What is 5.3: Trig graphs and exact values?

sin x and cos x have range [-1,1] and period 2π, while tan x has range R, period π and vertical asymptotes at x=π/2+kπ.

How many marks are 5.3 questions worth?

5.3 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 5.3 appeared in?

We have not yet indexed a verified past-paper appearance for 5.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

5.3

Understand and use the sine, cosine and tangent functions; their graphs, symmetries and periodicity; know and use exact values of sin, cos and tan for standard angles and their multiples.

Draft — not yet indexed

Trig graphs and exact values

Worked answers and methods for 5.3 on Edexcel A-level Maths 9MA0.

Explanation

$\sin x$ and $\cos x$ have range $[-1,1]$ and period $2\pi$ , while $\tan x$ has range $\mathbb{R}$ , period $\pi$ and vertical asymptotes at $x=\pi/2+k\pi$ .
Use $\sin(-x)=-\sin x$ , $\cos(-x)=\cos x$ and $\tan(-x)=-\tan x$ , then reduce an angle by a whole period before using its reference angle and quadrant.
Know exact sine and cosine values at $0$ , $\pi/6$ , $\pi/4$ , $\pi/3$ , $\pi/2$ and $\pi$ , and exact tangent values where defined.
Transformations such as $y=\cos(x+\pi/6)$ shift the graph, while $y=\tan2x$ halves its period.
In $a\cos(bx)+c$ , the amplitude is $|a|$ , period $2\pi/|b|$ , and $c$ moves the midline.

Two cycles of

y=\cos x

, showing range

[-1,1]

and period

2\pi

Worked example

For $y=3\cos(2x)-1$ , state the amplitude, period, maximum value and minimum value.

1.The coefficient outside cosine gives amplitude $|3|=3$ .
2.The factor $2$ inside gives period $2\pi/2=\pi$ .
3.Since $-1\leq\cos(2x)\leq1$ , multiplying by $3$ and subtracting $1$ gives $-4\leq y\leq2$ .

Answer: Amplitude $3$ Period $\pi$ Maximum $2$ Minimum $-4$

Common mistakes

Don't use the coefficient outside the trigonometric function to calculate the period instead of the amplitude.
Don't give tangent a period of $2\pi$ or omit its vertical asymptotes.

Exam tip

On a graph question, state amplitude, period, midline and range separately before sketching one complete cycle.

Worked practice

Tier 1 · Easy

Find the exact value of

\tan(5\pi/6)

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$-\dfrac{1}{\sqrt3}$ $-\dfrac{\sqrt3}{3}$	2
Notes
The reference angle is $\pi/6$ . Tangent is negative in the second quadrant, so $\tan(5\pi/6)=-\tan(\pi/6)=-1/\sqrt3=-\sqrt3/3$ .

(2 marks)

Tier 2 · Standard

Find the exact values of

\sin\left(\dfrac{7\pi}{6}\right)

\cos\left(\dfrac{5\pi}{3}\right)

and

\tan\left(\dfrac{3\pi}{4}\right)

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$-\dfrac12$ $\dfrac12$ $-1$	3
Notes
The reference angles are $\pi/6$ , $\pi/3$ and $\pi/4$ . The angle $7\pi/6$ is in quadrant III, so its sine is $-1/2$ . The angle $5\pi/3$ is in quadrant IV, so its cosine is $1/2$ . The angle $3\pi/4$ is in quadrant II, so its tangent is $-1$ .

(3 marks)

Tier 3 · Hard

Evaluate exactly

\sin(-11\pi/6)

\cos(13\pi/3)

and

\tan(7\pi/4)

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$\sin(-11\pi/6)=\dfrac12$ $\cos(13\pi/3)=\dfrac12$ $\tan(7\pi/4)=-1$	4
Notes
Add $2\pi$ to $-11\pi/6$ to get $\pi/6$ , so its sine is $1/2$ . Subtract $4\pi$ from $13\pi/3$ to get $\pi/3$ , so its cosine is $1/2$ . The angle $7\pi/4$ has reference angle $\pi/4$ in quadrant IV, where tangent is negative, giving $-1$ .

(4 marks)

Tier 1 · Easy

Find the exact value of

\cos(-7\pi/3)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\dfrac12$	2
Notes
Cosine is even and has period $2\pi$ . Thus $\cos(-7\pi/3)=\cos(7\pi/3)=\cos(\pi/3)=1/2$ .

(2 marks)

Tier 2 · Standard

For

y=2\sin(3x)-4

, state the amplitude and period, and find the exact coordinates of every maximum point for

0\leq x\leq2\pi

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Amplitude $2$ and period $2\pi/3$ Maximum points $(\pi/6,-2)$ , $(5\pi/6,-2)$ and $(3\pi/2,-2)$	4
Notes
The amplitude is $2$ and the period is $2\pi/3$ . A maximum occurs when $\sin(3x)=1$ , so $3x=\pi/2+2k\pi$ . Hence $x=\pi/6+2k\pi/3$ , giving $x=\pi/6,5\pi/6,3\pi/2$ in the interval. At each, $y=2(1)-4=-2$ .

(4 marks)

Tier 3 · Hard

The function

f(x)=a\cos(bx)+c

, where

b>0

, has maximum value

7

, minimum value

-3

and least positive period

4\pi/3

. Given that

f(0)=-3

, find

a

b

and

c

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$a=-5$ , $b=\dfrac32$ , $c=2$	5
Notes
The midline is $c=(7+(-3))/2=2$ and the amplitude is $\|a\|=(7-(-3))/2=5$ . Since $f(0)=a+c=-3$ , $a=-5$ . The period gives $2\pi/b=4\pi/3$ , so $b=3/2$ .

(5 marks)

Tier 2 · Standard

For

y=\tan(2x-\pi/4)

, state the period and find all zeros and vertical asymptotes for

0\leq x\leq\pi

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Period $=\pi/2$ Zeros at $x=\pi/8$ and $x=5\pi/8$ Vertical asymptotes at $x=3\pi/8$ and $x=7\pi/8$	4
Notes
The factor $2$ gives period $\pi/2$ . Zeros occur when $2x-\pi/4=k\pi$ , so $x=\pi/8+k\pi/2$ ; the values in the interval are $\pi/8$ and $5\pi/8$ . Vertical asymptotes occur when $2x-\pi/4=\pi/2+k\pi$ , so $x=3\pi/8+k\pi/2$ ; the values in the interval are $3\pi/8$ and $7\pi/8$ .

(4 marks)

Tier 3 · Hard

Let

f(x)=\sin^2x

. Without using a double-angle identity, show that

f

is even and has period

\pi

. State its range and give the exact coordinates of every maximum and minimum point for

0\leq x\leq2\pi

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$f(-x)=f(x)$ and $f(x+\pi)=f(x)$ Range $[0,1]$ Maximum points $(\pi/2,1)$ and $(3\pi/2,1)$ Minimum points $(0,0)$ , $(\pi,0)$ and $(2\pi,0)$	5
Notes
Since $\sin(-x)=-\sin x$ , $f(-x)=(-\sin x)^2=\sin^2x=f(x)$ , so $f$ is even. Also $\sin(x+\pi)=-\sin x$ , hence $f(x+\pi)=f(x)$ and $\pi$ is a period. No smaller positive period exists: $f(x)=0$ only at integer multiples of $\pi$ , so any period must map $0$ to another zero of $f$ and is therefore at least $\pi$ . From $-1\leq\sin x\leq1$ , the range of $\sin^2x$ is $[0,1]$ . The value $1$ occurs at $x=\pi/2,3\pi/2$ , and the value $0$ occurs at $x=0,\pi,2\pi$ in the stated interval.

(5 marks)

Tier 3 · Hard

For

f(x)=|2\cos x-1|

, state the least positive period and the range. Find the exact coordinates of every zero and every global maximum point for

0\leq x\leq2\pi

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Least positive period $=2\pi$ and range $[0,3]$ Zeros $(\pi/3,0)$ and $(5\pi/3,0)$ Global maximum point $(\pi,3)$	5
Notes
Since $-1\leq\cos x\leq1$ , the expression $2\cos x-1$ ranges from $-3$ to $1$ , so $0\leq f(x)\leq3$ . The zeros satisfy $\cos x=1/2$ , giving $x=\pi/3,5\pi/3$ . The maximum value $3$ occurs when $\cos x=-1$ , at $x=\pi$ . The absolute value does not halve the period here: the gaps between consecutive zeros alternate between $4\pi/3$ and $2\pi/3$ , so the zero pattern first repeats after $2\pi$ . Hence the least positive period is $2\pi$ .

(5 marks)

Q10

Tier 3 · Hard

10.

For integers

n\geq0

, define

u_n=2\cos(n\pi/3)+\sin(n\pi/2)

. Make a table of the exact values of

u_n

for

0\leq n\leq11

. Hence state the least positive integer period of the sequence and find how many integers

n

with

0\leq n\leq59

satisfy

u_n\leq-1

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$(u_0,u_1,\ldots,u_{11})=(2,2,-1,-3,-1,2,2,0,-1,-1,-1,0)$ Least positive integer period $12$ $30$ integers	6
Notes
Using exact values at multiples of $\pi/3$ and $\pi/2$ gives the stated $12$ -term table. Both component sequences repeat after $12$ integer steps, so $12$ is a period; the table has no repeated initial block with a smaller length dividing $12$ , so it is the least positive integer period. In one period, $u_n\leq-1$ for $n=2,3,4,8,9,10$ , giving $6$ qualifying terms. The $60$ integers from $0$ to $59$ form five complete periods, so the required number is $5\times6=30$ .

(6 marks)

Verified exam appearances

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