[" 18."cos A+cos B-sin C=4sin(C)/(2)],[qquad [sin((pi)/(4)-(A)/(2))sin((pi)/(4)-(B)/(2))]]

cos A + cos B + cos C = 1 + 4sin ((A) / (2)) sin ((B) / (2)) sin ((C) / (2))

sin A + sin B-sin C = 4sin ((A) / (2)) sin ((B) / (2)) cos ((C) / (2))

Cos((pi)/(4)+A).Cos((pi)/(4)-B)-Sin((pi)/(4)+A).Sin((pi)/(4)-B)=

Prove that: cos((pi)/(4)-A)cos((pi)/(4)-B)-sin((pi)/(4)-A)sin((pi)/(4)-B)=sin(A+B)

cos((pi)/(4)+A) cos((pi)/(4) - B) + sin ((pi)/(4) +A) * sin((pi)/4-B)=

If A, B, C are the angles in a triangle then prove that sin .(A)/(2)+ sin . (B)/(2)+ sin .(C)/(2) =1 +4 sin((pi-A)/(4)) sin ((pi-B)/(4)) sin((pi-C)/(4))

sin(pi/4+A).sin(pi/4-A)=1/2 cos2A

If A+B+C=pi then prove that cos A+cos B+cos C=1+4sin((A)/(2))*sin((B)/(2))*sin((C)/(2))

If A+B+C=pi then prove that sin((A)/(2))+sin((B)/(2))+sin((C)/(2))-1=4sin((pi-A)/(4))sin((pi-B)/(4))sin((pi-C)/(4))

sin (pi/2) = 2 sin(pi/4) cos (pi/4)