Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.4
ii.
[Note: Answer given in the textbook is sin + sin However, as per our calculation it is sin + sin
iii. 2cos 4θ cos 2θ = cos(4θ + 2θ)+cos (4θ – 2θ)
= cos 6θ + cos 2θ
iv. 2cos 35° cos75°
= cos(35° + 75°) + cos (35° – 75°)
= cos 110° + cos (-40)°
= cos 110° + cos 40° … [∵ cos(-θ) = cos θ]
ii. sin 6x + sin 4x – sin 2x = 4 cos x sin 2x cos 3x
Solution:
L.H.S. = sin 6x + sin 4x — sin 2x
= 2sin cos – 2 sin x cos x
= 2 sin 5x cos x — 2 sin x cos x
= 2 cos x (sin 5x — sin x)
= 2 cos
= 2 cos x (2 cos 3x sin 2x)
= 4 cos x sin 2x cos 3x
= R.H.S.
[Note: The question has been modified.]
iii. = cot 2x
Solution:
iv. sin 18° cos 39° + sin 6° cos 15° = sin 24° cos 33°
Solution:
L.H.S. = sin 18°.cos 39° + sin 6°.cos 15°
= (2 cos 39°sin 18° + 2.cos 15°.sin 6°)
= [sin(39° + 18°) — sin(39° — 18°) + sin (15° + 6°) — sin (15° — 6°)]
= (sin57° – sin21° + sin 21°- sin9°)
= (sin57° – sin9°)
= x 2. cos
= cos 33°.sin 24°
= sin 24°. cos 33°
= R.H.S.
v. cos 20° cos 40° cos 60°cos 80° = 1/16
Solution:
L.H.S. = cos 20°.cos 40°.cos 60°.cos 80°
= cos 20°.cos 40°..cos 80°
= (2 cos 40°.cos 20°).cos 80°
= [cos(40° + 20°) + cos(40°- 20°)].cos80°
= (cos 60° + cos 20°) cos 80°
=cos 60°. cos 80° + cos 20°. cos 80°
=
= cos 80° + [cos (80° + 20°) + cos (80° — 20°)]
= cos 80° + (cos 100° + cos 60°)
= cos 80° + cos 100° + cos 60°
= cos 80° = cos (180° – 80°) +
= cos 80° – cos 80° + … [∵ cos (180 – θ) = – cos θ]
= = R.H.S
vi. sin 20° sin 40° sin 60° sin 80° = 3/16
Solution:
