NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions Ex 3.3
Page No 73:
Question 1:
Answer:
L.H.S. = 
Question 2:
Prove that 
Answer:
L.H.S. = 
Question 3:
Prove that 
Answer:
L.H.S. =
Question 4:
Prove that 
Answer:
L.H.S =
Question 5:
Find the value of:
(i) sin 75°
(ii) tan 15°
Answer:
(i) sin 75° = sin (45° + 30°)
= sin 45° cos 30° + cos 45° sin 30°
[sin (x + y) = sin x cos y + cos x sin y]
(ii) tan 15° = tan (45° – 30°)
Question 6:
Prove that:
Answer:
Question 7:
Prove that: 
Answer:
It is known that 
∴L.H.S. =
Question 8:
Prove that
Answer:
Question 9:
Answer:
L.H.S. = 
Question 10:
Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Answer:
L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x
Question 11:
Prove that
Answer:
It is known that
.
.
∴L.H.S. = 
Question 12:
Prove that sin2 6x – sin2 4x = sin 2x sin 10x
Answer:
It is known that 
∴L.H.S. = sin26x – sin24x
= (sin 6x + sin 4x) (sin 6x – sin 4x)
= (2 sin 5x cos x) (2 cos 5x sin x)
= (2 sin 5x cos 5x) (2 sin x cos x)
= sin 10x sin 2x
= R.H.S.
Question 13:
Prove that cos2 2x – cos2 6x = sin 4x sin 8x
Answer:
It is known that 
∴L.H.S. = cos2 2x – cos2 6x
= (cos 2x + cos 6x) (cos 2x – 6x)
= [2 cos 4x cos 2x] [–2 sin 4x (–sin 2x)]
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= R.H.S.
Question 14:
Prove that sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x
Answer:
L.H.S. = sin 2x + 2 sin 4x + sin 6x
= [sin 2x + sin 6x] + 2 sin 4x
= 2 sin 4x cos (– 2x) + 2 sin 4x
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)
= 2 sin 4x (2 cos2 x – 1 + 1)
= 2 sin 4x (2 cos2 x)
= 4cos2 x sin 4x
= R.H.S.
Question 15:
Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
Answer:
L.H.S = cot 4x (sin 5x + sin 3x)
= 2 cos 4x cos x
R.H.S. = cot x (sin 5x – sin 3x)
= 2 cos 4x. cos x
L.H.S. = R.H.S.
Question 16:
Prove that 
Answer:
It is known that
∴L.H.S =
Question 17:
Prove that 
Answer:
It is known that
∴L.H.S. =
Question 18:
Prove that 
Answer:
It is known that
∴L.H.S. =
Question 19:
Prove that 
Answer:
It is known that
∴L.H.S. =
Question 20:
Prove that 
Answer:
It is known that
∴L.H.S. = 
Question 21:
Prove that 
Answer:
L.H.S. =
Page No 74:
Question 22:
Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
Answer:
L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x
= cot x cot 2x – cot 3x (cot 2x + cot x)
= cot x cot 2x – cot (2x + x) (cot 2x + cot x)
= cot x cot 2x – (cot 2x cot x – 1)
= 1 = R.H.S.
Question 23:
Prove that 
Answer:
It is known that
.
.
∴L.H.S. = tan 4x = tan 2(2x)
Question 24:
Prove that cos 4x = 1 – 8sin2 x cos2 x
Answer:
L.H.S. = cos 4x
= cos 2(2x)
= 1 – 2 sin2 2x [cos 2A = 1 – 2 sin2 A]
= 1 – 2(2 sin x cos x)2 [sin2A = 2sin A cosA]
= 1 – 8 sin2x cos2x
= R.H.S.
Question 25:
Prove that: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
Answer:
L.H.S. = cos 6x
= cos 3(2x)
= 4 cos3 2x – 3 cos 2x [cos 3A = 4 cos3 A – 3 cos A]
= 4 [(2 cos2 x – 1)3 – 3 (2 cos2 x – 1) [cos 2x = 2 cos2 x – 1]
= 4 [(2 cos2 x)3 – (1)3 – 3 (2 cos2 x)2 + 3 (2 cos2 x)] – 6cos2 x + 3
= 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3
= 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3
= 32 cos6x – 48 cos4x + 18 cos2x – 1
= R.H.S.
