NCERT Solutions for Chapter 3, Miscellaneous Exercise, Class 11, Maths

Miscellaneous Exercise
Trigonometric Functions
Question and Answers
Class 11 – Maths

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

Question (1) Prove that

Answer

Question (2)  Prove that (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

Answer

LHS = (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
LHS  = sin 3x sin x + sin2 x + cos 3x cos x – cos2 x
Taking out the common terms,
LHS = cos 3x cos x + sin 3x sin x – (cos² x – sin² x)
Using the formula
cos (A – B) = cos A cos B + sin A sin B
= cos (3x – x) – cos 2x
So, we get
= cos 2x – cos 2x
= 0
= RHS

Question (3) Prove that

Answer  

cos(x+y)=cos x cos y-sin x sin y
L.H.S=(cos x + cos y)²+(sin x – sin y)²
By expanding using the formula, we get
LHS = cos² x + cos² y + 2 cos x cos y + sin² x + sin² y – 2 sin x sin y
Grouping the terms,
= (cos² x + sin² x) + (cos² y + sin² y) + 2 (cos x cos y – sin x sin y)
Using the formula cos (A + B) = (cos A cos B – sin A sin B)
= 1 + 1 + 2 cos (x + y)
By further calculation,
= 2 + 2 cos (x + y)
Taking 2 as common
= 2 [1 + cos (x + y)] From the formula cos 2A = 2 cos² A – 1

Therefore , LHS =RHS
Hence Proved

Question (4) Prove that

Answer

LHS = (cos x – cos y) ² + (sin x – sin y) ²
By expanding using the formula,
= cos² x + cos² y – 2 cos x cos y + sin² x + sin² y – 2 sin x sin y
Grouping the terms,
= (cos² x + sin² x) + (cos² y + sin² y) – 2 (cos x cos y + sin x sin y)
Using the formula cos (A – B) = cos A cos B + sin A sin B
= 1 + 1 – 2 [cos (x – y)] By further calculation,
= 2 [1 – cos (x – y)] From formula cos 2A = 1 – 2 sin² A

LHS =RHS
Hence proved

Question (5) Prove that sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Answer

L.H.S. =sin x + sin 3x+sin 5x+sin 7x
LHS=(sin x + sin 5x)+(sin 3x+sin 7x)
LHS =(sin x + sin 5x)+(sin 3x+sin 7x)