# (cosx - cosy)2 + (sinx - siny)2 = 4 sin2(x-y)/2.

cosx-cosy= -2sin(x+y)/2. sin(x-y)/2 -----(1)

sinx-siny= 2 cos(x+y)/2 .sin(x-y)/2-----(2)

Therefore

(cosx-cosy)2+ (sinx-siny)2= 4sin2(x+y)/2 .sin2(x-y)/2 + 4cos2(x+y)/2.sin2(x-y)/2

= 4sin2(x-y)/2 {sin2(x+y)/2 + cos2(x+y)/2} =4sin(x-y)/2*1= 4sin2(x-y)/2 = RHs hence proved

• 17

other mathod u can just .cosx-cosy
cosx-cosy= -2sin(x+y)/2. sin(x-y)/2 -----(1)

sinx-siny= 2 cos(x+y)/2 .sin(x-y)/2-----(2)

Therefore

(cosx-cosy)2+ (sinx-siny)2= 4sin2(x+y)/2 .sin2(x-y)/2 + 4cos2(x+y)/2.sin2(x-y)/2

= 4sin2(x-y)/2 {sin2(x+y)/2 + cos2(x+y)/2} =4sin2 (x-y)/2*1= 4sin2(x-y)/2 = RHs hence proved

• 11

thank you girish and guruprasad...

• 1

if rhs=4 cos square(x+y/2)

• -1

@girish - wats da difference between my solution and urs.. cheater

• -5

hahaha..u r right guruprasad..

• 5

aaaaaaaaaaaaaaaaa

• -5
tq it was reaaly helpfull
• -1
tq it was really helpfull
• -5
Sin inverse sin100
• -3
tan-¹ x+2cot-1x=2x/3
• 2
Kya
• 3
Dinesh

• 0
Hii aparna
• 0
What are you looking for?