F.4(a MATHS)

1.因A+B+C+D=360度
L.H.S
=sin(A+C)sin(A+D)
=sin[360-(B+D]sin[360-(B+C)]
=-sin(B+D)[-sin(B+C)]
=sin(B+C)sin(B+D)
=R.H.S

2.因cos2A=2cos^2A-1=1-2sin^2A,cosA+cosB=2cos[(A+B)/2]cos[(A-B)/2]
L.H.S.
=1/2[cos2(x-y)+1]-1/2[1-cos2(x+y)]
=1/2[cos2(x-y)+1-1+cos2(x+y)]
=1/2[cos2(x-y)+cos2(x+y)]
=1/2(2cos{[2(x-y)+2(x+y)]/2}cos{[2(x-y)-2(x+y)]/2})
=cos2xcos(-2y)
=cos2xcos2y
=R.H.S.

1)Let A,B,C,D be the interior angles of a quadrilateral. Prove that sin(A+C)sin(A+D)=sin (B+C)sin(B+D).
A + B + C + D = 360
LHS
= sin(A + C)sin(A + D)
= sin[360 - (B + D)]．sin[360 - (B + C)]
= [-sin(B + D)]．[-sin(B + C)]
= sin(B + C)sin(B + D)
= RHS

2.Prove tha cos^2(x-y)-sin^2(x+y)=cos2xcos2y
LHS
cos^2(x - y) - sin^2(x + y)
= [cos(x - y) + sin(x + y)][cos(x - y) - sin(x + y)]
= (cosxcosy + sinxsiny + sinxcosy + sinycosx)(cosxcosy + sinxsiny - sinxcosy - sinycosx)
= [cosx(cosy + siny) + sinx(siny + cosy)][cosx(cosy- siny) - sinx(cosy - siny)]
= (cosx + sinx)(cosy + siny)(cosx - sinx)(cosy - siny)
= (cosx + sinx)(cosx - sinx)(cosy + siny)(cosy - siny)
= [cos^2x - sin^2x][cos^2y - sin^2y]
= cos2xcos2y
= RHS