Question

prove that (cosα - cosβ) whole square + (sinα-sinβ) whole square = 4sin square α-βdivide by 2

Answer 1

Answer:

To Prove:

( Cos A - Cos B )² + ( Sin A - Sin B )² = 4 Sin² ( A - B ) / 2

Considering LHS we get,

⇒ ( Cos²A + Cos²B - 2CosACosB + Sin²A + Sin²B - 2SinASinB )

⇒ ( Cos²A + Sin²A + Cos²B + Sin²B - 2 ( CosA.CosB + SinA.SinB )

⇒ [ 1  + 1 - 2 ( Cos ( A - B )) ]

⇒ [2 - 2 Cos ( A - B )]

⇒ 2 ( 1 - Cos ( A - B ))

Let ( A -B ) be denoted as X

⇒ 2 ( 1 - Cos ( X ) ) = 2 ( 2 Sin²(X/2) )      [ Reason: 1 - Cos 2x = 2Sin²x ]

⇒ 2 ( 2 Sin² ( A-B )/ 2 )

⇒ 4 Sin² ( A-B ) / 2

Hence Proved !!