Question

cos E-)c cos E-y -sin E-x sin 2-y =sin(x+y)4 4 4 4

Answer 1

Simplifying the L.H.S by multiplying and dividing by 2,

cos( π 4 −x )⋅cos( π 4 −y )−sin( π 4 −x )⋅sin( π 4 −y ) = 1 2 [ 2cos( π 4 −x )⋅cos( π 4 −y ) ]+ 1 2 [ −2sin( π 4 −x )⋅sin( π 4 −y ) ]

Use trigonometric identity 2cosA⋅cosB=cos( A+B )+cos(A−B) and −2sinA⋅sinB=cos( A+B )−cos( A−B ) in the above expression.

= 1 2 [ cos{ ( π 4 −x )+( π 4 −y ) } +cos{ ( π 4 −x )−( π 4 −y ) } ]+ 1 2 [ cos{ ( π 4 −x )+( π 4 −y ) } −cos{ ( π 4 −x )−( π 4 −y ) } ] = 1 2 cos{ ( π 4 −x )+( π 4 −y ) }+ 1 2 cos{ ( π 4 −x )−( π 4 −y ) } + 1 2 cos{ ( π 4 −x )+( π 4 −y ) }− 1 2 cos{ ( π 4 −x )−( π 4 −y ) } =2× 1 2 [ cos{ ( π 4 −x )+( π 4 −y ) } ] =cos[ π 2 −( x+y ) ] =sin( x+y )=R.H.S

L.H.S. = R.H.S.

Hence, the expression has been proved.