Mathematics • Year 9 • Unit 4 • Lesson 2

Trigonometric Ratios

Build fluency with the Pythagorean identity sin²θ + cos²θ = 1, the ratio identity tan θ = sin θ / cos θ, and complementary angles (sin θ = cos(90° − θ)). One step at a time, from a fully worked example through guided practice to independent problems.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every line. Each step has a short reason on the right so you can see why, not just what.

Problem. If sin θ = 3/5 and θ is acute, find cos θ and tan θ without finding θ itself.

Step 1 — Recall the Pythagorean identity.

sin²θ + cos²θ = 1 (true for every angle θ).

Reason: this is the key identity from card 1. It links sin and cos directly.

Step 2 — Substitute sin θ = 3/5.

(3/5)² + cos²θ = 1
9/25 + cos²θ = 1

Reason: sin²θ means (sin θ)², so (3/5)² = 9/25.

Step 3 — Solve for cos²θ.

cos²θ = 1 − 9/25 = 25/25 − 9/25 = 16/25

Reason: subtract 9/25 from both sides. Use a common denominator.

Step 4 — Take the positive square root (θ is acute).

cos θ = √(16/25) = 4/5

Reason: for an acute angle, cos θ is positive, so we take the positive root.

Step 5 — Find tan θ using the ratio identity.

tan θ = sin θ / cos θ = (3/5) / (4/5) = 3/5 × 5/4 = 3/4

Reason: tan = sin/cos, then dividing fractions = multiply by the reciprocal.

Answer: cos θ = 4/5, tan θ = 3/4. (This is the 3-4-5 right-angled triangle.)

Stuck? Revisit lesson § "The Pythagorean Identity" — it's the same idea as Pythagoras' theorem applied to a triangle with hypotenuse 1.

2. We do — fill in the missing steps

Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks

Problem. If cos θ = 8/17 and θ is acute, find sin θ and tan θ.

Step 1 — Use the identity: sin²θ + cos²θ = ____.

Step 2 — Substitute cos θ = 8/17:

sin²θ + (____ / ____)² = 1

Step 3 — Compute (8/17)²:

(8/17)² = ____ / ____

Step 4 — Solve for sin²θ, then sin θ:

sin²θ = 1 − ____ /289 = ____ /289 → sin θ = ____ / ____

Step 5 — Find tan θ:

tan θ = sin θ / cos θ = (____ /17) / (8/17) = ____ / ____

Stuck? Revisit lesson § "Worked Example" step 1, which solves a very similar problem with cos θ = 12/13.

3. You do — independent practice

Show your working in the space under each problem. The first four are foundation (single identity). The middle two are standard (combine two ideas). The last two are extension (proof / reasoning).

Foundation — single identity

3.1 Evaluate sin² 60° + cos² 60° using the exact values from Lesson 1. (Answer should be 1.) 1 mark

3.2 Complete: sin 25° = cos ____° (using complementary angles). 1 mark

3.3 If sin θ = 0.6 and θ is acute, find cos θ (use sin²θ + cos²θ = 1). 1 mark

3.4 If sin θ = 0.7 and cos θ = 0.7141, find tan θ using tan θ = sin θ / cos θ. Round to 2 dp. 1 mark

Standard — combine two ideas

3.5 If sin θ = 5/13 and θ is acute, find cos θ AND tan θ (no calculator). 2 marks

3.6 Show, without a calculator, that cos 30° = sin 60°. (Use complementary angles and the exact values from Lesson 1.) 2 marks

Extension — push your thinking

3.7 Prove that tan θ = sin θ / cos θ by writing sin and cos as ratios of sides in a right-angled triangle. 2 marks

3.8 Find all acute angles θ where sin θ = cos θ. Justify your answer. 2 marks

Stuck on 3.8? sin θ = cos θ means tan θ = 1 (divide both sides by cos θ). What acute angle has tangent equal to 1?

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (cos θ = 8/17)

Step 1: sin²θ + cos²θ = 1.
Step 2: sin²θ + (8/17)² = 1.
Step 3: (8/17)² = 64/289.
Step 4: sin²θ = 1 − 64/289 = 225/289 → sin θ = 15/17 (since √225 = 15, √289 = 17).
Step 5: tan θ = (15/17) / (8/17) = 15/8. (This is the 8-15-17 right-angled triangle.)

3.1 — sin² 60° + cos² 60°

sin 60° = √3/2, cos 60° = 1/2. So sin² 60° + cos² 60° = 3/4 + 1/4 = 4/4 = 1. ✓

3.2 — sin 25° = cos ?

sin 25° = cos(90° − 25°) = cos 65°.

3.3 — sin θ = 0.6, find cos θ

cos²θ = 1 − 0.6² = 1 − 0.36 = 0.64 → cos θ = √0.64 = 0.8.

3.4 — tan θ when sin θ = 0.7, cos θ = 0.7141

tan θ = 0.7 / 0.7141 ≈ 0.98.

3.5 — sin θ = 5/13

cos²θ = 1 − (5/13)² = 1 − 25/169 = 144/169 → cos θ = 12/13.
tan θ = sin θ / cos θ = (5/13) / (12/13) = 5/12.
This is the 5-12-13 Pythagorean triple.

3.6 — Show cos 30° = sin 60°

From Lesson 1 exact values: cos 30° = √3/2 and sin 60° = √3/2. They are equal.
Alternatively, by complementary angles: cos 30° = sin(90° − 30°) = sin 60°. ✓

3.7 — Prove tan θ = sin θ / cos θ

In a right-angled triangle with angle θ: sin θ = opp/hyp and cos θ = adj/hyp.
sin θ / cos θ = (opp/hyp) ÷ (adj/hyp) = (opp/hyp) × (hyp/adj) = opp/adj = tan θ. ✓
The hyp cancels — exactly the proof in card 2 of the lesson.

3.8 — sin θ = cos θ for acute θ

Divide both sides by cos θ: sin θ / cos θ = 1, so tan θ = 1. The only acute angle with tan = 1 is θ = 45°.
Check: sin 45° = cos 45° = √2/2 ≈ 0.707. ✓