Prove that:$$(cosx-cosy)^2+(sinx-siny)^2=4sin^2\dfrac{x-y}{2}$$

Asked by Pragya Singh | 1 year ago |  70

##### Solution :-

LHS = (cos x – cos y)2 + (sin x – sin y)2

By expanding using 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

$$2[1-\{1-2sin^2(\dfrac{x-y}{2})\}]$$

we get,

$$4sin^2(\dfrac{x-y}{2})$$

Therefore L.H.S = R.H.S

Hence proved.

Answered by Abhisek | 1 year ago

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