Sets, Relations, Functions, Trigonometry & Complex Numbers — Practice

Question 1

In a survey of 100 people, 60 read newspaper A, 40 read newspaper B, and 20 read both. Find (a) those who read only A, (b) those who read only B, (c) those who read neither.

Accepted Answer

(a) Only A = n(A) − n(A∩B) = 60 − 20 = 40. (b) Only B = n(B) − n(A∩B) = 40 − 20 = 20. (c) n(A∪B) = n(A) + n(B) − n(A∩B) = 60 + 40 − 20 = 80. Neither = total − n(A∪B) = 100 − 80 = 20. Summary: 40 read only A, 20 read only B, 20 read both, 20 read neither. Check: 40 + 20 + 20 + 20 = 100 ✓

Question 2

Prove that: (sin 3A − sin A)/(cos A − cos 3A) = cot 2A.

Accepted Answer

Use sum-to-product formulas. Numerator: sin 3A − sin A = 2 cos[(3A+A)/2] · sin[(3A−A)/2] = 2 cos 2A · sin A. Denominator: cos A − cos 3A = −2 sin[(A+3A)/2] · sin[(A−3A)/2] = −2 sin 2A · sin(−A) = −2 sin 2A · (−sin A) = 2 sin 2A · sin A. Therefore: (sin 3A − sin A)/(cos A − cos 3A) = (2 cos 2A · sin A)/(2 sin 2A · sin A) = cos 2A/sin 2A = cot 2A. ∎

Question 3

If 1, ω, ω² are the cube roots of unity, prove that (1 − ω + ω²)(1 + ω − ω²) = 4.

Accepted Answer

We know: 1 + ω + ω² = 0, so ω + ω² = −1, and ω³ = 1. Compute (1 − ω + ω²): use ω² = −1 − ω (from 1 + ω + ω² = 0). So 1 − ω + ω² = 1 − ω + (−1 − ω) = −2ω. Compute (1 + ω − ω²): 1 + ω − ω² = 1 + ω − (−1−ω) = 1 + ω + 1 + ω = 2 + 2ω = 2(1 + ω) = 2(−ω²) = −2ω². Therefore: (1 − ω + ω²)(1 + ω − ω²) = (−2ω)(−2ω²) = 4ω³ = 4 × 1 = 4. ∎

Practice — Sets, Relations, Functions, Trigonometry & Complex Numbers

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