यदि `x +y + z = .(pi)/(2)` तो सिध्द कीजिए कि:
`cos (x -y -z) + cos (y- x- z) + cos (z-x-y) =4 cos x cos y cos z`

Explanation

` :. , x + y + z= .(pi)/(2)`
तब `x+ y =.(pi)/(2)-z , y+ z = .(pi)/(2)- x ` तथा `z + y =.(pi)/(2) – y`
बायॉ पक्ष `=cos (x- y –z) + cos (y-x –z) + cos (z- x-y)`
`=cos [x- (y+z) ] + cos [y -(x +z) ] + cos [ z- (x+y)]`
` = cos [ x- ((pi)/(2)- x) ] + cos [y- ((pi)/(2) +2y ) ]+ cos (-(pi)/(2) -z)]`
`=cos (-(pi)/(2)+2x ) + cos (-(pi)/(2)+2y) + cos (-(pi)/(2)+ 2z)`
` =cos {-((pi)/(2)-2x) } +cos {-((pi)/(2)-2y) }+cos {-((pi)/(2)-2z)}`
` = cos ((pi)/(2) -2x) + cos ((pi)/(2)-2y) +cos ((pi)/(2) -2z) =sin 2x+ sin 2y + sin 2z`
`= 2 sin .(2x+2y)/(2) cos .(2x-2y)/(2) +sin 2z`
`=2 sin (x+y) cos (x-y) + sin 2z`
`= 2sin .((pi)/(2)-z) cos (x-y) + sin 2z`
` 2 cos z [ cos (x -y) +2 sin z cos z=2 cos z [ cos (x-y) + sin z]`
` 2 cos z [cos (x-y) + sin ((pi)/(2) -x +y)]`
` =2 cos z [ cos (x- y) + cos (x+y) ] = 2 cos z [ 2 cos x cos y]`
` 4 cos x cos y cos z =`दायाँ पक्ष।