Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
A-Level Maths: Core 3 for Edexcel C3.3 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 35 © Boardworks Ltd 2006 Contents The inverse trigonometric functions The inverse trigonometric functions The reciprocal trigonometric functions Trigonometric identities Examination-style question 2 of 35 © Boardworks Ltd 2006 The inverse of the sine function Suppose we wish to find θ such that sin θ = x In other words, we want to find the angle whose sine is x. This is written as θ = sin–1 x or θ = arcsin x In this context, sin–1 x means the inverse of sin x. This is not the same as (sin x)–1 which is the reciprocal of sin x, 1 . sin x Is y = sin–1 x a function? 3 of 35 © Boardworks Ltd 2006 The inverse of the sine function We can see from the graph of y = sin x between x = –2π and x = 2π that it is a manyto-one function: y y = sin x x y The inverse of this graph is not a function because it is one-to-many: y = sin–1 x x 4 of 35 © Boardworks Ltd 2006 The inverse of the sine function However, remember that if we use a calculator to find sin–1 x (or arcsin x) the calculator will give a value between –90° and 90° (or between – 2 ≤ x ≤ 2 if working in radians). There is only one value of sin–1 x in this range, called the principal value. So, if we restrict the domain of f(x) = sin x to – ≤ x ≤ we 2 2 have a one-to-one function: y y = sin x 1 2 2 x –1 5 of 35 © Boardworks Ltd 2006 The graph of y = sin–1 x Therefore the inverse of f(x) = sin x, – 2 ≤ x ≤ 2 , is also a one-to-one function: f –1(x) = sin–1 x y y = sin–1 x –1 The graph of y = sin x 2 1 is the reflection of y = sin x y = sin x in the line y = x: x (Remember the scale 1 2 2 –1 –1 used on the x- and y-axes 2 must be the same.) The domain of sin–1 x is the same as the range of sin x : –1 ≤ x ≤ 1 The range of sin–1 x is the same as the restricted domain of sin x : – 2 ≤ sin–1 x ≤ 2 6 of 35 © Boardworks Ltd 2006 The inverse of cosine and tangent We can restrict the domains of cos x and tan x in the same way as we did for sin x so that if f(x) = cos x then f –1(x) = cos–1 x And if f(x) = tan x then f –1(x) = tan–1 x for for for for 0≤x≤π –1 ≤ x ≤ 1. – 2 < x < 2 x . The graphs cos–1 x and tan–1 x can be obtained by reflecting the graphs of cos x and tan x in the line y = x. 7 of 35 © Boardworks Ltd 2006 The graph of y = cos–1 x y= cos–1 x y 2 1 –1 0 –1 1 2 x y = cosx The domain of cos–1 x is the same as the range of cos x : –1 ≤ x ≤ 1 The range of cos–1 x is the same as the restricted domain of cos x : 0 ≤ cos–1 x ≤ π 8 of 35 © Boardworks Ltd 2006 The graph of y = tan–1 x y y = tan tanxx y = tan–1 x 2 2 2 x 2 The domain of tan–1 x is the same as the range of tan x : x The range of tan–1 x is the same as the restricted domain of tan x : – 2 < tan–1 x < 2 9 of 35 © Boardworks Ltd 2006 Problems involving inverse trig functions Find the exact value of sin–1 3 2 in radians. To solve this, remember the angles whose trigonometric ratios can be written exactly: radians 0 6 4 3 degrees 0° 30° 45° 60° 90° sin 0 1 2 1 tan 0 3 2 1 3 3 2 1 2 1 cos 1 2 1 2 1 3 From this table sin–1 10 of 35 3 2 2 0 = 3 © Boardworks Ltd 2006 Problems involving inverse trig functions Find the exact value of cos–1 2 2 in radians. This is equivalent to solving the trigonometric equation cos θ = – 22 for 0 ≤ θ ≤ π this is the range of cos–1x We know that cos = 1 = 22 4 2 Sketching y = cos θ for 0 ≤ θ ≤ π : 1 From the graph, cos 34 = – 22 2 2 0 2 2 –1 11 of 35 4 2 3 4 θ So, cos–1 2 2 = 34 © Boardworks Ltd 2006 Problems involving inverse trig functions Find the exact value of cos (sin–1 Let sin–1 7 4 =θ so 7 4 ) in radians. 7 4 sin θ = Using the following right-angled triangle: 7 + a2 = 16 a=3 4 7 θ 3 The length of the third side is 3 so cos θ = But sin–1 7 4 = θ so cos (sin–1 12 of 35 3 4 7 ) 4 = 3 4 © Boardworks Ltd 2006 Contents The reciprocal trigonometric functions The inverse trigonometric functions The reciprocal trigonometric functions Trigonometric identities Examination-style question 13 of 35 © Boardworks Ltd 2006 The reciprocal trigonometric functions The reciprocal trigonometric functions are cosecant, secant and cotangent. They are related to the three main trigonometric ratios as follows: 1 cosec x = sin x This is short for cosecant. 1 sec x = cos x This is short for secant. 1 cot x = tan x This is short for cotangent. Notice that the first letter of sin, cos and tan happens to be the same as the third letter of the corresponding reciprocal functions cosec, sec and cot. 14 of 35 © Boardworks Ltd 2006 The graph of sec x 15 of 35 © Boardworks Ltd 2006 The graph of cosec x 16 of 35 © Boardworks Ltd 2006 The graph of cot x 17 of 35 © Boardworks Ltd 2006 Properties of the graph of sec x The properties of the graphs of sec x, cosec x and cot x can be summarized in the following table: function domain range odd period or even asymptotes f(x) = 0 at x = when x = f(x) = sec x x f(x) ≤ –1, x ≠ 90°+180n° f(x) ≥ 1 n 90°+180n°, n never 360° even f(x) = cosec x x x ≠ 180n° n f(x) ≤ –1, f(x) ≥ 1 180n°, n never 360° odd f(x) = cot x x x ≠ 180n° n f(x) 180n°, n 90°+180n°, n 180° odd 18 of 35 © Boardworks Ltd 2006 Transforming the graph of f(x) = sec x 19 of 35 © Boardworks Ltd 2006 Transforming the graph of f(x) = cosec x 20 of 35 © Boardworks Ltd 2006 Transforming the graph of f(x) = cot x 21 of 35 © Boardworks Ltd 2006 Problems involving reciprocal trig functions Use a calculator to find, to 2 d.p., the value of: a) sec 85° b) cosec 200° 1 a) sec 85° = cos85 c) cot –70° = 11.47 (to 2 d.p.) 1 b) cosec 220° = = –1.56 (to 2 d.p.) sin220 1 c) cot –70° = = –0.36 (to 2 d.p.) tan 70 22 of 35 © Boardworks Ltd 2006 Problems involving reciprocal trig functions Find the exact value of: a) cosec 6 a) sin 6 = b) tan 2 3 b) cot 2 3 1 2 = – tan 3 2 3 =– 3 ) c) cos – 34 = –cos(– 3 + π) =– 1 4 3 3 θ is in the 3rd quadrant so cos θ = –cos (θ + π) 4 2 so, 23 of 35 θ is in the 2nd quadrant so tan θ = tan (π – θ) cot 23 = – 1 = – 3 so, = – cos 4 cosec 6 = 2 so, = –tan (π – c) sec – 3 sec – 34 = – 2 © Boardworks Ltd 2006 Problems involving reciprocal trig functions Given that x is an acute angle and tan x = values of cot x, sec x and cosec x. 3 4 find the exact Using the following right-angled triangle: 5 3 The length of the hypotenuse is x 9 +16 = 5 4 So tan x = 3 4 cos x = 4 5 sin x = cot x = 4 3 sec x = 5 4 cosec x = 3 5 Therefore 24 of 35 5 3 © Boardworks Ltd 2006 Problems involving reciprocal trig functions Prove that tan x sin x sec x cos x. LHS = tan x sin x sin x = sin x cos x sin2 x = cos x 1 cos2 x = cos x Using sin2x + cos2x = 1 1 cos2 x = cos x cos x = sec x cos x = RHS 25 of 35 © Boardworks Ltd 2006 Problems involving reciprocal trig functions Solve sec (x + 20°) = 2 for 0 ≤ x ≤ 360°. sec( x + 20 ) = 2 1 =2 cos( x + 20 ) 1 cos (x + 20°) = 2 x + 20° = 60° or 300° x = 40° or 280° 26 of 35 © Boardworks Ltd 2006 Contents Trigonometric identities The inverse trigonometric functions The reciprocal trigonometric functions Trigonometric identities Examination-style question 27 of 35 © Boardworks Ltd 2006 Trigonometric identities Earlier in the course you met the following trigonometric identities: sin tan (cos 0) 1 cos sin2 cos2 1 2 We can write these identities in terms of sec θ, cosec θ and cot θ. Using 1 1 1 cos = cot = = sin tan sin cos So cos cot sin 28 of 35 (sin 0) © Boardworks Ltd 2006 Trigonometric identities sin2 cos2 1 2 Dividing 2 through by cos2θ gives sin2 cos2 1 + cos2 cos2 cos2 tan2 +1 sec 2 Dividing 2 through by sin2θ gives sin2 cos2 1 + 2 2 sin sin sin2 1+ cot 2 cosec 2 29 of 35 © Boardworks Ltd 2006 Trigonometric identities tan x Show that cot x. 2 1 sec x tan x LHS = 1 sec 2 x = tan x 1 tan2 x 1 Using sec2x = tan2x + 1 tan x = tan2 x = 1 tan x = cot x = RHS 30 of 35 © Boardworks Ltd 2006 Trigonometric identities Given that x is an obtuse angle and cosec x = 5, find the exact value of tan x. cosec x = 5 cosec2 x = 25 Using cosec2 x ≡ 1 + cot2 x, 1 + cot2 x = 25 cot2 x = 24 cot x = ±√24 = ±2√6 x is obtuse and so cot x is negative (since tan x is negative in the second quadrant). Therefore: 1 cot x = 2 6 tan x = 2 6 31 of 35 © Boardworks Ltd 2006 Trigonometric equations Solve 2sec 2 = 2 + tan for 0 360 Using sec 2 1+ tan2 : 2(1+ tan2 ) = 2 + tan 2 + 2tan2 2 tan = 0 2 tan2 tan = 0 tan (2tan 1) = 0 tan = 0 or θ = 0°, 180°, 360° tan = ½ θ = 26.6°, 206.6° (to 1 d.p.) The complete solution set is θ = 0°, 26.6°, 180°, 206.6°, 360°. 32 of 35 © Boardworks Ltd 2006 Contents Examination-style question The inverse trigonometric functions The reciprocal trigonometric functions Trigonometric identities Examination-style question 33 of 35 © Boardworks Ltd 2006 Examination-style question a) Prove that sec θ ≡ cos θ + sin θ tan θ. b) Hence solve the equation 2 cos θ = 3 cosec θ – 2 sin θ tan θ in the interval 0° < θ < 360°. Give all solutions in degrees to 1 decimal place. RHS = cos + sin a) sin cos cos2 sin2 = + cos cos 1 = cos = sec = LHS 34 of 35 © Boardworks Ltd 2006 The reciprocal trigonometric functions 2cos = 3cosec 2sin tan b) 2cos + 2sin tan = 3cosec 2(cos + sin tan ) = 3cosec Using the result given in part a): 2sec = 3cosec 2 3 = cos sin sin 3 = cos 2 tan = 32 = 56.3, 236.3 (to 1 d.p.) 35 of 35 © Boardworks Ltd 2006