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IGCSE FM Trigonometry Dr J Frost ([email protected]) Objectives: (from the specification) Last modified: 18th April 2016 Sin Graph What does it look like? -360 -270 -180 -90 ? 90 180 270 360 Sin Graph What do the following graphs look like? -360 -270 -180 -90 90 180 270 Suppose we know that sin(30) = 0.5. By thinking about symmetry in the graph, how could we work out: sin(150) = 0.5? sin(-30) = -0.5? sin(210) = -0.5 ? 360 Cos Graph What do the following graphs look like? -360 -270 -180 -90 ? 90 180 270 360 Cos Graph What does it look like? -360 -270 -180 -90 90 180 270 Suppose we know that cos(60) = 0.5. By thinking about symmetry in the graph, how could we work out: cos(120) = -0.5 ? cos(-60) = 0.5? cos(240) = -0.5 ? 360 Tan Graph What does it look like? -360 -270 -180 -90 ? 90 180 270 360 Tan Graph What does it look like? -360 -270 -180 -90 90 180 270 Suppose we know that tan(30) = 1/β3. By thinking about symmetry in the graph, how could we work out: tan(-30) = -1/β3 ? tan(150) = -1/β3 ? 360 Solving Trig Equations Solve sin π₯ = 0.6 in the range 0 β€ π₯ < 360 π = π¬π’π§βπ π. π =?ππ. ππ°, πππ.?ππ° Angle Law #1: π¬π’π§ π = π¬π’π§(πππ β π) 0.6 -360 -270 -180 -90 ? ππ. ππ° 90 180 270 360 Solving Trig Equations Solve 3cos π₯ = 2 in the range 0 β€ π₯ < 360 2 cos π₯ =? 3 π βπ π = πππ = ?ππ. ππ°, πππ.?ππ° π Angle Law #2: πππ π = πππ(πππ β π) π π -360 -270 -180 -90 ππ. ππ° 90 ? 180 270 360 Solving Trig Equations Solve sin π₯ = β0.3 in the range 0 β€ π₯ < 360 π = π¬π’π§βπ βπ. π = βππ. ? ππ°, πππ. ππ°,?πππ. ππ Angle Law #3: Sin and cos repeat every 360° βππ. ππ° -360 -270 -180 -90 90 -0.3 ? 180 ? 270 360 Laws of Trigonometric Functions ! sin π₯ = sin 180?β π₯ cos π₯ = cos 360?β π₯ ? sin and cos repeat every 360° ? tan repeats every 180° Test Your Understanding Solve cos π₯ = 0.9 in the range 0 β€ π₯ < 360 π₯ = cos β1 0.9 = ππ. ππ° ? 360 β 25.84° = πππ. ππ° Solve tan π₯ = 1 in the range 0 β€ π₯ < 360 π₯ = tanβ1 1 = ππ° ? 45° + 180° = πππ° Set 4 Paper 2 Q14 sin π = β0.68 β π = β42.84° ? = πππ. ππ° 180 β β42.84 β42.84 + 360 = πππ. ππ° Exercise 1 1 Solve the following in the range 0 β€ π₯ β€ 360 a sin π₯ = 0.5 β π = ππ°,?πππ° 3 b cos π₯ = β π = ππ°, πππ° ? 2 c tan π₯ = 3 ? β π = ππ°, πππ° β π = π. ππ°, πππ. d sin π₯ = 0.1 ? ππ° e 4 cos π₯ = 3 β ππ. ππ°,?πππ. ππ° 6 tan π₯ = 5 β ππ. ππ°,?πππ. ππ° f sin π₯ = 0.4 β ππ. ππ°,?πππ. ππ° g 2 Solve the following in the range 0 β€ π₯ β€ 360 a sin π = β0.5 β πππ°, πππ° ? b cos π = β0.5 β πππ°, πππ° ? c tan π = β1 β πππ°, πππ° ? d sin π = β0.4 β πππ. ππ°,?πππ. ππ° e cos π = β0.7 β πππ. π°, πππ. π° ? f tan π = β0.2 β πππ. ππ°,?πππ. ππ° Trigonometric Identities 1 Using basic trigonometry to find these two missing sidesβ¦ sin?π ο± cos? 1 2 Then πππ π½ = πππ π½ ? πππ π½ Pythagoras gives you... These two identities are all you will need for IGCSE FM. ππππ π½ + ππππ π½ = π ? sin2 π is a shorthand for sin π 2 . It does NOT mean the sin is being squared β this does not make sense as sin is not a quantity that we can square! Application #1: Solving Harder Trig Equations Solve sin π₯ = 2 cos π₯ in the range 0 β€ π₯ < 360° The problem here is that we have two different trig functions. Is there anything we could divide by to get just one trig function? sin π₯ 2 cos π₯ =? cos π₯ cos π₯ tan π₯? = 2 π₯ = 63.43°, ? 243.43° Bro Tip: In general, when you have a mixture of sin and cos, divide everything by cos. 1. 2. 3. 4. sin π₯ = sin 180 β π₯ cos π₯ = cos 360 β π₯ π ππ, πππ repeat every 360 π‘ππ repeats every 180 sin π₯ A. tan π₯ = cos π₯ B. sin2 π₯ + cos2 π₯ = 1 Test Your Understanding Solve 2 sππ π₯ = cos π₯ in the range 0 β€ π₯ < 360° 2 tan π₯ = 1 1 tan π₯ = ? 2 π₯ = 26.57°, 206.57° Solve cos π₯ = sin π₯ in the range 0 β€ π₯ < 360° 1 = tan π₯ π₯ = 45°, 225° ? 1. 2. 3. 4. sin π₯ = sin 180 β π₯ cos π₯ = cos 360 β π₯ π ππ, πππ repeat every 360 π‘ππ repeats every 180 sin π₯ A. tan π₯ = cos π₯ B. sin2 π₯ + cos2 π₯ = 1 Application #1: Solving Harder Trig Equations June 2013 Paper 2 Q22 Solve tan2 π + 3 tan π = 0 in the range 0 β€ π₯ < 360° This looks a bit like a quadratic. What would be our usual strategy to solve! tan π tan π + ?3 = 0 ? tan π = 0 ππ π = 0, 180, ? tan π = β3 108.43°, 288.4° 1. 2. 3. 4. sin π₯ = sin 180 β π₯ cos π₯ = cos 360 β π₯ π ππ, πππ repeat every 360 π‘ππ repeats every 180 sin π₯ A. tan π₯ = cos π₯ B. sin2 π₯ + cos2 π₯ = 1 More Examples Solve 2sin2 π β sin π = 0 in the range 0 β€ π₯ < 360° sin π 2 sin π β 1 = 0 1 sin π = 0 ππ? sin π = 2 π = 0, 180°, 30°, 150° 2 1 4 Solve cos π = in the range 0 β€ π₯ < 360° 1 1 cos π = ππ cos π = β 2 2 ? π = 60°, 300°, 120°, 240° Test Your Understanding Solve cos 2 π + cos π = 0 in the range 0 β€ π₯ < 360° cos π cos π + 1 = 0 cos π = 0 ππ? cos π = β1 π = 90°, 270°, 180° Expand and simplify (2π + 1)(π β 1). Hence or otherwise, solve 2 sin2 π β sin π β 1 = 0 for 0° β€ π < 360° 2 sin π + 1 sin π β 1 = 0 1 sin π = β ππ ? sin π = 1 2 π = 210°, 330°, 90° Exercise 2 1 Solve the following in the range 0° β€ π₯ < 360° a sin π = 3 cos π ? ππ° β π½ = ππ. ππ°, πππ. b 2 sin π = 3 cos π β π½ = ππ. ππ°, πππ. ? ππ° 2 a b c Solve the following by first factorising. 0° β€ π₯ < 360° cos 2 π β cos π = 0 β π½ = ππ, πππ, π? tan2 π β 3 tan π = 0 β π½ = π, πππ, ππ. ? ππ, πππ. ππ° sin π₯ cos π₯ + sin π₯ = 0 β π½ = π°, πππ° ? 3 Solve the following: 0° β€ π₯ < 360° 3 β π½ = ππ°, πππ°,?πππ°, πππ° a sin2 π = 4 3 b cos 2 π = β π½ = ππ°, πππ°, ? πππ°, πππ° 4 β π½ = ππ°, πππ°, ? πππ°, πππ° c tan2 π = 3 4 a b N By factorising these βquadraticsβ, solve in the range 0 β€ π₯ < 360 3 cos 2 π + 2 cos π β 1 = 0 β π½ = ππ. ππ°, πππ°, ? πππ. ππ° 6 sin2 π β sin π β 1 = 0 β π½ = ππ°, πππ°, πππ. ? ππ°, πππ. ππ° sin π cos π + sin π + cos π = β1 β π¬π’π§ π½ + π ππ¨π¬ π½ + π ? = π β π½ = πππ°, πππ° Review of what weβve done so far οΌ οΌ partly οΌ Application of identities #2: Proofs Prove that 1 β tan π sin π cos π β‘ cos 2 π Recall that β‘ means βequivalent toβ, and just means the LHS is always equal to the RHS for all values of π. sin π πΏπ»π = 1 β sin π cos π cos π ? sin2 π πππ π ? =1β cos π 2 = 1 β sin π ? = cos 2 π = π π»π ? 1. 2. 3. 4. sin π₯ = sin 180 β π₯ cos π₯ = cos 360 β π₯ π ππ, πππ repeat every 360 π‘ππ repeats every 180 π¬π’π§ π We want to use theseβ¦ A. πππ§ π = ππ¨π¬ π B. π¬π’π§π π + ππ¨π¬ π π = π Another Example June 2012 Paper 1 Q16 Prove that tan π + πΏπ»π = 1 tan π β‘ 1 sin π cos π sin π cos π + cos π ? sin π sin2 π cos 2 π = + sin π cos π ?sin π cos π sin2 π + cos 2 π = ?π sin π cos 1 = = π π»π sin π cos π? Bro Tip: Whenever you have a fraction in a proof question, always add the fractions. Test Your Understanding Prove that tan π₯ cos π₯ 1βcos2 π₯ β‘1 sin π₯ cos π₯ sin π₯ cos π₯ = =? =1 2 sin π₯ sin π₯ AQA Worksheet Prove that tan2 π β‘ 1 cos2 π β1 tan2 π sin2 π = cos 2 π 1 β cos 2 π =? cos 2 π 1 cos 2 π = β cos 2 π cos 2 π Exercise 3 1 Simplify 3 sin π₯ sin π₯ + 2 β 3 2 sin π₯ β cos 2 π₯ = π π¬π’π§π π + π π¬π’π§ π β π π¬π’π§ π + π ππ¨π¬ π π ?π = π π¬π’π§π π + ππ¨π¬ π π = 2 Write out the following in terms of sin π₯: 2 2 a) cos π₯ tan π₯ b) tan π₯ cos 3 π₯ c) cos π₯ 2 cos π₯ π¬π’π§π π = πππ π ππ¨π¬π π = π¬π’π§π π π¬π’π§ π = ππ¨π¬ π ππ¨π¬ π π = π¬π’π§ π ππ¨π¬ π π = β 3 tan π₯ = π ππ¨π¬ π π β π π¬π’π§ π π ? ? 3 Prove the following: a. tan π₯ 1 β sin2 π₯ β‘ sin π₯ b. c. d. e. f. 1βcos2 π₯ 1βsin2 π₯ β‘ tan2 π₯ 1 + sin π₯ 1 β sin π₯ β‘ cos 2 π₯ 2 sin π₯ cos π₯ 2π₯ β‘ 2 β 2 sin tan π₯ 1 cos π β cos π β‘ sin π tan π 2 sin π β cos π 2 + sin π + 2 cos π 2 β‘5 π¬π’π§ π π β π¬π’π§π π =π? π β π¬π’π§π π β π π¬π’π§ π sin/cos/tan of 30°, 45°, 60°, 90° You will frequently encounter angles of 30°, 60°, 45° in geometric problems. Why? We see these angles in equilateral triangles and right-angled isosceles triangles. ? You need to be able to calculate these in non-calculator exams. All you need to remember: ! Draw half a unit square and half an equilateral triangle of side 2. ?2 ?1 45° ? 1? sin 45° = cos 45° = 1 ? 2 1 tan 45° = 1 30° ? ?2 ? 1 ? 2 3 = ? 2 1 = ? 3 3 = ? 2 1 = ? 2 = ? 3 sin 30° = 2? tan 30° ?3 For 0°, 90° just think about the graphs of trig functions: sin 0° = 0 ? sin 90° = 1 ? cos 0° = 1 ? cos 90° = 0 ? tan 0° = 0 ? cos 30° 60° ? 1? sin 60° cos 60° tan 60° Example Exam Questions ? Mark Scheme Using triangles to change between sin/cos/tan 3 Given that sin π = and that π is 5 acute, find the exact value of: π a) cos π = ? b) tan π = π π π ? ?5 ?3 Represent as a triangle π 4 Test Your Understanding 1 5 13 Given that cos π = and that π is acute, find the value of: ππ a) sin π = ? b) tan π ππ ππ = ? π 2 5 Given that tan π = and that 2 π is acute, find the value of: a) sin π = b) cos π = π ? π π ? π