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(i) 2 sin3θ cosθ
= sin(3θ + θ) + sin(3θ - θ) [∵ 2sinA cosB = sin(A + B) + sin(A - B)]
= sin4θ + sin2θ
(ii) 2 cos3θ sin2θ
∵ 2 cosA sinB = sin(A + B) - sin(A - B)
⟹ 2 cos3θ sin2θ = sin(3θ + 2θ) - sin(3θ - 2θ)
= sin5θ - sinθ
(iii) 2 sin4θ sin3θ
∵ 2 sin4θ sin3θ = cos(4θ - 3θ) - cos(4θ + 3θ)
= cosθ - cos7θ
(iv) 2 cos7θ cos3θ
∵ 2 cosA cosB = cos(A + B) + cos(A - B)
⟹ 2 cos7θ cos3θ = cos(7θ + 3θ) + cos(7θ - 3θ)
= cos10θ + cos4θ
Now,
LHS = sin 25° cos 115°
We know that
2 sin A cos B = sin (A + B) + sin (A - B)
Also,
cos θ = sin(90° - θ)
cos 50° = sin(90°- 50°) = sin 40°
=1/2 [sin 40°- 1]
We have,
= -cos θ + [cos (θ + 2θ) + cos(2θ - θ)]
= -cos θ + cos 3θ + cos θ
= cos 3θ
= RHS
∴ LHS = RHS Hence Proved.
cos 10° cos 30° cos 50° cos 70° = 3/16
LHS = cos10° cos 30° cos 50° cos70°
= cos 30° cos 10° cos 50°cos 70°
⟹ 2cosA cos B = cos(A + B)+ cos(A - B) --- (i)
cos 40°cos 80°cos 160° = -1/8
LHS = cos 40°cos 80°cos 160°
= cos 80°cos 40°cos 160°
Multiplying and dividing by 2
=1/2 (cos 80°×(2cos 40°cos 160°))
2cosAcosB = cos(A + B) + cos(A - B)
=1/2 (cos 80°(cos(40°+160°)+cos(40° - 160°)))
=1/2 (cos 80°(cos(40° + 160°)+ cos(-120)))
=1/2 cos 80°(cos(180° + 20°)+ cos(180° - 60°))
=1/2 cos 80°(cos 20°+cos 60°)
=1/2 cos 80°cos 20°+1/2 cos 80°+cos 60°
= – 1/2 (2cos 80°cos 20°)+1/2 cos 80°cos 60°
= – 1/4[2 cos 80°cos 20° + cos 80°]
= – 1/4[cos(80° + 20°) + cos(80° - 20°)+ cos 80°]
= – 1/4[cos100° + cos 60° + cos 80°]
= – 1/4[cos(180° - 80°) + cos 60° + cos 80°]
= – 1/4[-cos 80° + cos 60° + cos 80°]
= – 1/4cos 60°
= – 1/4 × 1/2
= – 1/8 RHS
sin 20°sin 40°sin 80°
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Chapter 8: Transformation Formulae –...