These eight questions use new values and get harder, from a simple identity to equations and graph features. Write your own working first, then open the answer.
They follow the lessons in trigonometric functions. Give angles in degrees to one decimal place unless told otherwise.
Questions
Question 1. Simplify (1 − sin²A) ÷ (1 − cos²A).
Answer
Use sin²A + cos²A = 1 on both parts: 1 − sin²A = cos²A and 1 − cos²A = sin²A.
The fraction is cos²A ÷ sin²A = cot²A.
Question 2. Show that tan θ + cot θ = 2 cosec 2θ.
Answer
tan θ + cot θ = sin θ ÷ cos θ + cos θ ÷ sin θ = (sin²θ + cos²θ) ÷ (sin θ cos θ) = 1 ÷ (sin θ cos θ).
Since sin 2θ = 2 sin θ cos θ, we have sin θ cos θ = ½ sin 2θ. So the expression is 1 ÷ (½ sin 2θ) = 2 ÷ sin 2θ = 2 cosec 2θ.
Question 3. Solve 3 sin x = 2 for 0° ≤ x ≤ 360°.
Answer
sin x = 2/3, so the reference angle is 41.8°. Sine is positive in quadrants 1 and 2.
So x = 41.8° or x = 180° − 41.8° = 138.2°.
Question 4. Solve 2cos²x + cos x − 1 = 0 for 0° ≤ x ≤ 360°.
Answer
Factorise: (2cos x − 1)(cos x + 1) = 0, so cos x = ½ or cos x = −1.
Cos x = ½ gives x = 60° or 300°. Cos x = −1 gives x = 180°. The solutions are 60°, 180°, 300°.
Question 5. Solve tan 2x = 1 for 0° ≤ x ≤ 180°.
Answer
Stretch the interval: 0° ≤ 2x ≤ 360°. Tangent is positive in quadrants 1 and 3, so 2x = 45° or 225°.
Dividing by 2 gives x = 22.5° or 112.5°.
Question 6. Find the exact value of cos 15°.
Answer
Write 15° = 45° − 30°. Then cos 15° = cos 45° cos 30° + sin 45° sin 30° = (√2 ÷ 2)(√3 ÷ 2) + (√2 ÷ 2)(1 ÷ 2).
This gives (√6 + √2) ÷ 4, which is about 0.966.
Question 7. Given sin A = 8/17 with A acute, find tan 2A.
Answer
cos A = 15/17, so tan A = 8/15.
tan 2A = 2 tan A ÷ (1 − tan²A) = (16/15) ÷ (1 − 64/225) = (16/15) ÷ (161/225) = (16/15)(225/161) = 240/161.
Check: sin 2A = 240/289 and cos 2A = 161/289, so the ratio is also 240/161.
Question 8. For y = −2 sin x + 1 with 0° ≤ x ≤ 360°, state the maximum and minimum values, where each occurs, and the y-intercept.
Answer
The amplitude is 2 and the middle line is y = 1, so the maximum is 3 and the minimum is −1. The negative sign flips the sine curve.
The minimum occurs at x = 90° (where sin x = 1) and the maximum at x = 270° (where sin x = −1). When x = 0, y = 1, so the y-intercept is 1.
If you got these wrong
- Simplifying slips in Questions 1 and 2: revisit using identities to simplify expressions.
- Missing solutions in Questions 3 to 5: read solving equations over a stated interval.
- Formula slips in Questions 6 and 7: read addition and double-angle formulas.
- Graph slips in Question 8: read sketching transformed trigonometric graphs.
Log each slip in the mistake log, and try a timed practice session when you are ready. If you want a teacher to go through your working, see online one-to-one Additional Mathematics tuition.