Mathematics • Year 9 • Unit 4 • Lesson 2
Trig Identities — Mixed Challenge
Pull together the Pythagorean identity, the ratio identity tan = sin/cos, complementary angles, and the exact values from Lesson 1. You'll spot a mistake in someone else's working and tackle an open-ended challenge that connects identities to geometry.
1. Mixed problems — choose the right identity
Each question uses a different combination of the identities from Lessons 1-2. Decide which identity applies before you start writing. Show your working. 3 marks each
1.1 If tan θ = 3/4 and θ is acute, find sin θ and cos θ exactly (no decimals).
1.2 Simplify sin² 35° + cos² 35° + sin² 55° + cos² 55° without using a calculator.
1.3 Show that sin 80° = cos 10° using complementary angles, and verify on a calculator to 4 dp.
1.4 If sin θ = 0.28 and θ is acute, find cos θ and tan θ to 3 dp using the identities (not by finding θ first).
1.5 Find all acute angles θ where tan θ = √3. Then state sin θ and cos θ exactly.
1.6 Two acute angles α and β are complementary (α + β = 90°). Show that tan α × tan β = 1.
2. Find the mistake
Another student has tried to find cos θ given that sin θ = 0.6 and θ is acute. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — find cos θ given sin θ = 0.6 (θ acute):
Line 1: sin²θ + cos²θ = 1
Line 2: 0.6² + cos²θ = 1
Line 3: 0.36 + cos²θ = 1
Line 4: cos²θ = 1 − 0.36 = 0.64
Line 5: cos θ = 0.64
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? cos²θ = 0.64 means cos θ multiplied by itself equals 0.64. To get cos θ alone, what operation do you need?3. Open-ended challenge — find your own identity
This question has more than one valid answer. 4 marks
3.1 Find three different pairs of acute angles (α, β) where α + β = 90° (complementary) AND both α and β have rational values of sin (i.e. fractions, not surds).
For each pair you find:
(i) State α and β (e.g. α ≈ 36.9°, β ≈ 53.1°).
(ii) Show the Pythagorean triple (a, b, c) that gives those exact rational sines.
(iii) Write sin α, cos α, sin β, cos β as fractions and verify sin α = cos β.
Bonus: Your three pairs must come from THREE DIFFERENT Pythagorean triples (e.g. 3-4-5 used once, not three times).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — tan θ = 3/4
Build the triangle: opp = 3, adj = 4. By Pythagoras hyp = √(9 + 16) = 5.
sin θ = opp/hyp = 3/5. cos θ = adj/hyp = 4/5.
The 3-4-5 triangle.
1.2 — sin² 35° + cos² 35° + sin² 55° + cos² 55°
Pair them up: (sin² 35° + cos² 35°) + (sin² 55° + cos² 55°) = 1 + 1 = 2.
The Pythagorean identity holds for every angle, so each bracket = 1.
1.3 — sin 80° = cos 10°
Complementary angles: sin θ = cos(90° − θ). So sin 80° = cos(90° − 80°) = cos 10°.
Calculator check: sin 80° ≈ 0.9848 and cos 10° ≈ 0.9848. ✓
1.4 — sin θ = 0.28
cos²θ = 1 − 0.28² = 1 − 0.0784 = 0.9216 → cos θ = √0.9216 ≈ 0.960.
tan θ = sin θ / cos θ = 0.28 / 0.960 ≈ 0.292.
1.5 — tan θ = √3
From Lesson 1 exact values, tan 60° = √3, so θ = 60°.
Therefore sin 60° = √3/2 and cos 60° = 1/2.
1.6 — Show tan α × tan β = 1 when α + β = 90°
Since α + β = 90°, β = 90° − α. By complementary angles: sin β = cos α and cos β = sin α.
tan α × tan β = (sin α / cos α) × (sin β / cos β) = (sin α / cos α) × (cos α / sin α) = 1. ✓
So tan and "co-tan" (tan of the complementary angle) are reciprocals.
2 — Find the mistake
(a) The mistake is on Line 5.
(b) From cos²θ = 0.64 you must take the SQUARE ROOT to find cos θ. The student forgot the square root and just dropped the "²". The correct value is √0.64 = 0.8, not 0.64.
(c) Corrected working:
sin²θ + cos²θ = 1
0.36 + cos²θ = 1
cos²θ = 0.64
cos θ = √0.64 = 0.8 (positive root, since θ is acute).
This is the 3-4-5 triangle: sin = 0.6, cos = 0.8, hyp = 1 (scaled).
3 — Open-ended challenge (sample solutions)
Many valid answers; here are three using THREE DIFFERENT Pythagorean triples:
Pair 1 — from 3-4-5: α ≈ 36.9° (sin⁻¹(3/5)), β ≈ 53.1° (sin⁻¹(4/5)). sin α = 3/5, cos α = 4/5; sin β = 4/5, cos β = 3/5. Check sin α = cos β: 3/5 = 3/5 ✓.
Pair 2 — from 5-12-13: α ≈ 22.6° (sin⁻¹(5/13)), β ≈ 67.4° (sin⁻¹(12/13)). sin α = 5/13, cos α = 12/13; sin β = 12/13, cos β = 5/13. Check sin α = cos β: 5/13 = 5/13 ✓.
Pair 3 — from 8-15-17: α ≈ 28.1° (sin⁻¹(8/17)), β ≈ 61.9° (sin⁻¹(15/17)). sin α = 8/17, cos α = 15/17; sin β = 15/17, cos β = 8/17. Check sin α = cos β: 8/17 = 8/17 ✓.
Other valid triples: 7-24-25, 9-40-41, 20-21-29, 28-45-53.
Marking: 1 mark per valid pair with full working (triangle, sin/cos fractions, complementary check); 1 bonus mark for using THREE DIFFERENT triples. Up to 4 marks in total.