A-Level Maths: Core 3 for Edexcel C3.3 Trigonometry 1 These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. 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

The inverse trigonometric functions The reciprocal trigonometric functions Trigonometric identities Examination-style question Contents 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, . Stress that while we do write sin2x to mean (sin x)2, sin–1x does not mean (sin x)–1. Is y = sin–1 x a function?

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 many-to-one function: y y = sin x x x y The inverse of this graph is not a function because it is one-to-many: y = sin–1 x The inverse graph can be obtained by reflecting the original graph in the line y = x, effectively interchanging the role of x and y.

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 – ≤ x ≤ 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 have a one-to-one function: y 1 –1 x y = sin x

The graph of y = sin–1 x Therefore the inverse of f(x) = sin x, – ≤ x ≤ , is also a one-to-one function: f –1(x) = sin–1 x 1 –1 x y x y 1 –1 y = sin–1 x 1 –1 y = sin–1 x The graph of y = sin–1 x is the reflection of y = sin x in the line y = x: y = sin x (Remember the scale used on the x- and y-axes must be the same.) Stress that the graph of an inverse function will only be a reflection of the original function in y = x if the scale used on the axes is the same. For this reason we label the axes using radians rather than degrees, where π/2 is just over 1.5 radians. 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 : – ≤ sin–1 x ≤

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 for 0 ≤ x ≤ π f –1(x) = cos–1 x then for –1 ≤ x ≤ 1. – < x < And if f(x) = tan x for f –1(x) = tan–1 x then for x Remind students that tan x is undefined for x = –π/2 and x = π/2 and so these values are not included when restricting the domain. 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.

The graph of y = cos–1 x x y 1 –1 y = cos–1 x y 1 –1 y = cos–1 x 1 x –1 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

The graph of y = tan–1 x x y y = tan x y = tan–1 x y y = tanx y = tan–1 x x 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 : – < tan–1 x <

Problems involving inverse trig functions Find the exact value of sin–1 in radians. To solve this, remember the angles whose trigonometric ratios can be written exactly: tan cos sin 90° 60° 45° 30° 0° degrees radians 1 1 Remind students that these can be recalled quickly by sketching a right-angled triangle with acute angles equal to 45° and sides of length 1 and √2, or a right-angled triangle with acute angles equal to 30° and 60° and sides of length 1, 2 and √3. 1 From this table sin–1 =

Problems involving inverse trig functions Find the exact value of sin–1 in radians. This is equivalent to solving the trigonometric equation cos θ = – for 0 ≤ θ ≤ π this is the range of cos–1x We know that cos = = Sketching y = cos θ for 0 ≤ θ ≤ π : Alternatively, we could draw a quadrant diagram to show that cos 3π/4 is –√2/2 since it is in the second quadrant where cos θ is negative. –1 θ 1 From the graph, cos = – So, cos–1 =

Problems involving inverse trig functions Find the exact value of cos (sin–1 ) in radians. Let sin–1 = θ so sin θ = θ 4 Using the following right-angled triangle: 7 + a2 = 16 a = 3 3 The length of the third side is 3 so Alternatively, we could use the identity sin2 θ + cos2 θ = 1. This would give us 7/16 + cos2 θ = 1  cos2 θ = 9/16  cos θ = 3/4. cos θ = But sin–1 = θ so cos (sin–1 ) =

The reciprocal trigonometric functions The inverse trigonometric functions The reciprocal trigonometric functions Trigonometric identities Examination-style question Contents 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: cosec x = sec x = cot x = This is short for cosecant. This is short for secant. 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.

The graph of sec x Drag the point through the graph of cos x and as the graph of sec x is traced out make the following observations: When y = cos x = 1, y = sec x = 1. As y = cos x decreases from 1 towards 0, y = sec x increases from 1 towards +∞. When y = cos x = 0, y = sec x is undefined. As y = cos x decreases from 0 towards –1, y = sec x increases from –∞ towards –1. When y = cos x = –1, y = sec x = –1.

The graph of cosec x Drag the point through the graph of sin x and as the graph of cosec x is traced out make the following observations: When y = sin x = 0, y = cosec x is undefined. As y = sin x increases from 0 towards 1, y = cosec x decreases from +∞ towards 1. When y = sin x = 1, y = cosec x = 1. As y = sin x decreases from 1 towards 0, y = cosec x increases from 1 towards +∞. When y = sin x = –1, y = cosec x = –1.

The graph of cot x Drag the point through the graph of tan x and as the graph of cot x is traced out make the following observations: When y = tan x is 0, y = cot x is undefined. As y = tan x increases from 0 towards +∞, y = cot x decreases from +∞ towards 0. When y = tan x is undefined, y = cot x = 0. As y = tan x increases from –∞ towards 0, y = cot x decreases from 0 towards –∞.

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: f(x) = cot x f(x) = cosec x f(x) = sec x odd or even period f(x) = 0 when x = asymptotes at x = range domain function x x ≠ 90°+180n° n f(x) ≤ –1, f(x) ≥ 1 90°+180n°, n never 360° even x x ≠ 180n° n f(x) ≤ –1, f(x) ≥ 1 180n°, n never 360° odd x x ≠ 180n° n f(x) 180n°, n 90°+180n°, n 180° odd

Transforming the graph of f(x) = sec x

Transforming the graph of f(x) = cosec x

Transforming the graph of f(x) = cot x

Problems involving reciprocal trig functions Use a calculator to find, to 2 d.p., the value of: a) sec 85° b) cosec 200° c) cot –70° a) sec 85° = = 11.47 (to 2 d.p.) b) cosec 220° = = –1.56 (to 2 d.p.) Ensure that students have their calculators set to degrees before working through these problems. Note that only the cot of an angle can have values between –1 and 1 (the sec or cosec of an angle is either greater than 1 or less than –1). c) cot –70° = = –0.36 (to 2 d.p.)

Problems involving reciprocal trig functions Find the exact value of: a) cosec b) cot c) sec – a) sin = so, cosec = 2 θ is in the 2nd quadrant so tan θ = tan (π – θ) b) tan = –tan (π – ) = – tan = – so, cot = – = – c) cos – = –cos(– + π) θ is in the 3rd quadrant so cos θ = –cos (θ + π) = – cos = – so, sec – = –

Problems involving reciprocal trig functions Given that x is an acute angle and tan x = find the exact values of cot x, sec x and cosec x. Using the following right-angled triangle: x 4 3 5 The length of the hypotenuse is 5 So Using a right-angled triangle, we could define sec x as hypotenuse/adjacent, cosec x as hypotenuse/opposite and cot x as adjacent/opposite. tan x = cos x = sin x = Therefore cot x = sec x = cosec x =

Problems involving reciprocal trig functions Prove that Explain that this identity can be proved by rearranging the left-hand side (LHS) to give the right-hand side (RHS). Using sin2x + cos2x = 1

Problems involving reciprocal trig functions Solve sec (x + 20°) = 2 for 0 ≤ x ≤ 360°. cos (x + 20°) = When solving equations involving compound angles, care should be taken to insure that the solutions given lie within the correct range. For more difficult equations of this type it may be advisable to change the range to match the compound angle. x + 20° = 60° or 300° x = 40° or 280°

Trigonometric identities The inverse trigonometric functions The reciprocal trigonometric functions Trigonometric identities Examination-style question Contents 27 of 35 © Boardworks Ltd 2006

Trigonometric identities Earlier in the course you met the following trigonometric identities: 1 2 We can write these identities in terms of sec θ, cosec θ and cot θ. 1 Using If necessary, remind students of the difference between an equation and an identity. So

Trigonometric identities 2 Dividing through by cos2θ gives 2 Dividing through by sin2θ gives 2 These formulae are not given in the exam and should be learnt.

Trigonometric identities Using sec2x = tan2x + 1 Explain that this identity can be proved by rearranging the left-hand side (LHS) to give the right-hand side (RHS).

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 We could also solve this using a right angled triangle with sides of length 1, √24 and 5 although it must be remembered that since x is obtuse, it is in the second quadrant and tan x is therefore negative. cot x = ±√24 = ±2√6 x is obtuse and so cot x is negative (since tan x is negative in the second quadrant). Therefore: 

Trigonometric equations Solve Using : Remind students that tan is positive in the first and third quadrants. Using a calculator tan–1 ½ = 26.6°. Additional solutions in the range can therefore be found by adding on 180°. or ½ θ = 0°, 180°, 360° θ = 26.6°, 206.6° (to 1 d.p.) The complete solution set is θ = 0°, 26.6°, 180°, 206.6°, 360°.

Examination-style question The inverse trigonometric functions The reciprocal trigonometric functions Trigonometric identities Examination-style question Contents 33 of 35 © Boardworks Ltd 2006

Examination-style question Prove that sec θ ≡ cos θ + sin θ tan θ. 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. a)

The reciprocal trigonometric functions b) Using the result given in part a):