Question

Prove that cos2xcos2y+sin^2(x-y)-sin^2(x+y)= cos(2x+2y)

Answer 1

Answer:

To prove: cos 2θ cos2φ + sin^2 (θ – φ) – sin^2(θ + φ) = cos (2θ + 2φ)

Proof:

LHS = cos 2θ cos2φ + sin2 (θ – φ) – sin2(θ + φ)

= cos 2θ cos2φ + sin (θ – φ + θ + φ) sin(θ – φ - θ - φ) [Using: sin2 A - sin2 B = sin(A + B) sin (A - B)]

= cos 2θ cos2φ + sin 2θ sin(-2φ)

= cos 2θ cos2φ - sin 2θ sin 2φ

= cos (2θ + 2φ) [using: cos (A + B) = cos A cos B - sin A sin B]

= RHS [Hence Proved]

✨Hope it will help you.✨

Answer 2

Answer:The step by step explanation is given below

Step-by-step explanation:

Given: $cos2xcos2y+sin^2(x-y)-sin^2(x+y)= cos(2x+2y)$

To prove:We have to prove

             $cos2xcos2y+sin^2(x-y)-sin^2(x+y)= cos(2x+2y)$

Explanation:

Step 1: $LHS =$  $cos2xcos2y+sin^2(x-y)-sin^2(x+y)$

                    $= cos2xcos2y+[sin(x-y)]^{2}-[sin(x+y)]^{2}$

                  $= cos2xcos2y+(sinx.cosy-siny.cosx)^{2}-(sinx.cosy+siny.cosx)^{2}$

$=cos2xcos3y+(sixcosy)^{2}+(sinycosx)^{2}-2sinxcosxsinycosy-(sinxcosy)^{2}-(sinycosx)^{2}-2sinxsinycosxcosy$

$= cos2xcos2y-4sinxsinycosxcosy$

$=cos2xcos2y-(2sinxcox)(2cosysiny)$

$=cos2xcos2y-sin2ysin2x$

$=cos(2x+2y)$

$=RHS$

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