1. A-Level Maths: Core 3 for Edexcel
C3.3 Trigonometry 1
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2. The inverse trigonometric functions
The reciprocal trigonometric functions
Trigonometric identities
Examination-style question
Contents
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3. 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?

4. 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.

5. 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

6. 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 ≤

7. 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.

8. 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

9. 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 <

10. Problems involving inverse trig functions
Find the exact value of sin– 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– =

11. Problems involving inverse trig functions
Find the exact value of sin– 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– =

12. Problems involving inverse trig functions
Find the exact value of cos (sin–1 ) in radians.
Let sin– = 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– = θ so
cos (sin–1 ) =

13. The reciprocal trigonometric functions
The inverse trigonometric functions
The reciprocal trigonometric functions
Trigonometric identities
Examination-style question
Contents
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14. 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.

15. 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.

16. 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.

17. 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 –∞.

18. 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

19. Transforming the graph of f(x) = sec x

20. Transforming the graph of f(x) = cosec x

21. Transforming the graph of f(x) = cot x

22. 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° =
= (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.)

23. 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 – = –

24. 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 =

25. 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

26. 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°

27. Trigonometric identities
The inverse trigonometric functions
The reciprocal trigonometric functions
Trigonometric identities
Examination-style question
Contents
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28. 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

29. Trigonometric identities2
Dividing through by cos2θ gives 2
Dividing through by sin2θ gives 2
These formulae are not given in the exam and should be learnt.

30. 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).

31. 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: 

32. 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
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°.

33. Examination-style question
The inverse trigonometric functions
The reciprocal trigonometric functions
Trigonometric identities
Examination-style question
Contents
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34. 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)

35. The reciprocal trigonometric functionsb)
Using the result given in part a):