Question

Prove that
$\bf (cos \;x + cos \; y)^{2} + (sin \; x - sin \; y)^{2} = 4cos^{2}\dfrac{x +y}{2}(cosx+cosy)2+(sinx−siny)2=4cos22x+y​$
​

Answer 1

To prove:

(cosa + cosy)? + (sinæ – siny)? = 4 cos (")

Solution:

Consider,

(cosa + cosy)2 + (sinx - siny)?

Using

(a+b)² = a? +62 + 2ab =

(a b)? = a? + b? - = 2ab

(cos²x + cos²y + 2 cosx cosy) + (sin-x + siny - 2 sinx siny)

= (cos-z + sin’2) + (cos-y + sin’y) + 2(cosx cosy - sinx siny)

Using cose + sinºe

=1+1+2(cosx cosy - sinx siny)

= 2 + 2 cos(x + y)

= 2(1+ cos(x + y))

We know that

cos A = 2 cos² А2 -1= 1+ cos A = 2 cos

2(2cos?(프)) cOS ,2(x+y

= 4 cos (") x+y

2 (cosx + cosy)? + (sinx – siny)² = 4 cos

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Answer 2

Answer:

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