MathsISC Class 11 · Trigonometric Functions

Trigonometric Functions: Signs, Allied Angles, Periods, Graphs and Identities

Updated 17 October 2026 · Fact-checked

The trigonometric functions of any angle are defined on the unit circle: cosθ is the x-coordinate and sinθ the y-coordinate. Signs follow ASTC; allied angle formulae reduce any angle to an acute one; sin and cos have period 2π, tan period π; and the basic identities sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ are used to prove other identities.

Understand Trigonometric Ratios, Allied Angles, Graphs and Identities

Unit circle definition. For a point P(x, y) on the unit circle at angle θ from the positive x-axis, $cos θ = x$ and $sin θ = y$ , so $tan θ = y / x$ (x ≠ 0). The reciprocals are $sec θ = 1/ cos θ$ , $csc θ = 1/ sin θ$ and $cot θ = 1/ tan θ$ .

Signs (ASTC). All positive in the first quadrant, only sine (and cosec) in the second, only tangent (and cotangent) in the third, only cosine (and secant) in the fourth.

Standard values. For 0°, 30°, 45°, 60°, 90°: sin = 0, 1/2, 1/√2, √3/2, 1; cos = 1, √3/2, 1/√2, 1/2, 0; tan = 0, 1/√3, 1, √3, undefined.

Allied angles. $sin (- θ) = - sin θ$ , $cos (- θ) = cos θ$ ; $sin (π - θ) = sin θ$ , $cos (π - θ) = - cos θ$ ; $sin (π + θ) = - sin θ$ ; $sin (\frac{π}{2} - θ) = cos θ$ , $cos (\frac{π}{2} - θ) = sin θ$ . For angles like 570° or −330°, reduce by multiples of 360° first, then identify the quadrant and the sign.

Domain, range, period. sin and cos: domain ℝ, range [−1, 1], period 2π. tan: domain ℝ except odd multiples of π/2, range ℝ, period π. For $sin (k x)$ the period is 2π/|k|; $∣ sin x ∣$ and $sin^{2} x$ have period π. The graph of $y = a sin (k x)$ has amplitude |a|.

Identities. $sin^{2} θ + cos^{2} θ = 1$ ; $1 + tan^{2} θ = sec^{2} θ$ ; $1 + cot^{2} θ = csc^{2} θ$ . To prove an identity, start from the more complicated side, convert to sine and cosine, combine fractions and simplify to the other side.

Trigonometric Ratios, Allied Angles, Graphs and Identities: formulas to remember

Fundamental identities: $\sin^2θ + \cos^2θ = 1$ | $1 + \tan^2θ = \sec^2θ$ | $1 + \cot^2θ = \csc^2θ$; Pythagorean identities.
Allied angles: $\sin(\pi - θ) = \sin θ$ | $\cos(\pi - θ) = -\cos θ$ | $\sin(\frac{\pi}{2} - θ) = \cos θ$; Use ASTC for the sign.
Periods: $\sin x, \cos x$: $2\pi$ | $\tan x$: $\pi$ | $\sin kx$: $2\pi/|k|$; |sin x| has period π.
Standard values: $\sin 30° = \frac{1}{2}$ | $\cos 45° = \frac{1}{\sqrt{2}}$ | $\tan 60° = \sqrt{3}$; From the 30-60-90 and 45-45-90 triangles.

Trigonometric Ratios, Allied Angles, Graphs and Identities in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Height of a swing

Suppose a child on a swing has height above its lowest point given by $h = 0.5 (1 - cos θ)$ metres where θ is the angle of the rope with the vertical.

Find h when θ = 60° and when θ = 120°.

Try it first, then show the solutionSolution

θ = 60°: $cos 60° = 0.5$ , so $h = 0.5 (1 - 0.5) = 0.25$ m.
θ = 120°: $cos 120° = - cos 60° = - 0.5$ , so $h = 0.5 (1 + 0.5) = 0.75$ m.
The second quadrant gives a negative cosine, so the height rises.

Answer: 0.25 m and 0.75 m.

Memory hook: Use the allied angle to find cos 120°.

Tricks to remember Trigonometric Ratios, Allied Angles, Graphs and Identities

Short memory aids. Say each one aloud once, then use it on a question the same day.

ASTC
All, Sin, Tan, Cos: the positive functions in quadrants I to IV.
Use it when: Signs.
Even and odd
Cosine and secant are even; sine, tangent, cosecant, cotangent are odd.
Use it when: Negative angles.

Trigonometric Ratios, Allied Angles, Graphs and Identities: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Value of a sine · 1 mark · MCQBoard level

The value of sin 150° is:
(a) $\frac{1}{2}$
(b) $- \frac{1}{2}$
(c) $\frac{3}{2}$
(d) $- \frac{3}{2}$

Try it first, then show the solutionSolution

$sin 150° = sin (180° - 30°) = sin 30° = \frac{1}{2}$ .

Answer: (a) $\frac{1}{2}$

Example 2: Ratios from one ratio · 3 marks · Short answerBoard level

If $sin θ = \frac{3}{5}$ and θ lies in the second quadrant, find the other five trigonometric ratios.

Try it first, then show the solutionSolution

$cos^{2} θ = 1 - \frac{9}{25} = \frac{16}{25}$ , and in the second quadrant cos is negative: $cos θ = - \frac{4}{5}$ . (1 mark)
$tan θ = - \frac{3}{4}$ , $cot θ = - \frac{4}{3}$ . (1 mark)
$sec θ = - \frac{5}{4}$ , $csc θ = \frac{5}{3}$ . (1 mark)

Answer: cos −4/5, tan −3/4, cot −4/3, sec −5/4, cosec 5/3.

Example 3: Allied angles · 3 marks · Short answerHarder board level

Evaluate $cos 570° sin 510° + sin (- 330°) cos (- 390°)$ .

Try it first, then show the solutionSolution

$cos 570° = cos 210° = - \frac{3}{2}$ ; $sin 510° = sin 150° = \frac{1}{2}$ ; product $- \frac{3}{4}$ . (1 mark)
$sin (- 330°) = - sin 330° = \frac{1}{2}$ ; $cos (- 390°) = cos 30° = \frac{3}{2}$ ; product $\frac{3}{4}$ . (1 mark)
Sum $= 0$ . (1 mark)

Answer: 0

Example 4: Proving an identity · 4 marks · Short answerHarder board level

Prove that $sec^{2} θ + csc^{2} θ = sec^{2} θ csc^{2} θ$ .

Try it first, then show the solutionSolution

LHS $= \frac{1}{c o s ^{2} θ} + \frac{1}{s i n ^{2} θ}$ . (1 mark)
$= \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n ^{2} θ c o s ^{2} θ}$ . (1.5 marks)
$= \frac{1}{s i n ^{2} θ c o s ^{2} θ} = sec^{2} θ csc^{2} θ$ = RHS. (1.5 marks)

Answer: Proved.

Example 5: Finding sin θ from a condition · 5 marks · Long answerHOTS / competency

If $tan θ + sec θ = 2$ and θ is acute, find $sin θ$ , $cos θ$ and $tan θ$ . State the period of $y = 3 sin (2 x)$ and its amplitude.

Try it first, then show the solutionSolution

Using $sec^{2} θ - tan^{2} θ = 1$ , $(sec θ - tan θ) (sec θ + tan θ) = 1$ , so $sec θ - tan θ = \frac{1}{2}$ . (1.5 marks)
Adding: $2 sec θ = \frac{5}{2}$ , so $sec θ = \frac{5}{4}$ ; subtracting: $tan θ = \frac{3}{4}$ . (1.5 marks)
$cos θ = \frac{4}{5}$ and $sin θ = tan θ cos θ = \frac{3}{5}$ . (1 mark)
Period of $3 sin 2 x$ is $2 π /2 = π$ and amplitude 3. (1 mark)

Answer: sin 3/5, cos 4/5, tan 3/4; period π, amplitude 3.

Where students lose marks in Trigonometric Ratios, Allied Angles, Graphs and Identities

The slips examiners see most often, and what to write instead.

Mistake: Wrong sign in a quadrant
Taking cos 120° = +1/2.
Do this: Use ASTC: cos is negative in the second quadrant.
Usually costs: 1 to 2 marks.
Mistake: Dividing sin² by cos² by mistake
Cancelling terms in identities.
Do this: Combine fractions first, then simplify.
Usually costs: 1 to 2 marks.
Mistake: Wrong period of sin(kx)
Writing 2πk.
Do this: The period is 2π/|k|.
Usually costs: 1 mark.

Exam tips

Reduce large angles by subtracting multiples of 360°.
Write the quadrant and sign before evaluating.
In identities, state which side you start from.
Use the identities to find unknown ratios without calculators.

Practice questions on Trigonometric Ratios, Allied Angles, Graphs and Identities

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The period of $tan x$ is:
(a) $\frac{π}{2}$
(b) $π$
(c) $2 π$
(d) $4 π$
Hint
Repeats every π.
Answer
(b) $π$
Q2 · 2 marks · Short answerBoard level
Find the value of $tan 225°$ and of $sin (- 60°)$ .
Hint
Reduce to acute angles.
Answer
$tan 225° = tan 45° = 1$ ; $sin (- 60°) = - sin 60° = - \frac{3}{2}$ .
Q3 · 3 marks · Short answerHarder board level
If $cos θ = - \frac{5}{13}$ and θ lies in the third quadrant, find $sin θ$ and $tan θ$ .
Hint
Both sin and cos are negative.
Answer
$sin θ = - 1 - \frac{25}{169} = - \frac{12}{13}$ ; $tan θ = \frac{12}{5}$ .
Q4 · 3 marks · Short answerHarder board level
Prove that $\frac{1 - c o s θ}{s i n θ} = csc θ - cot θ$ .
Hint
Split the fraction.
Answer
LHS $= \frac{1}{s i n θ} - \frac{c o s θ}{s i n θ} = csc θ - cot θ$ .
Q5 · 3 marks · Short answerHarder board level
Find the domain and range of $y = 2 cos (3 x)$ and its period.
Hint
Scale the standard cosine.
Answer
Domain ℝ; range $[- 2, 2]$ ; period $2 π /3$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the tide height in a harbour (metres above mean level) is $h = 2 sin (\frac{π t}{6})$ with t in hours after midnight.
(i) Find the period. (1 mark)
(ii) Find the maximum height. (1 mark)
(iii) Find h at t = 1 and t = 9. (1 mark)
(iv) Find the first time after midnight when h = 1. (1 mark)
Hint
The period is 2π divided by the coefficient of t.
Answer
(i) $2 π / (π /6) = 12$ hours. (ii) 2 m. (iii) $h (1) = 2 sin 30° = 1$ m; $h (9) = 2 sin (270°) = - 2$ m. (iv) $sin (\frac{π t}{6}) = \frac{1}{2}$ , so $\frac{π t}{6} = \frac{π}{6}$ , giving $t = 1$ hour.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.sin²θ + cos²θ?Show answer
1.
2.Period of tan?Show answer
π.
3.Sign of tan in quadrant III?Show answer
Positive.
4.Range of cos?Show answer
[−1, 1].

Trigonometric Ratios, Allied Angles, Graphs and Identities: frequently asked questions

Why is tan 90° undefined?

tan θ = sin θ / cos θ and cos 90° = 0, so the quotient is not defined.

How do I find sin 570°?

Subtract 360° to get 210°; in the third quadrant sine is negative: -\sin 30° = -1/2.

What does amplitude mean?

The maximum displacement from the mean level: |a| for y = a \sin(kx).