Mathematics notes, chapter 6

Ratios and Graphs of Trigonometric Functions

Textbook Mathematics Form 5 (DLP), pages 172–195

What this chapter covers

Find the quadrant and corresponding reference angle of an angle between 0° and 360°, and relate sin, cos and tan of the angle to those of its reference angle (6.1)
Use the unit circle to define sin θ, cos θ and tan θ and state their signs in each quadrant (6.1)
Find the values of sine, cosine and tangent of angles in quadrants II, III and IV, with a calculator and without one for 30°, 45° and 60° (6.1)
Find the angle when the value of sine, cosine or tangent is given, and solve problems (6.1)
Describe the graphs of y = sin x, y = cos x and y = tan x for 0° ≤ x ≤ 360° (6.2)
Explain the effects of a, b and c on y = a sin bx + c, y = a cos bx + c and y = a tan bx + c, sketch the graphs and solve problems (6.2)

6.1Sine, cosine and tangent for 0° < θ < 360°

A unit circle has radius 1 unit and its centre at the origin O. A point P moves round it and θ is the angle from the positive x-axis to OP, measured anticlockwise. The x-axis and y-axis split the circle into quadrants I, II, III and IV.

A unit circle with the radius OP at angle θ above the x-axis; P has coordinates cos θ and sin θ — For a point P on the unit circle: cos θ is the x-coordinate and sin θ is the y-coordinate.

sin θ = y-coordinate of P
cos θ = x-coordinate of P
tan θ = yx = y-coordinate ÷ x-coordinate (undefined when x = 0, at 90° and 270°)

Quadrant	Angle θ	Reference angle α
I	0° < θ < 90°	α = θ
II	90° < θ < 180°	α = 180° − θ
III	180° < θ < 270°	α = θ − 180°
IV	270° < θ < 360°	α = 360° − θ

The corresponding reference angle α is the acute angle between OP and the x-axis. It is always less than 90°.

The four quadrants with the sign that is positive in each and the formula for the reference angle — Quadrant I: all positive. II: sin. III: tan. IV: cos. (textbook p. 176)

Remember the signs as A-S-T-C, going anticlockwise from quadrant I: All, Sin, Tan, Cos. Only the named ratio is positive in that quadrant, the other two are negative.

The value of a ratio is the value for the reference angle, with the sign of the quadrant.

Quadrant	sin θ	cos θ	tan θ
II	+ sin α	− cos α	− tan α
III	− sin α	− cos α	+ tan α
IV	− sin α	+ cos α	− tan α

Example 1 and 2 (textbook p. 176-177): quadrant, reference angle and relationship

138° is in quadrant II, α = 180° − 138° = 42°.
239° is in quadrant III, α = 239° − 180° = 59°.
312° is in quadrant IV, α = 360° − 312° = 48°.
So sin 167° = sin 13°, cos 258° = − cos 78° and tan 349° = − tan 11°.

Do not use θ − 180° in quadrant II or 180° − θ in quadrant III. Sketch the quadrant first, then choose the formula. For an angle above 360°, subtract 360° first (498° is the same direction as 138°).

Finding a value from the reference angle

Find the quadrant and the reference angle α.
Decide the sign from A-S-T-C.
Calculate the ratio of α with the calculator in Deg mode.

Example 4 (textbook p. 179): values with a calculator

sin 104° = + sin 76° = 0.9703
cos 208.2° = − cos 28.2° = −0.8813
tan 318°17′ = − tan 41°43′ = −0.8915

On the calculator, a minute sign is entered with the degrees key. Always check that the display shows Deg and not Rad.

Special angles without a calculator

Angle	sin	cos	tan
30°	12	√32	1√3
45°	1√2	1√2	1
60°	√32	12	√3

Example 5 (textbook p. 180): without a calculator

cos 150° = − cos 30° = −√32
tan 225° = + tan 45° = 1
sin 300° = − sin 60° = −√32

Quick check. Without a calculator, find sin 210° and cos 315°.

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210° is in quadrant III, α = 30°, sin is negative: sin 210° = −12.
315° is in quadrant IV, α = 45°, cos is positive: cos 315° = 1√2 = √22.

Finding the angle from a value

The sign tells you the two quadrants.
Get α from the inverse key, using the value without its sign.
Write both angles using the quadrant formulas.

Value	Quadrants	Angles
sin θ = 0.6157	I and II	38° or 180° − 38° = 142°
cos θ = −0.4226	II and III	180° − 65° = 115° or 180° + 65° = 245°
tan θ = −1.4826	II and IV	180° − 56° = 124° or 360° − 56° = 304°

Values from Example 6, textbook p. 181.

If you type sin⁻¹ (−0.5) you get a negative angle. Use the positive value to get α, then place α in the right quadrants.

Problems with sin, cos and tan

Draw the situation, use Pythagoras' theorem to complete the right-angled triangle, then give the ratio of the reference angle with the sign of the quadrant.

Example 7 (textbook p. 182): signs in a diagram

QRS is a straight line with QS = 8 cm and QR = RS. P is 3 cm from Q and PQ is perpendicular to QS.
QR = 8 ÷ 2 = 4 cm, so PR = √(3² + 4²) = 5 cm.
Angle x is in quadrant IV, so cos x = + cos ∠QPR = + 35.
Angle y is in quadrant II, so tan y = − tan ∠PRQ = − 34 (negative because y is in quadrant II).

Quick check. Given sin θ = 35 and 90° < θ < 180°, find cos θ and tan θ.

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The triangle is 3, 4, 5, so α has cos α = 45 and tan α = 34. In quadrant II, cos and tan are negative: cos θ = −45, tan θ = −34.

6.2Graphs of sine, cosine and tangent

The sine graph starting at 0 rising to 1 at 90 degrees and the cosine graph starting at 1, for 0 to 360 degrees — y = sin x and y = cos x for 0° ≤ x ≤ 360°. The shapes are the same wave, shifted by 90°.

The tangent graph for 0 to 360 degrees with broken vertical lines at 90 and 270 degrees — y = tan x. The dashed lines at 90° and 270° are values where tan x is undefined.

	y = sin x	y = cos x	y = tan x
Maximum value	1 at 90°	1 at 0° and 360°	none (∞)
Minimum value	−1 at 270°	−1 at 180°	none (−∞)
x-intercepts	0°, 180°, 360°	90°, 270°	0°, 180°, 360°
y-intercept	0	1	0
Period	360°	360°	180°

Characteristics from textbook p. 185-186 and the summary on p. 192.

A periodic function repeats its graph over a fixed interval, the period. The amplitude is the greatest distance from the centre line: 12 (maximum − minimum).

y = a sin bx + c (a > 0, b > 0)

Constant	What it does
a	Sets the amplitude = a. The maximum and minimum change. (Tangent has no amplitude; a changes how steep it is.)
b	Changes the period. For sine and cosine, period = 360°b. For tangent, period = 180°b. A larger b squeezes the graph.
c	Moves the graph up (c > 0) or down (c < 0) by \|c\| units. The centre line is y = c.

Four sine graphs: y = sin x, y = 2 sin x with a larger height, y = sin 2x with two cycles, and y = sin x + 1 shifted up — Changing a, b and c in turn (Example 9, textbook p. 189).

Example 10 (textbook p. 189): sketch y = 2 cos x + 1

a = 2, so amplitude is 2. b = 1, so the period is 360°. c = +1, so the centre line is y = 1.
Maximum = 1 + 2 = 3 at 0° and 360°. Minimum = 1 − 2 = −1 at 180°. y-intercept = 3.

Maximum and minimum values are c + a and c − a, not a and −a. The period of y = sin 2x is 180°, not 720°: divide, do not multiply.

To write a model y = a sin bx + c from a graph: a = 12(max − min), c = 12(max + min), then b = 360°period.

Example 11 (textbook p. 190-191): breathing cycle

Air flow velocity has amplitude 0.6 and repeats every 5 seconds, with no vertical shift. a = 0.6, b = 360 ÷ 5 = 72, c = 0, so y = 0.6 sin 72x. At x = 7: y = 0.6 sin 504° = 0.6 sin 144° = 0.35 litres per second.

Quick check. State the amplitude, period, maximum and minimum of y = 3 cos 2x − 1.

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Amplitude 3. Period = 360° ÷ 2 = 180°. Maximum = −1 + 3 = 2. Minimum = −1 − 3 = −4.