Chapter 3 Trigonometry – II Ex 3.3 Solutions

Maharashtra State Board 11th Maths Solutions Chapter 3 Trigonometry – II Ex 3.3

ii. We know that, cos²θ =
Substituting θ = , we get

ii. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
Solution:
L.H.S. = (sin 3x + sin x) sin x + (cos 3x – cosx)cosx
= sin 3x sin x + sin2 x + cos 3x cos x – cos²x
= (cos 3x cos x + sin 3x sin x)
— (cos²x — sin²x)
= cos (3x – x) – cos 2x
= cos 2x – cos 2x
= 0
= R.H.S.

iii. (cos x + cos y )²+ (sin x + sin y)²= 4cos²
Solution:
L.H.S. = (cos x + cos y)²+ (sin x + sin y)²
= cos²x + cos²y + 2cos x.cos y + sin²x + sin²y + 2sin x.siny
= (cos²x + sin²x) + (cos²y + sin²y) + 2(cos x.cos y + si x.sin y)
= 1 + 1 +2cos(x – y)
= 2 + 2 cos (x – y)
= 2[1 + cos(x – y)]
= 2[2cos²[()] … [∵ 1 + cos θ = 2 cos²]
= 4 cos²()
= R.H.S.
[ Note: The question has been modified]

iv. (cos x – cos y)²+ (sin x – sin y)²= 4sin²
Solution:

L.H.S. = (cos x – cos y)²+ (sin x – sin y)²
= cos²x + cos²y + 2cos x.cos y + sin²x + sin²y + 2sin x.siny
= (cos²x + sin²x) + (cos²y + sin²y) – 2(cos x.cos y + sin x.sin y)
= 1 + 1 – 2cos(x – y)
= 2 – 2 cos (x – y)
= 2[1 – cos(x – y)]
= 2[2sin²[()] … [∵ 1 – cos θ = 2 sin²]
= 4 sin²()
= R.H.S.

v. tan x + cot x = 2 cosec 2x
Solution:
L.H.S. = tan x + cot x

vi. = 2 tan 2x
Solution:

vii. = 2 cos x
Solution:

= 2 cos x
= R.H.S.
[Note : The question has been modified.]

viii. 16 sin θ cos θ cos 2θ cos 4θ cos 8θ = sin 16θ
Solution:
L.H.S. = 16 sin θ cos θ cos 2θ cos 4θ cos 8θ
= 8(2sinθ cosθ) cos2θ cos 4θ cos 8θ
= 8sin 2θ cos 2θ cos 4θ cos 8θ
= 4(2sin 2θ cos 2θ) cos 4θ cos 8θ
= 4sin 4θ cos 4θ cos 8θ
= 2(2sin 4θ cos 4θ) cos 8θ
= 2sin 8θ cos 8θ
= sin 16θ
= R.H.S.

ix. = 2 cot 2x
Solution:

x. ={ ≤ft()-1}{ ≤ft(+1}
Solution:

xi.
Solution:

xii. = cot 2A
Solution:

xiii. cos 7° cos 14° cos 28° cos 56°
Solution:
L.H.S. = cos 7° cos 14° cos 28° cos 56°
= (2sin 7°cos 7°)cos 14°cos 28°cos 56°
= (sin 14° cos 14° cos 28° cos 56°)
…[∵ 2sinθ cosθ = sin 2θ]
= [{1}{2≤ft(2 7°)}latex] (2sin 14° cos 14°) cos 28° cos 56°
= (sin 28° cos 28° cos 56°)
= (2 sin 28° cos 28°) cos 56°
= (sin 56° cos 56°)
= (2 sin 56° cos 56°)
= (sin 112°)
=
=
= R.H.S.

xiv. = = sec²20°
Solution:

xv. = (2 cos x – 1)(2 cos 2x – 1)
Solution:

xvi. = cos²x + cos²(x + 120°) + cos²(x – 120°) =
Solution:
L.H.S = cos²x + cos²(x + 120°) + cos²(x – 120°) =

[cos 2x + cos(2x + 240°) + cos(2x 240°)]
= (cos 2x + cos 2x cos 240°— sin 2x sin 240° + cos 2x cos 240° + sin 2x sin 240°)
= (cos 2x + 2 cos 2x cos 240°)
= [cos 2x + 2 cos 2x cos( 180° + 60°)]
= [cos 2x + 2cos 2x(-cos 600)]
= [cos 2x —2 cos 2x()]
= ( cos 2x – cos 2x)
= (0)
= = R.H.S.

xvii. 2 cosec 2x + cosec x = sec cot
Solution:

xviii. 4 cos x cos ( + x) cos ( – x) = cos 3x

Solution:

= cos³x — 3cos x.sin²x
= cos³x — 3cos x (1— cos²x)
= cos³x — 3cos x + 3 cos³x
=4 cos³x — 3cos x
= cos 3x = R.H.S.
INote: The question has been modijìed.I

xix. sin x tan + 2cos x =
Solution:
L.H.S. = sin x tan (x/2)+ 2cos x
= + 2cos x
= + 2 cos x
= 2sin²x/2 + 2cosx
= 1 – cosx + 2cosx
= 1 + cos x
=2cos²x/2
= =R.H.S.