Report

C1: Chapters 8 & 10 Trigonometry Dr J Frost (jfrost@tiffin.kingston.sch.uk) Last modified: 28th September 2013 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/√3. By thinking about symmetry in the graph, how could we work out: tan(-30) = -1/√3 ? tan(150) = -1/√3 ? 360 Laws of Trigonometric Functions We saw for example sin(30) = sin(150) and cos(30) = cos(330). It’s also easy to see by looking at the graphs that cos(40) = sin(50). What laws does this give us? ! sin(x) = sin(180? – x) cos(x) = cos(360? – x) sin and cos repeat every 360? tan repeats every 180? sin(x) = cos(90?– x) Bro Tip: These 5 things are pretty much the only thing you need to learn from this Chapter! Practice Find all the values in the range 0 to 360 for which sin/cos/tan will be the same. 1 ? sin(30) = sin(150) ? cos(70) = cos(290) 2 ? cos(30) = cos(330) 3 sin(-10) = sin(190) ? = sin(350) ? 4 cos(-40) = cos(40) ? = cos (320) 5 sin(20) = cos(70) ? 6 sin(80) = sin(100) ? 7 8 11 cos(-25) = cos(25) ? = cos(335) ? 9 10 cos(80) = sin(10) ? ? sin(15) = sin(165) sin(-60) = sin(240) ? = sin(300) ? 12 ? tan(80) = sin(260) Dr Frost’s technique for remembering trig values (once described by a KGS tutee of mine as ‘the Holy Grail of teaching’) I literally picture this table in my head when I’m trying to remember my values. 0 45 All the values in this square are over 2. 90 30 60 sin 0 _1_ √2 1 _1_ 2 √3 2 cos 1 _1_ √2 0 √3 2 _1_ 2 tan 0 1 _1_ √3 √3 All the surds in this block are √2 The diagonals starting from the top left are rational. The other values in the square are not. All the surds in this block are √3 I remember that out of tan(30) and tan(60), one is 1/√3 and the other √3. However, by considering the graph of tan, clearly tan(30) < tan(60), so tan(30) must be the smaller one, 1/√3 Practice 0 45 90 30 60 sin 0 ? _1_ ? √2 1 ? _1_ ?2 √3 ?2 cos ?1 _1_ ? √2 ?0 √3 ?2 _1_ ?2 tan ? 0 ? _1_ ? √3 √3 1 ? ‘Magic Triangles’ You can easily work out sin(45), cos(45), sin(30), tan(30) etc. if you were ever to forget. ? √2 1 45 1 _1_ sin(45) = ? √2 30 2 ? √3 60 1? _√3_ cos(30) = ? 2 Angle quadrants Unlike bearings, angles are generally measured anticlockwise starting from the x-axis. y S A 124° x T C By labelling the 4 quadrants ASTC (mnemonic: Alan Sugar Talks Crap), we can tell with Sin, Cos, Tan, or All the trigonometric functions will give a positive value for that angle. sin(124) will be: positive ? cos(34) will be: positive ? tan(100) will be: negative ? cos(213) will be: negative ? tan(213) will be: positive ? Angle quadrants Given that sin α = 2/5, and that α is obtuse, find (without a calculator) the exact value of cos α. cos ф = √21? / 5 5? Therefore thinking about ASTC: 2? ? /5 cos α = -√21 ф ? √21 Imagine working instead with the acute angle ф such that sin ф = 2/5 (We can alternatively think about the graphs of sin and cos) Angle quadrants 1 Given that tan α = 5/12, and that α is acute, find the exact value of sin α and cos α. ? ? sin α = 5/13, cos α = 12/13 2 Given that cos α = -3/5, and that α is obtuse, find the exact value of sin α and tan α. ? tan α = -4/3 ? sin α = 4/5, 3 Given that tan α = -√3, and that α is reflex, find the if tan α is negative, then is our reflex exact value of sin α and tan α. Hint: angle between 180 and 270, or 270 and 360? ? sin α = -√3/2, cos α = 1/2? Onwards to Chapter 10... The only 2 identities you need this chapter... r ? y = r sin ? x = r cos 1 2 sin = y/r and cos = x/r and tan = y/x Pythagoras gives you... sin = tan? cos ? 2=1 sin2 + cos Examples of use 1 Simplify sin2 3 + cos2 3 =1 ? Simplify 5 – 5sin2 = 5cos2? 3 Given that p = 3 cos and q = 2 sin , show that 4p2 + 9q2 = 36. 2 Show that: cos 4 − sin4 2 ≡ 1 − tan 2 cos This box is intentionally left blank. Solving Trigonometric Equations Edexcel May 2013 (Retracted) = 123.44, ?176.57 Bro Tips for solving: 1. If 0 ≤ < 180, then what range does 2 – 30 have? 2. Immediately after the point at which you do sin-1 of both sides, list out the other possible angles in the above adjusted range. Recall that sin(x) = sin(180-x) and that sin repeats every 360. Solving Trigonometric Equations Edexcel June 2010 a b tan = 0.4? tan 2x = 0.4 0 ≤ 2x < 720 2x = 21.801, 201.801,? 381.801, 561.801, x = 10.9, 100.9, 190.9, 280.9 Solving Trigonometric Equations Edexcel Jan 2010 (2sin x – 1)(sin x + 3) = 0 sin x = 0.5 or? sin x = -3 x = 30°, 150° Bro Tip: In general, when you have sin and cos, and one is squared, change the squared term to be consistent with the other. Exercises Edexcel Jan 2009 Edexcel Jun 2009 ? ? = −45°, 135°, 23.6°, 156.4° = 41.4°, 318.6° Edexcel Jun 2008 284.5, 435.5, ? 644.5 Edexcel Jan 2008 65, 155 ? 40 80 160 ? 200 280 320 Edexcel Jan 2013 41.2, 85.5, 161.2 ? θ = 230.785, 309.23152, 50.8, 129.2 ? Things to remember 1 If you square root both sides, don’t forget the +-. You’ll probably lose 2 marks otherwise. sin2 3x = 1/2 sin 3x = 1/√2 2 Don’t forget solutions. If you have sin, you’ll always be able to get an extra solution by using 180 – x. If you have cos you can get an extra one using 360-x. 3 Remember that tan repeats every 180, sin/cos every 360. 4 If you had sin2x and cos x, you’d replace the sin2 x with 1 – cos2 x. You’d then have a quadratic in terms of cos x which you can factorise. 5 Check whether the question expects you to give your answers in degrees or radians. If they say 0 ≤ x < , then clearly they want radians.