 109.5°, 250 .5° © Boardworks Ltd 2005