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You need to know the shapes of \(y = \sin x\), \(y = \cos x\) and \(y = \tan x\) in degrees, their symmetries and periods, and how transformations change them. Sketches, related angles and the maximum or minimum of a transformed graph all appear in the exam.
Key facts
sin and cos: period 360° = 2π, range \(-1 \le y \le 1\)
tan: period 180° = π, asymptotes at \(x = 90^\circ + 180^\circ n\)
\(\sin(180^\circ - x) = \sin x\)
\(\cos(-x) = \cos x\); \(\sin(-x) = -\sin x\)
\(\sin(90^\circ - x) = \cos x\)
\(y = a\sin(bx + c) + d\) (b > 0): max \(d + |a|\), min \(d - |a|\), period \(\dfrac{360^\circ}{b}\)
\(y = \sin x\): starts at (0, 0), has a maximum of 1 at 90° \(\left(\dfrac{\pi}{2}\right)\), is back to 0 at 180° (π) and has a minimum of −1 at 270° \(\left(\dfrac{3\pi}{2}\right)\). Period 360° (2π).
\(y = \cos x\): starts at (0, 1), is 0 at 90°, −1 at 180°, 0 at 270° and 1 at 360°. Period 360° (2π).
sin starts at 0, cos starts at 1; both repeat every 360° and stay between −1 and 1.
Both have range \(-1 \le y \le 1\). The cos graph is the sin graph translated 90° to the left.
\(y = \tan x\): passes through (0, 0) and (180°, 0) and has vertical asymptotes at x = 90°, 270°, … \(\left(x = \dfrac{\pi}{2} + n\pi\right)\). Period 180° (π). It takes every real value.
tan x repeats every 180° and has asymptotes where cos x = 0.
Signs: all three ratios are positive between 0° and 90°, only sin between 90° and 180°, only tan between 180° and 270°, and only cos between 270° and 360° (the CAST diagram).
CAST: which ratios are positive in each quadrant, counting anticlockwise from 0°.
Transformations
\(y = a\sin x\): vertical stretch with scale factor a. The amplitude is |a|.
\(y = \sin bx\): horizontal stretch with scale factor \(\dfrac{1}{b}\). The period becomes \(\dfrac{360^\circ}{b}\) (or \(\dfrac{2\pi}{b}\)); for tan it becomes \(\dfrac{180^\circ}{b}\)
\(y = \sin(x + c)\): translation c units to the left. \(y = \sin x + d\): translation d units up.
To find the key points of \(y = \sin(bx + c)\), set \(bx + c\) equal to 0°, 90°, 180°, … and solve for x.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
Sketch a transformed graph and give where it meets the axes
3 marksGrade C
Start from the basic graph and apply the transformation (+c inside the bracket moves the graph left).
Find the y-intercept by putting x = 0.
Find the x-intercepts by setting the bracket equal to 90°, 270°, … and solving.
Only draw, and only list points, inside the given interval.
Example. Sketch the graph of \(y = \cos(x + 30^\circ)\) for \(0 \le x \le 360^\circ\). Give the coordinates of the points where the graph meets the coordinate axes.
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A cosine curve translated 30° to the left: it starts at \(\left(0, \dfrac{\sqrt{3}}{2}\right)\), has a minimum at (150°, −1) and a maximum at (330°, 1)
B1
y-intercept: \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\), so \(\left(0, \dfrac{\sqrt{3}}{2}\right)\)
B1
x-intercepts: \(x + 30^\circ = 90^\circ\) or \(270^\circ\), so (60°, 0) and (240°, 0)
B1
Use symmetry to write related values
3 marksGrade C
Sketch the graph or use the CAST diagram to see whether the new angle gives the same sign or the opposite sign.
Related angles give the same size of ratio.
Decide the sign from the quadrant.
Example. Given that \(\cos x = p\), where \(0^\circ \lt x \lt 90^\circ\), write each of the following in terms of p. (a) \(\cos(180^\circ - x)\) (b) \(\sin(90^\circ + x)\) (c) \(\cos(x + 180^\circ)\)
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(a) \(-p\)
B1
(b) \(\sin(90^\circ + x) = \cos x = p\)
B1
(c) \(-p\)
B1
Find constants from the features of a graph
3 marksGrade A
Maximum = C + A and minimum = C − A: solve these simultaneously.
Use where the maximum (or minimum) occurs to find the horizontal translation.
Check that your values give the stated point.
Example. The curve \(y = A\cos(x - B) + C\), where A > 0 and \(0^\circ \lt B \lt 180^\circ\), has a maximum point at (40°, 7) and a minimum value of −1. Find the values of A, B and C.
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\(C + A = 7\) and \(C - A = -1\)
M1
A = 4, C = 3
A1
cos has its maximum when \(x - B = 0\), so B = 40°
B1
Don’t lose marks
Translating the wrong way: \(y = \sin(x + 30^\circ)\) moves left, not right.
Thinking \(y = \sin 2x\) has period 720°; it is 180°, because the graph is squashed.
Giving tan a period of 360°, or drawing its asymptotes in the wrong places.
Drawing curves with pointed peaks, or not reaching 1 and −1.
More tips
Memory tricks
The sin and cos graphs have a key point every quarter period: mark these on the x-axis first.
Inside the bracket, '+' moves the graph left and '−' moves it right.
Use your calculator's table function to check points on a sketch.
Symmetry: the sin graph is symmetrical about x = 90°, and the cos graph about x = 0° and x = 180°.
Exam technique
Label every intercept, turning point and asymptote with coordinates or equations.
Stick to the interval given in the question.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Sketch the sin, cos and tan graphsMark the key points: where they cross the axes, the maximum and minimum values, and the asymptotes of tan.
Grade D
State the period of each graphsin and cos repeat every 360° (2π); tan repeats every 180° (π).
Grade C
Use symmetry to find related anglese.g. \(\sin(180^\circ - x) = \sin x\), \(\cos(-x) = \cos x\) and \(\tan(x + 180^\circ) = \tan x\)
Grade C
Sketch translated and stretched trig graphse.g. \(y = \cos(x + 30^\circ)\) is \(y = \cos x\) translated 30° to the left.
Grade B
Find maximum, minimum and period after transformationsFor \(y = a\sin(bx + c) + d\) with b > 0: maximum \(d + |a|\), minimum \(d - |a|\), period \(\dfrac{360^\circ}{b}\)
Grade A
Find constants from a graph's featuresUse the maximum, the minimum and the position of a turning point to find the unknown constants.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
Given that \(\tan 25^\circ = k\), write down, in terms of \(k\), the value of (i) \(\tan 205^\circ\) (ii) \(\tan 565^\circ\) (iii) \(\tan 385^\circ\)[3]
Show the answer and mark scheme
Answer: (i) \(k\), (ii) \(k\), (iii) \(k\)
B1 for (i) \(\tan 205^\circ = k\)
B1 for (ii) \(\tan 565^\circ = k\)
B1 for (iii) \(\tan 385^\circ = k\)
Worked solution: Use the symmetries of the graphs, with \(\theta = 25^\circ\): (i) \(\tan(180^\circ + \theta) = \tan\theta\) (the period is \(180^\circ\)), so \(\tan 205^\circ = k\). (ii) \(\tan(540^\circ + \theta) = \tan\theta\) (the period is \(180^\circ\)), so \(\tan 565^\circ = k\). (iii) \(\tan(360^\circ + \theta) = \tan\theta\) (the period is \(180^\circ\)), so \(\tan 385^\circ = k\).
Question 2Medium7 marks
The diagram shows the curve with equation \(y = a\cos(x - \alpha) + b\), \(0 \le x \le 360^\circ\), where \(a\), \(b\) and \(\alpha\) are constants, \(a \gt 0\) and \(0 \lt \alpha \lt 90^\circ\). The curve has a maximum point at \((50^\circ, 9)\) and a minimum point at \((230^\circ, -3)\).
[object Object]
(a) Find the values of \(a\), \(b\) and \(\alpha\).[3]
(b) Find the \(y\)-coordinate of the point where the curve meets the \(y\)-axis, giving your answer to 3 significant figures.[1]
(c) Find the values of \(x\), for \(0 \le x \le 360^\circ\), for which \(y = 0\).[3]
The function \(\mathrm{f}\) is defined by \[\mathrm{f}(x) = 2\sin(x + 45^\circ) + 2, \quad 0 \le x \le 450^\circ.\]
(a) Sketch the graph of \(y = \mathrm{f}(x)\) for \(0 \le x \le 450^\circ\), showing the coordinates of any turning points and of the end points of the graph.[3]
(b) Find the set of values of \(k\) for which the equation \(\mathrm{f}(x) = k\) has exactly two solutions in the interval \(0 \le x \le 450^\circ\).[3]
Show the answer and mark scheme
(a)Answer: A sine-shaped curve with a maximum at \((45^\circ, 4)\) and a maximum at \((405^\circ, 4)\) and a minimum at \((225^\circ, 0)\); end points \((0, 2 + \sqrt{2})\) and \((450^\circ, 2 + \sqrt{2})\).
B1 for a sine-shaped curve over the correct interval
B1 for the turning points correct: a maximum at \((45^\circ, 4)\), a maximum at \((405^\circ, 4)\), a minimum at \((225^\circ, 0)\)
B1 for the end points \((0, 2 + \sqrt{2})\) and \((450^\circ, 2 + \sqrt{2})\)
Worked solution: The graph is \(y = \sin x\) translated 45° to the left, stretched vertically by scale factor 2 and translated 2 units up. Over the interval it has a maximum at \((45^\circ, 4)\) and a maximum at \((405^\circ, 4)\) and a minimum at \((225^\circ, 0)\); it starts at \((0, 2 + \sqrt{2})\) and ends at \((450^\circ, 2 + \sqrt{2})\).
(b)Answer: \(0 \lt k \lt 2 + \sqrt{2} \text{ or } k = 4\)
M1 for using the critical values \(0, \ 2 + \sqrt{2}, \ 4\) (turning values and end values)
A1 for one correct piece of the set
A1 for \(0 \lt k \lt 2 + \sqrt{2} \text{ or } k = 4\) with correct strict and non-strict inequalities
Worked solution: Counting the intersections of horizontal lines \(y = k\) with the graph: \(k = 0\): 1; \(0 \lt k \lt 2 + \sqrt{2}\): 2; \(k = 2 + \sqrt{2}\): 4; \(2 + \sqrt{2} \lt k \lt 4\): 4; \(k = 4\): 2. So exactly two solutions when \(0 \lt k \lt 2 + \sqrt{2} \text{ or } k = 4\).