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AS-Level Maths: Core 2 for Edexcel C2.4 Trigonometry 1 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 the 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 is 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 is 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 angles 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 in 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 Multiple 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° (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°, 228.2° 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