NCERT Solutions for Exercise 3.3, Class 11, Maths

Exercise 3.3 Trigonometric Functions
Question and Answers
Class 11 – Maths

Class	Class 11
Subject	Mathematics
Chapter Name	Trigonometric Functions
Chapter No.	Chapter 3
Exercise	Exercise 3.3
Category	Class 11 Maths NCERT Solutions

Question (1) Prove that 

Answer

Question (2) Prove that

Answer 

Question 3 Prove that

Answer 

Question 4 Prove that

Answer

Question 5 Find the value of
(i) Sin 75°

Answer

Sin 75°= Sin ( 45°+ 30°)
Sin 75°= Sin 45° Cos 30° + Cos 45° Sin 30°
sin(x+y) = sin x cos y + cos x sin y
Therefore we have,

Question (5) Find the value of
(ii) tan 15°

Answer
tan 15°= tan ( 45°- 30° )

Question (6) Prove the following:

Answer  

cos(x+y) = cos x cos y – sin x sin y

Question (7 ) Prove that

Answer

Question (8) Prove that

Answer

Observe that cos x repeats the same value after an interval
2 π  and sin x repeats the same value after an interval  2 π

Question 9

Answer 

We know that cot x  repeats the same value after an interval 2π

Question (10) Prove that sin (n+1) x sin ( n+2 ) x + cos (n+1) x cos (n+2) x = cos x

Answer

cos(x-y) = cos x cos y + sin x sin y
L.H.S.= sin (n+1) x sin (n+2) x + cos (n+1) x cos (n+2) x
=cos [(n+1) x – ( n+2 ) x] = cos [ (n+1) x – (n+2) x] =cos (-x)
=cos x
=  R.H.S

Question (11) Prove that

Answer

Question (12) Sin²6x – sin ² 4x = Sin 2x sin 10x 

Answer

Question (13)  cos² 2x – cos² 6x = sin 4x sin 8x

Answer

Question (14) sin² x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x

Answer

Question (15) cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Answer

Question (16)

Answer

Question (17)

Answer

Question (18)

Answer

Question (19) Prove that

Answer

Question (20) Prove that

Answer

Question (21)

Answer

Question (22) cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Answer

Question (23)

Answer

Question (24) cos 4x = 1 – 8sin² x cos² x

Answer
cos 2x = 1-2sin²x

And sin 2x=2sin x cos x
L.H.S.=cos 4x
=cos 2(2x)
=1-2sin²2x
=1-2(2sin x cos x)²
L.H.S =1- 8sin²x cos²x
= R.H.S.
Hence proved.

Question (25) cos 6x = 32 cos⁶ x – 48cos⁴ x + 18 cos² x – 1

Answer  

cos 3A = 4cos³A – 3cosA
cos 2x= 1 – 2 sin2x
cos 2x = 2 cos² x – 1
L.H.S. = Cos 6x = Cos  cos 3(2x)
= 4 cos³ 2x – 3 cos 2x
= 4 [(2 cos² x – 1)³ – 3 (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