Trig Graphs: Shifts and Counting Solutions

Horizontal shifts of sin and cos: maxima, minima and zeros; finding the shift from a given point

In the exam for Edexcel International AS Maths · 9 real past-paper questions below

Shifted and flipped waves

You can already read the amplitude, period and midline of a trig graph. AS questions add three things: the wave is moved sideways, or flipped upside down, and you must find a key point many waves away.

Rule

In $y = a sin (x + B)$ or $y = a cos (x + B)$ , the whole wave slides left by $B$ . Its height and period do not change.

Memory trick

Inside the brackets, things go the opposite way. $+ 20$ moves the wave 20 to the left, $- 30$ moves it 30 to the right.

You do not need to slide anything in your head. Use the inside of the brackets: $sin$ has its top when the inside is $9 0^{\circ}$ , its bottom when the inside is $27 0^{\circ}$ , and its zeros when the inside is $0^{\circ}$ or $18 0^{\circ}$ . For $cos$ , the top is at $0^{\circ}$ and the bottom at $18 0^{\circ}$ .

Worked example

1
The curve $y = A sin (x + 20)^{\circ}$ has a maximum point at $(70, 6)$ . Find $A$ , the minimum point with the smallest positive $x$ , and the maximum point with $700 < x < 800$ .
2
$A = 6$ — A maximum of sin is $1 \times A$ , so $A$ is the top height.
3
$x + 20 = 270 \Rightarrow x = 250$ — The bottom of sin is where the inside is $27 0^{\circ}$ .
4
$(250, - 6)$ — Write the point as $(x, y)$ .
5
$70, 430, 790$ — The period is $36 0^{\circ}$ : keep adding it to the first maximum.
6
$(790, 6)$ — The one between 700 and 800.

Sometimes they give you the point and you find the shift. Work backwards from the inside.

Worked example

1
The curve $y = 6 sin (x + B)$ , with $x$ in radians and $B > 0$ , has its minimum with the smallest positive $x$ at $(\frac{π}{3}, A)$ . Find $A$ and the smallest possible $B$ .
2
$A = - 6$ — A minimum of sin is $- 1 \times 6$ .
3
$\frac{π}{3} + B = \frac{3 π}{2}$ — At a minimum the inside is $\frac{3 π}{2}$ , which is $27 0^{\circ}$ .
4
$B = \frac{3 π}{2} - \frac{π}{3} = \frac{7 π}{6}$ — Taking away $2 π$ makes it negative, so this is the smallest positive $B$ .

A minus sign in front flips the wave. $y = - 2 cos x$ starts at the bottom: bottom, middle, top, middle, bottom.

Worked example

1
For $y = - 2 cos x + k$ , with $x$ in radians, the greatest value of $y$ is 7. Find $k$ and the minimum point with the smallest positive $x$ .
2
$(0, - 2), (\frac{π}{2}, 0), (π, 2), (\frac{3 π}{2}, 0), (2 π, - 2)$ — Key points of $y = - 2 cos x$ : the cos pattern, upside down.
3
$2 + k = 7 \Rightarrow k = 5$ — Adding $k$ lifts the top from 2 to 7.
4
$(2 π, 3)$ — Bottoms at $x = 0$ and $2 π$ , now at height $- 2 + 5 = 3$ . Zero is not positive, so take $2 π$ .

Rule

These are “State” or “Write down” marks: no working needed, so the answer itself must be right. Give $(x, y)$ in that order, with exact values such as $\frac{5 π}{2}$ . If $x$ is in radians, answer in radians: degrees are not always accepted.

Common mistake

Moving the wave the wrong way. In $sin (x + 20)^{\circ}$ the top is at $x = 70$ , not $x = 110$ . Always solve “inside $= 90$ ”.

Key idea

High tide comes about 50 minutes later each day. Same wave, same height, just shifted along: that is a horizontal shift.

Key idea

Find the angle where the basic curve has the feature, set the inside equal to it, then add whole periods.

Solutions in terms of α

Some questions give no numbers. They say “ $sin α = p$ ” and ask for other angles in terms of $α$ . The graph's symmetry gives them.

Graph	same value as at $α$	opposite value
$sin$	$18 0^{\circ} - α$ , $α + 36 0^{\circ}$	$- α$ , $18 0^{\circ} + α$ , $α - 18 0^{\circ}$
$cos$	$- α$ , $36 0^{\circ} - α$	$18 0^{\circ} - α$ , $18 0^{\circ} + α$

Memory trick

Sin is a mirror at its top, $9 0^{\circ}$ : $α$ and $18 0^{\circ} - α$ are the same height. Cos is a mirror at the $y$ -axis: $α$ and $- α$ match. Tan just repeats every $18 0^{\circ}$ .

Worked example

1
Given $cos β = q$ , where $0 < β < 9 0^{\circ}$ , write in terms of $q$ : $cos (18 0^{\circ} - β)$ , $4 + cos (36 0^{\circ} - β)$ and $cos (β - 18 0^{\circ})$ .
2
$cos (18 0^{\circ} - β) = - q$ — Either side of $9 0^{\circ}$ , cos has the same size but the opposite sign.
3
$4 + cos (36 0^{\circ} - β) = 4 + q$ — Cos is the same at $β$ and $36 0^{\circ} - β$ .
4
$cos (β - 18 0^{\circ}) = - q$ — Half a period away: the opposite value.

Then a “Hence” part asks you to solve an equation with a stretched angle. Find all the angles for the inside first, then divide.

Worked example

1
Hence find, in terms of $β$ , the solutions of $cos 2 x = q$ for $0 < x < 18 0^{\circ}$ .
2
$0 < 2 x < 36 0^{\circ}$ — Double the range, because the inside is $2 x$ .
3
$2 x = β or 2 x = 36 0^{\circ} - β$ — The two angles in one turn with cos equal to $q$ .
4
$x = \frac{β}{2}, x = 18 0^{\circ} - \frac{β}{2}$ — Halve both. Both lie in the range.

On a stretched curve, use the period of that curve. The mirror that was at $9 0^{\circ}$ moves to a quarter period.

Worked example

1
The smallest positive solution of $sin \frac{x}{4} = k$ , where $k > 0$ and $x$ is in radians, is $α$ . Find the next positive solution and the negative solution closest to zero.
2
$period = \frac{2 π}{\frac{1}{4}} = 8 π$ — Half a period is $4 π$ .
3
$x = 4 π - α$ — Same height, mirrored inside the first half-wave: like $18 0^{\circ} - α$ .
4
$x = - 4 π - α$ — Same height again, one full period back from $4 π - α$ .

Rule

Mark schemes usually give a mark for the first solution and a method mark for using symmetry correctly. The last accuracy mark needs every solution in the range and no extra ones.

Common mistake

Dividing the range too early. For $cos 2 x$ on $0 < x < 18 0^{\circ}$ , find both values of $2 x$ in $0$ to $36 0^{\circ}$ first, then halve.

Key idea

One angle plus the graph's symmetry gives you every other angle.

Counting solutions

“State the number of solutions” over a huge range is not a lot of work. The wave repeats, so count one period and multiply. Then check the ends.

Steps

1
1. Rearrange to $sin x = \dots$ or $cos x = \dots$ and look at the value.
2. Between $- 1$ and $1$ : the line cuts each period twice. Exactly $1$ or $- 1$ : it touches a top or bottom, once per period. Outside: none.
3. Number of periods $=$ length of the range $\div$ period.
4. Check the ends: $<$ leaves an end out, $\leq$ lets it in.

Worked example

1
Using $y = 3 sin x$ , with $x$ in degrees, state the number of solutions of (i) $3 sin x = 2$ for $0 < x \leq 360 0^{\circ}$ , (ii) $5 + 3 sin x = 8$ for $- 72 0^{\circ} < x \leq 72 0^{\circ}$ , (iii) $3 sin x = 0$ for $0 \leq x \leq 72 0^{\circ}$ .
2
$sin x = \frac{2}{3} : 2 \times 10 = 20$ — Two per period, and $3600 \div 360 = 10$ periods.
3
$sin x = 1 : 1 \times 4 = 4$ — It only touches the tops: $x = - 630, - 270, 90, 450$ .
4
$x = 0, 180, 360, 540, 720 : 5$ — Zeros every $18 0^{\circ}$ , and both ends are allowed. That is $720 \div 180 + 1$ .

Tan has one branch per period, and each full branch runs from far below to far above. So a line usually cuts each full branch once. Watch the branch through $O$ and any half-branch at the ends.

Worked example

1
State the number of solutions of $tan x = 3 x$ for $- 3 π < x < 3 π$ , and for $0 < x < 50 π$ .
2
$central branch: 3$ — They meet at $O$ . Just right of $O$ , tan is close to $x$ (try $tan 0.1 = 0.1003 \dots$ ), so it is below $3 x$ . Then tan shoots up and crosses the line. Same on the left.
3
$4 full branches: 4$ — Each runs from far below to far above, so it crosses the line once.
4
$3 + 4 = 7$ — Beyond $\pm \frac{5 π}{2}$ only half-branches are left, where tan and the line have opposite signs: no more crossings.
5
$0 < x < 50 π : 1 + 49 = 50$ — One on the right half of the central branch, then one in each of the 49 full branches.

Common mistake

Two solutions per period when the value is $1$ or $- 1$ . The line $y = 1$ only touches $sin x$ at each top, so that is one per period.

Common mistake

Forgetting an end. With $0 \leq x \leq 72 0^{\circ}$ the zeros at 0 and 720 both count. With $0 < x < 72 0^{\circ}$ neither does.

Key idea

Count one period, multiply, then fix the ends.

Crossing lines and curves

The solutions of $f (x) = g (x)$ are where $y = f (x)$ and $y = g (x)$ meet. Sketch both on the same axes and count.

Rule

When the question says “giving a reason”, write: “there are 3 solutions because the graphs intersect 3 times.” Saying where a graph crosses the $x$ -axis is not the reason.

Worked example

1
Using a sketch of $y = 2 sin x$ for $0 \leq x \leq 2 π$ , state the number of solutions of $2 sin x = 2 - \frac{2 x}{π}$ , giving a reason.
2
$(0, 2), (π, 0), (2 π, - 2)$ — Three points on the line $y = 2 - \frac{2 x}{π}$ . It slopes down.
3
$x = π : 2 sin π = 0$ — Both are 0 here: one meeting point on the $x$ -axis.
4
$x = \frac{π}{2} : 2 > 1$ — The curve is above the line here but below it at 0, so they cross once in between.
5
$x = \frac{3 π}{2} : - 2 < - 1$ — The curve is below the line, but above it at $2 π$ : one more crossing.
6
$3 solutions$ — Because the graphs intersect 3 times.

Two trig curves work the same way. Count in one shared period, then use the repeat.

Worked example

1
$cos x = sin 2 x$ has 4 solutions with $0 \leq x < 2 π$ . How many are there with $0 \leq x < 10 π$ , and with $0 \leq x < 11 π$ ?
2
$x = \frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}, \frac{3 π}{2}$ — The 4 crossings in the first $2 π$ .
3
$2 π repeats$ — Both curves repeat every $2 π$ , so the picture does too.
4
$4 \times 5 = 20$ — $10 π$ is 5 repeats.
5
$20 + 3 = 23$ — The extra $π$ adds the crossings in $0 \leq x < π$ : $\frac{π}{6}, \frac{π}{2}, \frac{5 π}{6}$ .

Common mistake

Drawing the line with the wrong slope or through the wrong point. Plot two exact points first, such as $x = 0$ and $x = π$ .

Key idea

Sketch both, count where they meet, and say so in words.

Which method? Read the question

The wording tells you which tool to pick up.

You see	You do
$sin (x + B)$ or $cos (x - B)$ , find a point	Set the inside to $9 0^{\circ}$ , $27 0^{\circ}$ , … then add periods
A point given, find $B$	Inside at that point $=$ the feature's angle; solve for $B$
$- a cos x$ or $- a sin x$	Flip the pattern: cos starts at the bottom
“In terms of $α$ ”	Symmetry: $18 0^{\circ} - α$ , $18 0^{\circ} + α$ , $- α$ ; use the stretched period
“State the number of solutions” over a big range	Count one period, multiply, check the ends and any touching
$f (x) = g (x)$ , “giving a reason”	Sketch both, count intersections, say so

Common mistake

Mixing units. If $x$ is in radians, every coordinate and period must be in radians too.

Key idea

Every question here is the same wave: shifted, flipped or crossed by something. Find its key points and the rest follows.

Practise trig graphs: shifts and counting solutions now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on trig graphs: shifts and counting solutions

Each with a full worked solution.

All 9 questions on this topic →

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.