What is 5.4: Reciprocal and inverse trig functions?

The reciprocal functions are x=1/cos x, x=1/sin x and x=cos x/sin x; their graphs inherit zeros of the denominator as vertical asymptotes. Secant and cosecant have range (-,-1][1,) and period 2π, while cotangent has range R and period π.

How many marks are 5.4 questions worth?

5.4 practice drills on this page are worth 1–6 marks each; marks in a real exam vary by question.

Which past exam papers has 5.4 appeared in?

We have not yet indexed a verified past-paper appearance for 5.4 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

5.4

Understand and use the definitions of secant, cosecant and cotangent and of arcsin, arccos and arctan; their relationships to sine, cosine and tangent; understanding of their graphs; their ranges and domains.

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Reciprocal and inverse trig functions

Worked answers and methods for 5.4 on Edexcel A-level Maths 9MA0.

Explanation

The reciprocal functions are $\sec x=1/\cos x$ , $\cosec x=1/\sin x$ and $\cot x=\cos x/\sin x$ ; their graphs inherit zeros of the denominator as vertical asymptotes. Secant and cosecant have range $(-\infty,-1]\cup[1,\infty)$ and period $2\pi$ , while cotangent has range $\mathbb{R}$ and period $\pi$ .
The inverse graphs are restrictions reflected in $y=x$ .
Their principal ranges are $-\pi/2\leq\arcsin x\leq\pi/2$ , $0\leq\arccos x\leq\pi$ and $-\pi/2<\arctan x<\pi/2$ ; $\arcsin x$ and $\arccos x$ require $-1\leq x\leq1$ .
Angles may be in degrees or radians.
The notation $\sin^{-1}x$ means $\arcsin x$ , not the reciprocal $\cosec x$ .

Worked example

Give the principal values, in radians, of $\arcsin(-1/2)$ and $\arctan(-1)$ .

1.Within the principal sine-inverse range, $\sin(-\pi/6)=-1/2$ , so $\arcsin(-1/2)=-\pi/6$ .
2.Within the principal tangent-inverse range, $\tan(-\pi/4)=-1$ , so $\arctan(-1)=-\pi/4$ .

Answer: $\arcsin(-1/2)=-\pi/6$ $\arctan(-1)=-\pi/4$

Common mistakes

Don't read $\sin^{-1}x$ as the reciprocal $1/\sin x$ instead of the inverse function $\arcsin x$ .
Don't give an inverse-trigonometric answer outside the stated principal range.

Exam tip

Check the input domain and principal output range before giving an inverse-trigonometric value.

Worked practice

Tier 1 · Easy

Given

\cos\theta=-4/5

, write down

\sec\theta

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\sec\theta=-\dfrac54$	1
Notes
Secant is the reciprocal of cosine, so $\sec\theta=1/(-4/5)=-5/4$ .

(1 mark)

Tier 2 · Standard

For

0\leq x\leq2\pi

, state where

\sec x

is undefined and solve

\sec x=-2

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Undefined at $x=\dfrac{\pi}{2},\dfrac{3\pi}{2}$ $x=\dfrac{2\pi}{3},\dfrac{4\pi}{3}$	4
Notes
Since $\sec x=1/\cos x$ , it is undefined when $\cos x=0$ , namely at $x=\pi/2$ and $3\pi/2$ . Also $\sec x=-2$ is equivalent to $\cos x=-1/2$ , which occurs in the interval at $x=2\pi/3$ and $4\pi/3$ .

(4 marks)

Tier 3 · Hard

For

y=2\sec x-1

, state the period and range, then give all vertical asymptotes in

-\pi\leq x\leq\pi

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Period $2\pi$ Range $y\leq-3$ or $y\geq1$ Vertical asymptotes $x=-\pi/2$ and $x=\pi/2$	5
Notes
Secant has period $2\pi$ , so the transformation does not change the period. Since $\sec x\leq-1$ or $\sec x\geq1$ , multiplying by $2$ and subtracting $1$ gives $y\leq-3$ or $y\geq1$ . Vertical asymptotes occur where $\cos x=0$ , namely $x=\pi/2+k\pi$ ; in the stated interval these are $x=-\pi/2$ and $x=\pi/2$ .

(5 marks)

Tier 1 · Easy

Find the principal value, in radians, of

\arccos(-\sqrt3/2)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\dfrac{5\pi}{6}$	2
Notes
The principal range of arccos is $[0,\pi]$ . In this range, $\cos(5\pi/6)=-\sqrt3/2$ , so the principal value is $5\pi/6$ .

(2 marks)

Tier 2 · Standard

Let

f(x)=\arcsin(2x-1)

. State the domain and range of

f

, and solve

f(x)=\pi/6

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Domain $0\leq x\leq1$ and range $-\pi/2\leq f(x)\leq\pi/2$ $x=\dfrac34$	4
Notes
For arcsin, the input must satisfy $-1\leq2x-1\leq1$ , giving $0\leq x\leq1$ . Its principal range is $[-\pi/2,\pi/2]$ . If $\arcsin(2x-1)=\pi/6$ , then $2x-1=\sin(\pi/6)=1/2$ , so $x=3/4$ .

(4 marks)

Tier 3 · Hard

Show that

\arcsin x+\arccos x=\pi/2

for

-1\leq x\leq1

. Hence solve

3\arcsin x-\arccos x=0

exactly.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$\arcsin x+\arccos x=\pi/2$ $x=\dfrac{\sqrt{2-\sqrt2}}{2}$	5
Notes
Let $y=\arcsin x$ , so $-\pi/2\leq y\leq\pi/2$ and $\sin y=x$ . Since $\cos(\pi/2-y)=\sin y=x$ and $0\leq\pi/2-y\leq\pi$ , the principal value is $\arccos x=\pi/2-y$ . Hence $\arcsin x+\arccos x=\pi/2$ . The equation becomes $3\arcsin x-(\pi/2-\arcsin x)=0$ , so $\arcsin x=\pi/8$ . Therefore $x=\sin(\pi/8)=\sqrt{2-\sqrt2}/2$ .

(5 marks)

Tier 2 · Standard

Let

f(x)=\arccos(1/x)

. State the domain and range of

f

, and solve

f(x)=2\pi/3

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Domain $(-\infty,-1]\cup[1,\infty)$ Range $[0,\pi/2)\cup(\pi/2,\pi]$ $x=-2$	4
Notes
The input to arccos must satisfy $-1\leq1/x\leq1$ , which gives $x\leq-1$ or $x\geq1$ . On this domain, $1/x$ takes every value in $[-1,0)\cup(0,1]$ , so applying the decreasing principal arccos function gives the range $[0,\pi/2)\cup(\pi/2,\pi]$ ; $\pi/2$ is excluded because $1/x$ cannot equal zero. Finally, $\arccos(1/x)=2\pi/3$ gives $1/x=\cos(2\pi/3)=-1/2$ , so $x=-2$ .

(4 marks)

Tier 3 · Hard

For

0\leq x\leq3\pi

, express

f(x)=\arcsin(\sin x)

as a piecewise function. State the range of

f

on this interval and hence solve

f(x)=\pi/6

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$f(x)=x$ for $0\leq x\leq\pi/2$ ; $f(x)=\pi-x$ for $\pi/2\leq x\leq3\pi/2$ ; $f(x)=x-2\pi$ for $3\pi/2\leq x\leq5\pi/2$ ; $f(x)=3\pi-x$ for $5\pi/2\leq x\leq3\pi$ Range $[-\pi/2,\pi/2]$ $x=\pi/6,\ 5\pi/6,\ 13\pi/6,\ 17\pi/6$	6
Notes
The principal range of arcsin is $[-\pi/2,\pi/2]$ . On the first quarter-cycle the principal angle is $x$ ; reflection across $\pi/2$ gives $\pi-x$ until $3\pi/2$ . Over the next principal branch it is $x-2\pi$ , followed by the reflection $3\pi-x$ . These branches attain every value from $-\pi/2$ to $\pi/2$ . For $f(x)=\pi/6$ , equivalently $\sin x=1/2$ with principal output $\pi/6$ , the four values in the stated interval are $\pi/6,5\pi/6,13\pi/6,17\pi/6$ .

(6 marks)

Tier 3 · Hard

For

x\neq0

, determine the value of

\arctan x+\arctan(1/x)

separately for

x>0

and

x<0

. Hence solve

\arctan x+\arctan(1/x)=-\pi/2

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\arctan x+\arctan(1/x)=\pi/2$ for $x>0$ and $-\pi/2$ for $x<0$ The solution is every $x<0$ .	5
Notes
For $x>0$ , let $\alpha=\arctan x$ , so $0<\alpha<\pi/2$ . The angle $\pi/2-\alpha$ is also in the principal range of arctan and has tangent $1/x$ , hence $\arctan(1/x)=\pi/2-\alpha$ . The sum is therefore $\pi/2$ . If $x<0$ , apply the oddness of arctan to $-x>0$ : both inverse-tangent terms change sign, so the sum is $-\pi/2$ . Consequently the equation holds exactly when $x<0$ .

(5 marks)

Q10

Tier 3 · Hard

10.

For

f(x)=3-2\cosec x

, find the exact coordinates of the maximum point on

0<x<\pi

and the minimum point on

\pi<x<2\pi

. Hence solve

f(x)=7

for

0<x<2\pi

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Maximum point $(\pi/2,1)$ on $0<x<\pi$ Minimum point $(3\pi/2,5)$ on $\pi<x<2\pi$ $x=7\pi/6,\ 11\pi/6$	5
Notes
On $0<x<\pi$ , $0<\sin x\leq1$ , so $\cosec x\geq1$ and $f(x)\leq1$ . Equality occurs only when $\sin x=1$ , giving the maximum point $(\pi/2,1)$ . On $\pi<x<2\pi$ , $-1\leq\sin x<0$ , so $\cosec x\leq-1$ and $f(x)\geq5$ . Equality occurs only when $\sin x=-1$ , giving the minimum point $(3\pi/2,5)$ . Finally, $f(x)=7$ gives $\cosec x=-2$ , or $\sin x=-1/2$ . The complete solution set in the stated interval is $x=7\pi/6,11\pi/6$ .

(5 marks)

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