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Trigonometry So far you work has been mainly restricted to using trigonometry for right-angled triangles. Both acute angles i.e. < 90° You have found the sine, cosine or tangent of acute angles in order to find unknown sides. You have also used known sides to find the acute angles. Trigonometric Graphs From drawing the graphs of y = sin θ, y = cos θ and y = tan θ, you have discovered that sine, cosine and tangent of larger angles also exists: y 1 0.8 0.6 y = sinx 0.4 0.2 -360 -270 -180 -90 90 -0.2 -0.4 -0.6 -0.8 -1 180 270 360 x y 1 0.8 0.6 0.4 y = cosx 0.2 -360 -270 -180 -90 90 180 270 x 360 -0.2 -0.4 -0.6 -0.8 -1 y 10 y = tanx 5 -360 -270 -180 -90 90 -5 -10 180 270 360 x The sine and cosine graphs repeat their pattern every 360 degrees so sin (70°) = sin (70° + 360°), for example. The tangent graph repeats its pattern every 180 degrees so tan (70°) = tan (70° + 180°). What other things do you notice? sin (-x) = - sin (x) -1 ≤ sin θ ≤ 1 cos (-x) = cos (x) -1 ≤ cos θ ≤ 1 tan (-x) = - tan (x) The Unit Circle 1 2nd Quadrant -1 y Angles are measured from the positive x axis in an anticlockwise direction. 1st Quadrant 3rd Quadrant 4th Quadrant 1 x In 3A we will only concern ourselves with the first two quadrants: 240° 35° 0° ≤ θ ≤ 180° -1 Finding the sine of an angle on the unit circle The radius of a unit circle is 1 A A is the coordinate point (x, y) y 𝑦 θ sin θ = 1 1 θ y i.e. sin θ = y so sin θ is the y coordinate of the point A. When the unit circle is drawn onto an accurate scale, the y coordinate can be estimated. Sadler 3A Exercise 1A It also works for angles between 90° and 180° (i.e. the second quadrant). Sin 135° is the y coordinate of the point A below: A (x, y) y 1 45° 135° sin 45° = y = 1 √2 1 45° 135° on calculator sin 45° = 0.707 (also, on calculator, sin 135° = 0.707). The general rule is sin (180° - θ) = sin θ when 0° ≤ θ ≤ 90° Finding the cosine of an angle on the unit circle 𝑥 A (x, y) 1 1 x A(x, y) Cos 70° = 1 Cos 70° = x 70° x So cos 70° is the x coordinate of the point A Cos 160° = the x coordinate of the point A – WHICH IS A NEGATIVE VALUE 1 160° If 0° ≤ θ ≤ 90° then cos θ > 0 If 90° ≤ θ ≤ 180° then cos θ < 0 The following general rule applies: Cos ( 180° - θ) = - cos θ when 0° ≤ θ < 90° Alternate Formula for the Area of a Triangle We already know that: h h b h b 1 2 Area = × base × b perpendicular height However, sometimes the perpendicular height is not given. Instead the lengths of the sides of the triangle and some angles are given. E.g. B B a A C b Acute angled triangle a C A C b Obtuse angled triangle If the known angle is between the two known sides then: Area = 𝑎 ×𝑏 × sin 𝐶 2 See Sadler page 27 for the proof of this area formula and then see page 28 for examples and complete Exercise 1B Proof: Finding θ, given sin θ, when 0° ≤ θ ≤ 180° E.g. sin θ = 1 2 On Unit Circle: 1 Find 2 on the y-axis and draw a horizontal line to the circle. 0.5 Where it crosses the circle, join these points to the origin (0, 0) and estimate the size of the angles. Two possible angles are 30° and 150° Remember that the unit circle is symmetrical so Also remember that 150° 30° 180° - 30° = 150° sin θ = sin (180° - θ ) 30° On the graph of y = sin θ: y 1 0.5 30° 90 150° 180 x There are e-activities for this area formula available for your classpad. Ensure that you obtain a copy from your teacher. Alternatively, try to write one yourself. The Sine Rule for non-right triangles It is not necessary to memorise this proof. However, it is worthwhile understanding it. Examples: Find the value of x in the following: a) 70° 16.4 m b) 120° 7.1 cm 60° θ x 𝑥 sin 70° x = = 10.3 cm 16.4 sin 60° sin 𝜃 7.1 16 × sin 70° sin 60° = sin 𝜃 = sin 120 10.3 7.1 × sin 120° 10.3 𝜃 = sin−1 ( x = 17.79 m (2 d.p.) 7.1 × sin 120° ) 10.3 θ = 36.7° (1 d.p.) Using solve facility on classpad solve ( sin 𝜃 7.1 = gives and sin 120 10.3 As angles in a triangle add up to 180° , θ ) ǀ 0 ≤ θ ≤ 180 θ = 36.65312298 θ = 143.346877 Why two answers? There are always two angles between zero and 180 degrees that have the same sine value. (Look back at the unit circles and the y coordinates) However, we can discard 143.35° for this question. 120° WHY? We already have 120° so 143° is too big as angles In a triangle add up to 180°. Θ See Sadler 3A Example 5 on page 33 and then do Exercise 1C questions 1 to 9 The Cosine Rule It is not necessary to memorise this proof. However, it is worthwhile understanding it. c2 = a2 + b2 - 2ab cos C → cos 𝐶 = 𝑎2 + 𝑏2 − 𝑐 2 2𝑎𝑏 Examples: θ a) 14.6 cm b) x 5.2 m 6.9 m 50° 16.8 cm 𝑥 2 = 14.62 + 16.82 − 2 × 14.6 × 16.8 × cos 50° 8.3 m cos 𝜃 = 5.22 +6.92 −8.32 2×5.2×6.9 𝑥 = √180.074 cos 𝜃 = 0.08 𝑥 = 13.42 𝑐𝑚 (2 𝑑. 𝑝. ) 𝜃 = 85.4° See Sadler Example 8 on page 36 and then do the rest of Exercise 1C Sadler 3A Misc Exercise 1