Question

What is the value of cos[(180° – θ)/2].cos[(180° – 9θ)/2] + sin[(180° – 3θ)/2].sin[(180° – 13θ)/2]?

• A. sin2θ.sin4θ
• B. cos2θ.cos6θ
• C. sin2θ.sin6θ
• D. cos2θ.cos4θ

Answer 1

Answer:

B. cos2θ.cos6θ

Solution: Formula used: cos(-θ) = cos θ 2 sin A.sin B = cos(A – B) – cos(A + B) 2 cos A.cos B = cos(A + B) + cos(A – B) cos C + cos D = 2 cos(C + D)/2} . cos {(C - D)/2} Calculation: cos[(180° – θ)/2].cos[(180° – 9θ)/2] + sin[(180° – 3θ)/2].sin[(180° – 13θ)/2] ⇒ cos(90 - θ/2) cos(90 - 9θ/2) + sin(90 - 3θ/2) sin(90 - 13θ/2) ⇒ sin θ/2.sin 9θ/2 + cos 3θ/2.cos 13θ/2 ⇒ 1/2 [cos(θ/2 - 9θ/2) - cos(θ/2 + 9θ/2)] + 1/2 [cos(3θ/2 + 13θ/2) + cos(3θ/2 - 13θ/2)] ⇒ 1/2 [cos (-8θ/2) - cos(10θ/2)] + 1/2 [cos(16θ/2) + cos(-10θ/2)] ⇒ 1/2 [cos (-4θ) - cos(5θ)] + 1/2 [cos(8θ) + cos(-5θ)] ⇒ 1/2 cos 4θ - 1/2 cos 5θ + 1/2 cos 8θ + 1/2 cos 5θ ⇒ 1/2 cos 4θ + 1/2 cos 8θ ⇒ 1/2 [cos 4θ + cos 8θ] ⇒ 1/2 × 2 [cos{(4θ + 8θ)/2}. cos{(4θ - 8θ)/2} ⇒ [cos (12θ/2) . cos (-4θ/2)] ⇒ cos 6θ . cos 2θ ∴ The correct answer is option (B).