Write down the exact value of 30°.

Recall the special angles: Key angles: 0°, 30°, 45°, 60°, 90° State the exact value: 30° = (1)/(2) Write the final answer: 30° = (1)/(2) Answer: (1)/(2)

Write down the coordinates of the maximum point of y= x for 0° ≤ x ≤ 360°.

Recall the maximum value of sine: x ≤ 1 Find where the peak occurs: 90° = 1 Write the coordinates: (90°, 1) Answer: (90°, 1)

Solve x = √(3)2 for 0° ≤ x ≤ 360°.

Find the principal value: x = ^-1(32) = 30° Decide where cosine is positive: 1st and 4th quadrants Use symmetry for the second solution: x = 360° - 30° Evaluate the second solution: x = 330° Check the range: 0° ≤ x ≤ 360° List all solutions: x = 30°, 330° Answer: x = 30°, 330°

Solve 2^2 x + x - 1 = 0 for 0° ≤ x ≤ 360°.

Recognise the hidden quadratic: let c = x Write the quadratic: 2c^2 + c - 1 = 0 Identify the coefficients: a = 2, b = 1, c_0 = -1 Factorise: (2c - 1)(c + 1) = 0 Solve each bracket: c = 12 or c = -1 First value: find the acute angle: ^-1(12) = 60° First value: quadrant solutions: x = 12 x = 60°, 300° Second value: solve cosine = -1: x = -1 x = 180° Combine the solutions: x = 60°, 180°, 300° Check the factorisation: (2c-1)(c+1) = 2c^2 + c - 1 Verify a solution: 2^2 60° + 60° - 1 = 2(14) + 12 - 1 = 0 Verify the fourth-quadrant solution: 2^2 300° + 300° - 1 = 2(14) + 12 - 1 = 0 Explain why three, not four: x = -1 gives only one angle Interpret using the graph: lines y=12, y=-1 meet the curve Check the range: 0° ≤ x ≤ 360° List all solutions: x = 60°, 180°, 300° Answer: x = 60°, 180°, 300°

A-Level Trigonometric graphs Practice Questions

Free A-Level Trigonometric graphs practice questions with full step-by-step worked solutions. Covers exact values, sine, cosine, tangent. Practise exam-style problems and check your method.

exact valuessinecosinetangentradiansperiod

A-Level70 questionsStep-by-step solutions

Question 1

1 markeasy

Write down the exact value of

\sin 30^\circ

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Worked solution

Recall the special angles
$\text{Key angles: } 0^\circ,\,30^\circ,\,45^\circ,\,60^\circ,\,90^\circ$
There are five special angles whose sine, cosine and tangent we are expected to know by heart. The value of $\sin 30^\circ$ is one of these, so we do not need a calculator.
State the exact value
$\sin 30^\circ = \frac{1}{2}$
From the standard results, $\sin 30^\circ$ is exactly one half. A quick check: on the sine graph the height at $30^\circ$ is halfway up to the maximum of 1.
Write the final answer
$\sin 30^\circ = \frac{1}{2}$
So the exact value is $\tfrac{1}{2}$ . Keeping it as a fraction (not a rounded decimal) is what 'exact' means.

Answer

\frac{1}{2}

Question 2

2 markseasy

Write down the coordinates of the maximum point of

y=\sin x

for

0^\circ \le x \le 360^\circ

Show worked solution

Worked solution

Recall the maximum value of sine
$\sin x \le 1$
The sine curve never rises above 1, so its maximum height is $y=1$ . We now need the $x$ -value where this happens.
Find where the peak occurs
$\sin 90^\circ = 1$
Reading the graph, the peak of the first wave is at $x=90^\circ$ . This is one quarter of the way through the full $360^\circ$ cycle.
Write the coordinates
$(90^\circ,\ 1)$
So the maximum point is $(90^\circ,1)$ .

Answer

(90^\circ,\ 1)

Question 3

3 marksintermediate

Solve

\cos x = \dfrac{\sqrt{3}}{2}

for

0^\circ \le x \le 360^\circ

Show worked solution

Worked solution

Find the principal value
$x = \cos^{-1}\left(\tfrac{\sqrt3}{2}\right) = 30^\circ$
Taking the inverse cosine of $\tfrac{\sqrt3}{2}$ gives the basic angle $30^\circ$ . This is the first solution.
Decide where cosine is positive
$\text{1st and 4th quadrants}$
The value is positive, and cosine is positive in the first and fourth quadrants.
Use symmetry for the second solution
$x = 360^\circ - 30^\circ$
The cosine graph is symmetric about $x=180^\circ$ in a way that gives a partner solution at $360^\circ$ minus the principal value.
Evaluate the second solution
$x = 330^\circ$
This gives $x=330^\circ$ , which is inside the range.
Check the range
$0^\circ \le x \le 360^\circ$
Both $30^\circ$ and $330^\circ$ lie in range; no others do.
List all solutions
$x = 30^\circ,\ 330^\circ$
So the solutions are $30^\circ$ and $330^\circ$ .

Answer

x = 30^\circ,\ 330^\circ

Question 4

4 markshard

For the curve

y=5\cos 3x - 2

, which statement correctly gives its amplitude, period and range?

Show worked solution

Worked solution

Identify the amplitude
$|5| = 5$
The number multiplying the cosine is 5, so the amplitude is 5. The curve rises 5 above and falls 5 below its centre line.
Identify the period
$\text{period} = \frac{360^\circ}{3} = 120^\circ$
The 3 multiplying $x$ divides the usual period, giving $\tfrac{360^\circ}{3}=120^\circ$ .
Find the centre line
$y = -2$
The $-2$ shifts the whole curve down, so it oscillates about the line $y=-2$ .
Work out the maximum
$y_{\max} = -2 + 5 = 3$
The highest value is centre plus amplitude, $-2+5=3$ .
Work out the minimum
$y_{\min} = -2 - 5 = -7$
The lowest value is centre minus amplitude, $-2-5=-7$ .
State the range
$-7 \le y \le 3$
So the range is from $-7$ to $3$ .
Reject the wrong-period option
$\text{period } 360^\circ \text{ ignores the } 3$
An option giving period $360^\circ$ has forgotten to divide by the 3 multiplying $x$ , so it is wrong.
Reject the swapped option
$\text{amplitude } 3 \text{ confuses } b \text{ with } a$
An option with amplitude 3 has mixed up the stretch factor 5 with the frequency 3, so it is wrong.
Reject the wrong-shift option
$c = -2 \text{ gives centre } y=-2$
An option with range starting at $-2$ has ignored that the whole curve drops by 2, so it is wrong.
Choose the correct statement
$\text{amplitude } 5,\ \text{period } 120^\circ,\ -7 \le y \le 3$
The correct option lists amplitude 5, period $120^\circ$ and range $-7\le y\le3$ .

Answer

Amplitude 5, period 120 degrees, range -7 <= y <= 3.

Question 5

7 markschallenging

Solve

2\cos^2 x + \cos x - 1 = 0

for

0^\circ \le x \le 360^\circ

Show worked solution

Worked solution

Recognise the hidden quadratic
$\text{let } c = \cos x$
The equation is a quadratic in $\cos x$ . Writing $c=\cos x$ turns it into $2c^2+c-1=0$ , which we can factorise like any quadratic.
Write the quadratic
$2c^2 + c - 1 = 0$
Now it looks exactly like a standard quadratic equation in $c$ .
Identify the coefficients
$a = 2,\ b = 1,\ c_0 = -1$
Comparing with $ac^2+bc+c_0=0$ , the coefficients are $a=2$ , $b=1$ and constant $-1$ . We look for factors of $a\times(-1)=-2$ that add to $1$ : these are $+2$ and $-1$ .
Factorise
$(2c - 1)(c + 1) = 0$
We look for two brackets that multiply to give $2c^2+c-1$ . Checking: $(2c-1)(c+1)=2c^2+2c-c-1=2c^2+c-1$ , correct.
Solve each bracket
$c = \tfrac{1}{2} \ \text{ or } \ c = -1$
Setting each factor to zero gives $\cos x=\tfrac12$ or $\cos x=-1$ .
First value: find the acute angle
$\cos^{-1}\left(\tfrac12\right) = 60^\circ$
For $\cos x=\tfrac12$ the basic angle is $60^\circ$ .
First value: quadrant solutions
$\cos x = \tfrac12 \Rightarrow x = 60^\circ,\ 300^\circ$
Cosine is positive in the first and fourth quadrants, giving $60^\circ$ and $360^\circ-60^\circ=300^\circ$ .
Second value: solve cosine = -1
$\cos x = -1 \Rightarrow x = 180^\circ$
Cosine reaches $-1$ only at the bottom of its cycle, at $x=180^\circ$ .
Combine the solutions
$x = 60^\circ,\ 180^\circ,\ 300^\circ$
Putting both cases together gives three solutions.
Check the factorisation
$(2c-1)(c+1) = 2c^2 + c - 1$
Expanding the brackets returns the original quadratic, so the factorisation was correct — a quick safeguard against slips.
Verify a solution
$2\cos^2 60^\circ + \cos 60^\circ - 1 = 2\left(\tfrac14\right) + \tfrac12 - 1 = 0$
Substituting $x=60^\circ$ gives $\tfrac12+\tfrac12-1=0$ , confirming this solution.
Verify the fourth-quadrant solution
$2\cos^2 300^\circ + \cos 300^\circ - 1 = 2\left(\tfrac14\right) + \tfrac12 - 1 = 0$
Since $\cos 300^\circ=\tfrac12$ as well, substituting $x=300^\circ$ also gives $\tfrac12+\tfrac12-1=0$ , confirming this solution.
Explain why three, not four
$\cos x = -1 \text{ gives only one angle}$
The value $\cos x=\tfrac12$ gives two angles, but $\cos x=-1$ gives only one (at $180^\circ$ ), so there are three solutions rather than four.
Interpret using the graph
$\text{lines } y=\tfrac12,\ y=-1 \text{ meet the curve}$
The line $y=\tfrac12$ cuts the cosine graph twice and $y=-1$ just touches it at the bottom, giving three intersection points.
Check the range
$0^\circ \le x \le 360^\circ$
All three values lie in the range.
List all solutions
$x = 60^\circ,\ 180^\circ,\ 300^\circ$
So the solutions are $60^\circ$ , $180^\circ$ and $300^\circ$ .

Answer

x = 60^\circ,\ 180^\circ,\ 300^\circ

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