IF SINA+SINB=C AND COSA+COSB=D, THEN THE VALUE OF SIN(A+B)=?

i just substituted the values sinA =C-SINB   AND  cosA=D-cosB

 

SINCE  sinA+B =SinAcosS+CosAsinB

 

SO THAT THE ANSWER BECOMES = C cosB + D sinB -2 sinBcosB  = C cosB + D sinB - Sin2B;

IF U CAN FURTHER SIMPLIFY IT THEN TRY IT.

 

(sinA+sinB)(cosA+cosB)=cd

sinAcosA+sinAcosB+sinBcosA+sinBcosB=CD

(sin2A/2)+(sin2B/2)+sin(A+B)=CD

1/2(sin2A+sin2B)+sin(A+B)=cd

1/2(2sin(A+B)cos(A-B))+sin(A+B)=CD

sin(A+B)[cos(A-B)+1)=CD

sin(A+B)=2CD/C²+D²

because  by squaring and adding the two given equations...

we get...[cos(A+B)+1]=C²+D²/2

HERE , cot(x+y)=0

SO, ON SOLVING WE GET ,TANxTANy=1.

so, SINxSINy=COSxCOSy

 

 

NOW , SIN(x+2y)=SINxCOS2y+COSxSIN2y

=SINx(1-2SINySINy)+COSx2SINyCOSy

=SINx-2SINySINySINx+2SINySINySINx

=SINx

please aproove it if u like.......................

Given SinA+sinB=C -- 1

cosA+cosB=D ---2

By using transformation,

from 1,

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

and from 2,

2cos(A+B/2)cos(A-B/2)=D

Now dividing 1 and 2, we get

tan(A+B/2)=C/D

Now, sin(A+B)= 2tan(A+B/2) whole divided by 1+tan^2 (A+B/2)

= 2CD/(C^2+D^2)

Given SinA+sinB=C (1)cosA+cosB=D (2)By using Substitution,from 1,2sin(A+B/2)cos(A-B/2)=Cand from 2,2cos(A+B/2)cos(A-B/2)=DNow dividing 1 and 2, we get;tan(A+B/2)=C/DNow, sin(A+B)= 2tan(A+B/2) whole divided by 1+tan^2 (A+B/2)= 2CD/(C^2+D^2)Hope you liked!!!😉😉😉