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

Question 1 (a) , Exercise 3.3

Answer

Exercise 3.3 , question 1 (a) , answers

Question (2) Prove that

Exercise 3.3 , question 2

Answer

Exercise 3.3 , question 2 answer

Question 3 Prove that

Exercise 3.3 , Question 3
Answer

Exercise 3.3 , Question 3 ans

Question 4 Prove that

Exercise 3.3 , question 4

Answer

Exercise 3.3 , Answer 4

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(a) , Exercise 3.3

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

Answer
tan 15°= tan ( 45°- 30° )

Exercise 3.3 , question 5(b)

Question (6) Prove the following:

Exercise 3.3 , Question 6

Answer

$cos (x+y) =cos xcos y-sin xsin y$

Exercise 3.3 , Answer 6

Question (7 ) Prove that

Exercise 3.3 , Question 7
Answer

Exercise 3.3 , Answer 7

Question (8) Prove that

Exercise 3.3 , Question 8

Answer

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

Exercise 3.3 , Answer 8

Question 9

Exercise 3.3 , Question 9

Answer

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

Exercise 3.3 , Answer 9

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
Sin A + Sin B

Exercise 3.3 , Answer 11
Question (12) Sin²6x – sin ² 4x = Sin 2x sin 10x

Answer
Exercise 3.3 , Answer 12

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

Answer

Exercise 3.3 , Answer 13

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

Answer

Exercise 3.3 , Answer 14

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

Answer

Exercise 3.3 , Answer 15

Question (16)

Exercise 3.3 ,Question 16

Answer

Exercise 3.3 , Answer 16

Question (17)

Exercise 3.3 , Question 17

Answer

Exercise 3.3 , Answer 17

Question (18)

Exercise 3.3 , Question 18

Answer

Exercise 3.3 , Answer 18

Question (19) Prove that

Exercise 3.3 , Question 19

Answer

Exercise 3.3 , Answer 19

Question (20) Prove that

Exercise 3.3 ,Question 20

Answer

Exercise 3.3 , Answer 20

Question (21)

Exercise 3.3 , Question 21

Answer

Exercise 3.3 , Answer 21

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

Answer

Exercise 3.3 , Answer 22

Question (23)

Exercise 3.3 , Question 23

Answer

Exercise 3.3 , Answer 23

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

Answer
$cos 2x=1-2si n^{2} x$
And $sin 2x=2sin x cos x$ L.H.S. $=cos 4x$ $=cos 2 (2x)$ $=1-2si n^{2} 2x$ $=1-2 {(2sin x cos x)}^{2}$ L.H.S $=1-8si n^{2} xco s^{2} x$ = 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 cos2x
= 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

Exercise 3.3 Trigonometric Functions Question and Answers Class 11 – Maths

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Exercise 3.3 Trigonometric Functions
Question and Answers
Class 11 – Maths