Pure Maths - Chapter 5 - Radians Student Booklet.pdf

A Level Mathematics Year 2 - Pure RADIANS EDEXCEL A-LEVEL SPECIFICATION  5.1 Radian Measure 2𝜋 𝑅𝑎𝑑𝑖𝑎𝑛𝑠 = 360° 2𝜋 𝑐 = 360° 𝜋 𝑅𝑎𝑑𝑖𝑎𝑛𝑠 = 180° ÷ 2𝜋 ÷ 2𝜋...

A Level Mathematics Year 2 - Pure RADIANS EDEXCEL A-LEVEL SPECIFICATION  5.1 Radian Measure 2𝜋 𝑅𝑎𝑑𝑖𝑎𝑛𝑠 = 360° 2𝜋 𝑐 = 360° 𝜋 𝑅𝑎𝑑𝑖𝑎𝑛𝑠 = 180° ÷ 2𝜋 ÷ 2𝜋 𝑐 180° 𝜋 𝑅𝑎𝑑𝑖𝑎𝑛𝑠 = 90° 1 = 2 𝜋 𝜋 180 𝑅𝑎𝑑𝑖𝑎𝑛𝑠 = 60° × 3 𝜋 𝜋 𝑅𝑎𝑑𝑖𝑎𝑛𝑠 = 45° 4 3𝜋 𝑅𝑎𝑑𝑖𝑎𝑛𝑠 𝐷𝑒𝑔𝑟𝑒𝑒𝑠 𝑅𝑎𝑑𝑖𝑎𝑛𝑠 = 270° 2 180 ÷ 𝜋 EXAMPLE 1 Convert the following angles into degrees. 2𝜋 a) 𝑟𝑎𝑑 8 4𝜋 b) 𝑟𝑎𝑑 15 EXAMPLE 2 Convert the following angles into radians. Leave your answer in terms of . a) 150° b) 110° EXAMPLE 3 Find the answers to 2 decimal places. a) sin(0. 3𝑐 ) b) tan(2𝑐 )  5.1 Radian Measure EXAMPLE 4 Sketch the graph of y = sin 𝑥 for 0 ≤ 𝑥 ≤ 2𝜋 𝑦 1 0.5 𝑥 𝜋 𝜋 3𝜋 2𝜋 2 2 -0.5 -1 EXAMPLE 5 Sketch the graph of y = cos (𝑥 + 𝜋) for 0 ≤ 𝑥 ≤ 2𝜋 𝑦 1 0.5 𝑥 𝜋 𝜋 3𝜋 2𝜋 2 2 -0.5 -1  5.1 Radian Measure YOU NEED TO LEARN THESE EXACT VALUES 𝜋 1 𝜋 3 𝜋 3 sin = cos = tan = 6 2 6 2 6 3 (30°) (30°) (30°) 𝜋 3 𝜋 1 𝜋 sin = cos = tan = 3 3 2 3 2 3 (60°) (60°) (60°) 𝜋 2 𝜋 2 𝜋 sin = cos = tan =1 4 2 4 2 4 (45°) (45°) (45°) EXAMPLE 6 Find the exact values of: 4𝜋 −7𝜋 (a) cos (b) sin 3 6 EXERCISE 5A page 116 and EXERCISE 5B page 118  5.1 Radian Measure PRACTICE  5.2 Arc Length 𝒍 𝒍 = 𝒓𝜽 𝑙 = 𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ 𝒓 𝒓 𝑟 = 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑜𝑟 𝜽 𝜃 = 𝑎𝑛𝑔𝑙𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 EXAMPLE 7 𝑨 The diagram shows a sector AOB. The perimeter of the sector is twice the 𝑶 length of the arc AB. Find the size of angle AOB. 𝑩  5.2 Arc Length EXAMPLE 8 The diagram shows a triangular garden, PQR, with PQ = 12m, PR = 7m and ∠QPR=0.5 radians. The curve SR is a small path separating the shaded patio area and lawn and in an arc of a circle with centre at P and radius 7m. Find: a) The length of the path SR. b) The perimeter of the shaded patio, giving your answer to 3 significant figures.  5.2 Arc Length EXAMPLE 9 The border of a garden pond consists of a straight edge AB of length 2.4m and a curved part C as shown in the diagram. The curved part is an arc of a circle, centre O and radius 2m. Find the length of C.  5.2 Arc Length PRACTICE EXERCISE 5C pages 120 - 122  5.2 Arc Length PRACTICE EXERCISE 5C pages 120 - 122  5.3 Area of Sectors and Segments 1 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑠𝑒𝑐𝑡𝑜𝑟 = 𝑟 2 𝜃 2 𝒓 𝒓 𝑟 = 𝑟𝑎𝑑𝑖𝑢𝑠 𝜽 𝜃 = 𝑎𝑛𝑔𝑙𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 1 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑚𝑖𝑛𝑜𝑟 𝑠𝑒𝑐𝑡𝑜𝑟 = 𝑟 2 𝜃 2 1 = 𝑟2𝑥 2 1 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑚𝑎𝑗𝑜𝑟 𝑠𝑒𝑐𝑡𝑜𝑟 = 𝑟 2 𝜃 2 1 = 𝑟 2 (2𝜋 − 𝑥) 2 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 (shaded region) 1 1 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 = 𝑟 2 𝜃 − 𝑟 2 𝑠𝑖𝑛𝜃 2 2 𝑟 1 𝑟 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑠𝑒𝑔𝑚𝑒𝑛𝑡 = 𝑟 2 (𝜃 − 𝑠𝑖𝑛𝜃) 2 𝜃  5.3 Area of Sectors and Segments EXAMPLE 10 A plot of land is in the shape of a sector of a circle of radius 55m. The length of fencing that is erected along the edge of the plot to enclose the land is 176m. Calculate the area of the plot of land 𝑨 55𝒎 𝜽 𝑶 55𝒎 𝑩 EXAMPLE 11 𝜃 In the diagram above, OAB is a sector of a circle, radius 4m. The chord AB is 5m long. Find the area of the shaded segment.  5.3 Area of Sectors and Segments EXAMPLE 12 In the diagram, AD and BC are arcs of circles with centre O, such that 𝑂𝐴 = 𝑂𝐷 = 𝑟 𝑐𝑚, 𝐴𝐵 = 𝐷𝐶 = 8 𝑐𝑚 and ∠BOC =  radians. 6 a) Given that the area of the shaded region is 48𝑐𝑚2 , show that 𝑟 = − 4 𝜃 b) Given also that 𝑟 = 10θ, calculate the perimeter of the shaded region. B 8 𝑐𝑚 A 𝑟 𝑐𝑚 O 𝜃 𝑟 𝑐𝑚 D 8 𝑐𝑚 C  5.3 Area of Sectors and Segments PRACTICE EXERCISE 5D pages 125-128  5.4 Solving Trigonometric Equations In Year 1, you learned how to solve trigonometric equations. You can solve trigonometric equations in radians the same way. EXAMPLE 13 Find the solutions of these equations in the interval 0 ≤ θ ≤ 2π: a) sin θ = 0.3 b) 4 cos θ = 2 c) 5 tan θ + 3 = 1 (90°) 𝜋 a) 2 S A y 1 𝜋 0 (180°) 2𝜋 (360°) 0 π π 3π 2π θ T 3𝜋 C 2 2 2 (270°) −1 b) (90°) 𝜋 2 S A 𝜋 0 (180°) 2𝜋 (360°) T 3𝜋 C 2 (270°)  5.4 Solving Trigonometric Equations c) 𝑦 𝑦 = tan 𝜃 π π 3π −2 2 π 2 2π θ (90°) 𝜋 2 S A 𝜋 0 (180°) 2𝜋 (360°) T 3𝜋 C 2 (270°)  5.4 Solving Trigonometric Equations EXAMPLE 14 Solve the equation 17 cos 𝜃 + 3 sin2 𝜃 = 13 in the interval 0 ≤ 𝜃 ≤ 2𝜋. (90°) 𝜋 2 S A 𝜋 0 (180°) 2𝜋 (360°) T 3𝜋 C 2 (270°) y 𝑦 = cos 𝜃 1 0 π π 3π 2π θ 2 2 −1  5.4 Solving Trigonometric Equations EXAMPLE 15 3 Solve the equation sin 3𝜃 = , in the interval 0 ≤ 𝜃 ≤ 2𝜋 2 y 0 π 2π 3π 4π 5π 6π X y = sin X (90°) 𝜋 2 S A 𝜋 0 (180°) 2𝜋 (360°) T 3𝜋 C 2 (270°)  5.4 Solving Trigonometric Equations PRACTICE EXERCISE 5E page 131-132  5.4 Solving Trigonometric Equations PRACTICE  5.5 Small Angle Approximation If  is small (is close to zero) and is measured in radians, then: sin θ ≈ θ sin3θ ≈ 3θ (5θ)2 cos5θ ≈ 1 − 2 θ2 sin10θ ≈ 10θ cos θ ≈ 1 − 2 (9θ)2 tan2θ ≈ 2θ cos9θ ≈ 1 − 2 tan θ ≈ θ tan10θ ≈ 10θ etc. etc. EXAMPLE 16 If  is small, find the approximation of: 𝑐𝑜𝑠θ−1 2 1−cosθ −1 a) b) θ tan 2θ tan θ−1  5.5 Small Angle Approximation EXAMPLE 17 7 + 2 cos 2𝜃 a) If  is small, show that the expression can be written as 3 − 2𝜃. tan 2𝜃+3 7 + 2 cos 2𝜃 b) Hence write down the value of when  is small. tan 2𝜃+3  5.5 Small Angle Approximation EXAMPLE 18 a) Write θ is small, show that the equation 32 cos 5θ + 203 tan 10θ = 182 can be written as 40θ2 − 203θ + 15 = 0 b) Hence, find the solutions of the equation 32 cos 5θ + 203 tan 10θ = 182 c) Comment on the validity of your solutions. EXAMPLE 19 Given that θ is small, use the small angle approximations to show that: θ 4sin + 3𝑐𝑜𝑠 2 θ ≈ 𝑎 + 𝑏θ + 𝑐θ2 2 where a, b and c are integers to be found.  5.5 Small Angle Approximation EXAM QUESTION The figure show a sector OABCO of a circle centre O. 𝐵 Given that: OA = OC = 12cm Angle AOC =  radians 𝐴 𝐶 Area triangle OAC : Area segment ABC = 3:1 Show that: 12 𝑐𝑚 12 𝑐𝑚 𝜃 3 - 4sin = 0 𝑂  5.5 Small Angle Approximation PRACTICE EXERCISE 5F page 134 EXAM QUESTIONS MIXED EXERCISE 5 pages 135-140