`cos(x+y)*cos(x-y)=cos^2x-sin^2y` ``
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`cos(x+y)*cos(x-y)= cos^2 x - sin^2 x`
`` We will use trigonometric identities to prove the identities.
We know that:
`cosa*cosb= (1/2)(cos(a-b)+cos(a+b)) `
`==gt cos(x+y)*cos(x-y)= (1/2)(cos(x+y-x+y)+cos(x+y+x-y)) `
`==gt cos(x+y)*cos(x-y)= (1/2)(cos(2y) + cos(2x)) `
`==gt cos(x+y)*cos(x-y)= (1/2)((1-2sin^2 y)+(2cos^2x -1)) `
`==gt cos(x+y)*cos(x-y)= (1/2)( 2cos^2 x - 2sin^2y +1 -1) `
`==gt cos(x+y)*cos(x-y)= (1/2)2(cos^2 x-sin^2y) `
`==gt cos(x+y)*cos(x-y)= cos^2 x - sin^2 y.`
``
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first cos(a+b)=cosacosb-sinasinb and cos(a-b)=cosacosb+sinasinb
therefore, cos(x+y).cos(x-y)=(cosxcosy-sinxsiny)(cosxcosy+sinxsiny)
=cos^2 xcos^2 y - sinxsinycosxcosy + sinxsinycosxcosy - sin^2 xsin^2 y
=cos^2 xcos^2 y-sin^2 xsin^2 y
sin^2 y+cos^2 y = 1
= cos^2 x (1-sin^2 y)-sin^2 y (1-cos^2 x)
= cos^2 x - sin^2 y cos^2 x - sin^2 y +sin^2 y cos^2 x
= cos^2 x - sin^2 y
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