InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Prove that:(i) 2 sin 5π/12 sin π/12 = 1/2(ii) 2 cos 5π/12 cos π/12 = 1/2(iii) 2 sin 5π/12 cos π/12 = (√3 + 2)/2 |
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Answer» (i) 2 sin 5π/12 sin π/12 = 1/2 On using the formula, 2 sin A sin B = cos (A – B) – cos (A + B) 2 sin 5π/12 sin π/12 = cos (5π/12 – π/12) – cos (5π/12 + π/12) = cos (4π/12) – cos (6π/12) = cos (π/3) – cos (π/2) = cos (180°/3) – cos (180°/2) = cos 60° – cos 90° = 1/2 – 0 = 1/2 Thus proved. (ii) 2 cos 5π/12 cos π/12 = 1/2 On using the formula, 2 cos A cos B = cos (A + B) + cos (A – B) 2 cos 5π/12 cos π/12 = cos (5π/12 + π/12) + cos (5π/12 – π/12) = cos (6π/12) + cos (4π/12) = cos (π/2) + cos (π/3) = cos (180°/2) + cos (180°/3) = cos 90° + cos 60° = 0 + 1/2 = 1/2 Thus proved. (iii) 2 sin 5π/12 cos π/12 = (√3 + 2)/2 On using the formula, 2 sin A cos B = sin (A + B) + sin (A – B) 2 sin 5π/12 cos π/12 = sin (5π/12 + π/12) + sin (5π/12 – π/12) = sin (6π/12) + sin (4π/12) = sin (π/2) + sin (π/3) = sin (180°/2) + sin (180°/3) = sin 90° + sin 60° = 1 + √3 = (2 + √3)/2 = (√3 + 2)/2 Thus proved. |
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| 2. |
The value of sin 78° – sin 66°– sin 42°+ sin 6°is A. 1/2 B. -1/2 C. -1 D. None of these |
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Answer» Option : (B) sin 78° - sin 66° - sin 42° + sin 6° = (sin 78° - sin 42°) – (sin 66° - sin 6°) = 2cos 60° sin 18° - 2cos 36° sin 30° = 2 ⋅ 1/2 ⋅ sin 18° - 2 ⋅ 1/2 ⋅ cos 36° = sin 18° - cos 36° = \(\frac{\sqrt 5 -1}{4}\) - \(\frac{\sqrt 5 +1}{4}\) = - 1/2 (Ans) |
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| 3. |
If sin α + sin β = a and cos α – cos β = b, then tan(α - β)/2 = A. - (a/b)B. - (b/a)C. √(a2+b2)D. None of these |
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Answer» Option : (B) sin α + sin β = a ⇒ 2 sin\(\frac{\alpha+\beta}{2}\)cos\(\frac{\alpha-\beta}{2}\) = a …(I) cos α – cos β = b ⇒ - 2sin\(\frac{\alpha+\beta}{2}\)sin\(\frac{\alpha-\beta}{2}\) = b …(II) Dividing equation (II) by equation (I) – - tan\(\frac{\alpha-\beta}{2}\) = \(\frac{b}{a}\) Or, tan\(\frac{\alpha-\beta}{2}\) = \(-\frac{b}{a}\) |
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| 4. |
cos 40°+ cos 80°+ cos 160°+ cos 240 °= A. 0 B. 1 C. 1/2 D. -1/2 |
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Answer» Option : (D) cos 40° + cos 80° + cos 160° + cos 240° = (cos 40° + cos 80°) + (cos 160° + cos 240°) = 2cos \(\frac{80°+40°}{2}\) cos \(\frac{80°-40°}{2}\)+ 2cos \(\frac{240°+160°}{2}\) cos\(\frac{240°-160°}{2}\) = 2cos 60° cos 20° + 2cos 200° cos 40° = 2⋅ 1/2 ⋅cos 20° + 2cos (180 + 20)° cos 40° = cos 20° - 2cos 20° cos 40° = cos 20° - (cos (40° + 20°) + cos (40° - 20°)) = cos 20° - (cos 60° + cos 20°) = - cos 60° = - 1/2 (Ans) |
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| 5. |
Prove that :(i) sin 38° + sin 22° = sin 82°(ii) cos 100° + cos 20° = cos 40° |
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Answer» (i) sin 38° + sin 22° = sin 82° Let us consider the LHS sin 38° + sin 22° On using the formula, sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2 sin 38° + sin 22° = 2 sin (38° + 22°)/2 cos (38° – 22°)/2 = 2 sin 60°/2 cos 16°/2 = 2 sin 30° cos 8° = 2 × 1/2 × cos 8° = cos 8° = cos (90° – 82°) = sin 82° (since, {cos (90° – A) = sin A}) = RHS Thus proved. (ii) cos 100° + cos 20° = cos 40° Let us consider the LHS: cos 100° + cos 20° On using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2 cos 100° + cos 20° = 2 cos (100° + 20°)/2 cos (100° – 20°)/2 = 2 cos 120°/2 cos 80°/2 = 2 cos 60° cos 4° = 2 × 1/2 × cos 40° = cos 40° = RHS Thus proved. |
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| 6. |
The value of sin 50° – sin 70° + sin 10° is equal to A. 1 B. 0 C. 1/2 D. 2 |
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Answer» Option : (B) sin 50° - sin 70° + sin 10° = (sin 50° + sin 10°) – sin 70° = 2sin 30° cos 20° - sin (90°- 20°) = 2 ⋅ 1/2 ⋅ cos 20°- cos 20° = cos 20° - cos 20° = 0 |
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| 7. |
Prove that:(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A(iii) sin A + sin 2A + sin 4A + sin 5A = 4 cos A/2 cos 3A/2 sin 3A |
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Answer» (i) Given as cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A Let us consider the LHS cos 3A + cos 5A + cos 7A + cos 15A Therefore now, (cos 5A + cos 3A) + (cos 15A + cos 7A) On using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2 (cos 5A + cos 3A) + (cos 15A + cos 7A) = [2 cos (5A + 3A)/2 cos (5A - 3A)/2] + [2 cos (15A + 7A)/2 cos (15A - 7A)/2] = [2 cos 8A/2 cos 2A/2] + [2 cos 22A/2 cos 8A/2] = [2 cos 4A cos A] + [2 cos 11A cos 4A] = 2 cos 4A (cos 11A + cos A) Again on using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2 2 cos 4A (cos 11A + cos A) = 2 cos 4A [2 cos (11A + A)/2 cos (11A - A)/2] = 2 cos 4A [2 cos 12A/2 cos 10A/2] = 2 cos 4A [2 cos 6A cos 5A] = 4 cos 4A cos 5A cos 6A = RHS Thus proved. (ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A Let us consider the LHS cos A + cos 3A + cos 5A + cos 7A Therefore now, (cos 3A + cos A) + (cos 7A + cos 5A) On using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2 (cos 3A + cos A) + (cos 7A + cos 5A) = [2 cos (3A + A)/2 cos (3A - A)/2] + [2 cos (7A + 5A)/2 cos (7A - 5A)/2] = [2 cos 4A/2 cos 2A/2] + [2 cos 12A/2 cos 2A/2] = [2 cos 2A cos A] + [2 cos 6A cos A] = 2 cos A (cos 6A + cos 2A) Again on using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2 2 cos A (cos 6A + cos 2A) = 2 cos A [2 cos (6A + 2A)/2 cos (6A - 2A)/2] = 2 cos A [2 cos 8A/2 cos 4A/2] = 2 cos A [2 cos 4A cos 2A] = 4 cos A cos 2A cos 4A = RHS Thus proved. (iii) Given as sin A + sin 2A + sin 4A + sin 5A = 4 cos A/2 cos 3A/2 sin 3A Let us consider the LHS sin A + sin 2A + sin 4A + sin 5A Therefore now, (sin 2A + sin A) + (sin 5A + sin 4A) On using the formula, sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2 (sin 2A + sin A) + (sin 5A + sin 4A) = [2 sin (2A + A)/2 cos (2A - A)/2] + [2 sin (5A + 4A)/2 cos (5A - 4A)/2] = [2 sin 3A/2 cos A/2] + [2 sin 9A/2 cos A/2] = 2 cos A/2 (sin 9A/2 + sin 3A/2) Again on using the formula, sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2 2 cos A/2 (sin 9A/2 + sin 3A/2) = 2 cos A/2 [2 sin (9A/2 + 3A/2)/2 cos (9A/2 – 3A/2)/2] = 2 cos A/2 [2 sin ((9A + 3A)/2)/2 cos ((9A - 3A)/2)/2] = 2 cos A/2 [2 sin 12A/4 cos 6A/4] = 2 cos A/2 [2 sin 3A cos 3A/2] = 4 cos A/2 cos 3A/2 sin 3A = RHS Thus proved. |
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