Q(4)  Prove that 

\small (\cos x-\cos y)^{2} + (\sin x - \sin y)^{2} = 4\sin^{2}\left ( \frac{x-y}{2} \right )

Answers (1)

We know that (a+b)^{2} = a^{2} + 2ab + b^{2}
                         and 
                        (a-b)^{2} = a^{2} - 2ab + b^{2}
We use these two in our problem
 
(\sin x-\sin y)^{2} = \sin^{2}x - 2\sin x\sin y + \sin^{2}y
    and 
(\cos x-\cos y)^{2} = \cos^{2}x - 2\cos x\cos y + \cos^{2}y

\small (\cos x - \cos y)^{2} + (\sin x - \sin y)^{2} = \cos^{2}x - 2\cos x\cos y + \cos^{2}y +   \sin^{2}x - 2\sin x\sin y + \sin^{2}y
                                                                  = 1 - 2cosxcosy + 1 - 2sinxsiny                           \left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )
                                                                  = 2 - 2(cosxcosy + sinxsiny)
                                                                  = 2 - 2cos(x - y)                                                   \small (\because \cos(x-y) =\cos x \cos y + \sin x \sin y)          
                                                                  =  2(1 - cos(x - y) )
    Now we can write
                           cos(x + y) = 1 -2sin^{2}\frac{(x + y)}{2}                                             \left ( \because \cos2x = 1 - 2\sin^{2}x \ \Rightarrow \cos x = 1 - 2\sin^{2}\frac{x}{2} \right )

                                     so

                           2(1 - cos(x - y) ) = 2(1 - ( 1 -2sin^{2}\frac{(x + y)}{2})) 
                  
                                             
                                           = 4sin^{2}\frac{(x - y)}{2}   =  R.H.S.

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions