Convert 5π6 radians to degrees. [1 mark]

State the period and amplitude of y = 4cos(12x). [2 marks]

ScotlandMathsSyllabus dot point

How do you measure angles in radians, read exact trigonometric values, and sketch and transform trigonometric graphs?

Radian measure and the conversion between degrees and radians, exact values of sine, cosine and tangent for the standard angles, the graphs of the trigonometric functions, and the transformations that change their amplitude, period and phase.

A focused answer to the SQA Higher Mathematics trigonometry and radians content, covering radian measure and degree conversion, exact values of sine, cosine and tangent, the trigonometric graphs, and the amplitude, period and phase transformations that reshape them.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Radians and conversion
Exact values
Graphs and transformations
Examples in context
Try this

What this dot point is asking

The SQA wants you to measure angles in radians and convert to and from degrees, recall the exact values of the trigonometric ratios for the standard angles, sketch the trigonometric graphs, and apply transformations that change their amplitude, period and phase.

Radians and conversion

A radian is the angle subtended at the centre of a circle by an arc whose length equals the radius. Because the full circumference is $2\pi r$ , a complete turn is $2\pi$ radians, so $2\pi$ radians $= 360^\circ$ and therefore $\pi$ radians $= 180^\circ$ . This single fact drives every conversion.

Paper 1 is non-calculator, so you must work in exact radian multiples of $\pi$ rather than decimals.

Convert between degrees and radians

Practise both directions of conversion using the single key fact $\pi \text{ rad} = 180^\circ$ .

Step 1: Convert $135^\circ$ to radians

To go from degrees to radians, multiply by $\dfrac{\pi}{180}$ , since $1^\circ = \dfrac{\pi}{180}$ radians. Simplify by cancelling common factors:

135 \times \dfrac{\pi}{180} = \dfrac{135\pi}{180} = \dfrac{3\pi}{4}

So $135^\circ = \dfrac{3\pi}{4}$ radians. On Paper 1 you must give an exact multiple of $\pi$ , not a decimal.

Step 2: Convert $\dfrac{7\pi}{6}$ radians to degrees

To go from radians to degrees, multiply by $\dfrac{180}{\pi}$ . The $\pi$ cancels:

\dfrac{7\pi}{6} \times \dfrac{180}{\pi} = \dfrac{7 \times 180}{6} = 210^\circ

Final answer: $135^\circ = \dfrac{3\pi}{4}$ radians and $\dfrac{7\pi}{6}$ radians $= 210^\circ$ .

Exact values

These come from two triangles. The $45$ - $45$ - $90$ triangle has sides $1$ , $1$ , $\sqrt{2}$ , giving the $45^\circ$ ratios. The $30$ - $60$ - $90$ triangle has sides $1$ , $\sqrt{3}$ , $2$ , giving the $30^\circ$ and $60^\circ$ ratios. Use the CAST diagram or the symmetry of the graphs to extend these to angles in any quadrant: in each quadrant you find the related acute angle, take its exact value, then attach the correct sign.

Graphs and transformations

For $y = a\sin(bx + c)$ the amplitude is $|a|$ , the period is $\dfrac{2\pi}{b}$ , and the graph is shifted horizontally so that the starting feature occurs where $bx + c = 0$ . A vertical shift $y = a\sin(bx) + d$ raises the whole curve by $d$ , so the midline becomes $y = d$ and the maximum and minimum are $d + |a|$ and $d - |a|$ .

State the amplitude and period of

y = 3\sin(2x)

Identify the key features of the graph of $y = 3\sin(2x)$ by reading off the parameters.

Step 1: Identify the amplitude from the coefficient outside the sine

The general form is $y = a\sin(bx)$ . Here $a = 3$ , and the amplitude is $|a| = 3$ . This means the graph stretches from $-3$ to $3$ on the $y$ -axis: three times the height of the standard sine curve.

Step 2: Identify the period from the coefficient inside the sine

The coefficient inside is $b = 2$ . The period is $\dfrac{2\pi}{b} = \dfrac{2\pi}{2} = \pi$ radians. A larger $b$ compresses the graph horizontally, so the wave completes one full cycle every $\pi$ radians rather than every $2\pi$ .

Final answer: The amplitude is $3$ and the period is $\pi$ radians, so two full cycles fit within the $2\pi$ width of the standard display window.

Examples in context

Many natural cycles are modelled by $y = a\sin(bx) + d$ . A tide that rises to $5$ m and falls to $1$ m every $12$ hours has midline $d = 3$ , amplitude $a = 2$ , and period $12$ , so $b = \dfrac{2\pi}{12} = \dfrac{\pi}{6}$ . The model $y = 2\sin\!\left(\dfrac{\pi}{6}t\right) + 3$ then predicts the depth $y$ metres after $t$ hours, exactly the kind of periodic model that the amplitude-period-midline reading lets you build from a verbal description.

Try this

Q1. Convert $\dfrac{5\pi}{6}$ radians to degrees. [1 mark]

Cue. $\dfrac{5\pi}{6} \times \dfrac{180}{\pi} = 150^\circ$ .

Q2. State the period and amplitude of $y = 4\cos\left(\dfrac{1}{2}x\right)$ . [2 marks]

Cue. Amplitude $4$ ; period $\dfrac{2\pi}{1/2} = 4\pi$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20193 marksEvaluate

\sin\dfrac{2\pi}{3} + \cos\dfrac{5\pi}{6}

, giving your answer as an exact value. (Non-calculator.)

Show worked answer →

Place each angle. $\dfrac{2\pi}{3} = 120^\circ$ is in the second quadrant, where sine is positive; its related angle is $\dfrac{\pi}{3}$ , so $\sin\dfrac{2\pi}{3} = \sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$ (1 mark).

$\dfrac{5\pi}{6} = 150^\circ$ is in the second quadrant, where cosine is negative; its related angle is $\dfrac{\pi}{6}$ , so $\cos\dfrac{5\pi}{6} = -\cos\dfrac{\pi}{6} = -\dfrac{\sqrt{3}}{2}$ (1 mark).

Add: $\dfrac{\sqrt{3}}{2} + \left(-\dfrac{\sqrt{3}}{2}\right) = 0$ (1 mark). Markers reward both exact values with correct quadrant signs and the simplified sum.

SQA Higher 20213 marksPart of the graph of

y = a\sin(bx) + c

is shown to have a maximum value of

7

, a minimum value of

-1

, and to complete two full cycles between

x = 0

and

x = 2\pi

. Determine the values of

a

b

and

c

Show worked answer →

The amplitude is half the gap between maximum and minimum: $a = \dfrac{7 - (-1)}{2} = 4$ (1 mark).

The vertical shift is the midline: $c = \dfrac{7 + (-1)}{2} = 3$ (1 mark).

Two cycles in $2\pi$ means the period is $\dfrac{2\pi}{2} = \pi$ . Since period $= \dfrac{2\pi}{b}$ , we have $\pi = \dfrac{2\pi}{b}$ , so $b = 2$ (1 mark). Markers reward the amplitude from the range, the midline for $c$ , and $b$ from the period.

Related dot points

Sources & how we know this

SQA Higher Mathematics Course Specification — SQA (2018)