Prove that-cos x-cos y2-sin x-sin y2-4 cos 2x-y-2

We have L.H.S = (cos x + cos y )^2 + (sin x sin y )^2 = [ 2 cos ( x+y/2) cos ( x y/2 ) ]^2 + [ 2 cos (x+y/2) sin ( x y/2 ) ]^2 = 4 cos^2 ( x+y/2) cos^2 ( ...

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Standard XII

Mathematics

Properties Derived from Trigonometric Identities

Prove that:co...

Question

Prove that:

$(cos x + cos y)^{2} + (sin x - sin y)^{2}$ = $4 {cos}^{2} (\frac{x + y}{2})$

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Solution

We have L.H.S

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

= ${[2 cos (\frac{x + y}{2}) cos (\frac{x - y}{2})]}^{2} + {[2 cos (\frac{x + y}{2}) sin (\frac{x - y}{2})]}^{2}$

= $4 {cos}^{2} (\frac{x + y}{2}) {cos}^{2} (\frac{x - y}{2}) + 4 {cos}^{2} (\frac{x + y}{2}) {sin}^{2} (\frac{x - y}{2})$

= $4 {cos}^{2} (\frac{x + y}{2})$ $[{cos}^{2} (\frac{x - y}{2}) + {sin}^{2} (\frac{x - y}{2})]$

= $4 {cos}^{2} (\frac{x + y}{2})$ = R.H.S.

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$(cos x - cos y)^{2} + (sin x - sin y)^{2} = 4 {sin}^{2} \frac{x - y}{2}$

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(cos x + cos y)^{2} + (sin x - sin y)^{2} = 4 {cos}^{2} \frac{x + y}{2}

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Prove that: (cos x+cos y)2+(sin x−sin y)2 = 4cos2(x+y2)

We have L.H.S = (cos x+cos y)2+(sin x−sin y)2 = [2cos(x+y2)cos(x−y2)]2+[2cos(x+y2)sin(x−y2)]2 = 4cos2(x+y2)cos2(x−y2)+4cos2(x+y2)sin2(x−y2) = 4cos2(x+y2)[cos2(x−y2)+sin2(x−y2)] = 4cos2(x+y2) = R.H.S.

Prove that:

$(cos x + cos y)^{2} + (sin x - sin y)^{2}$ = $4 {cos}^{2} (\frac{x + y}{2})$