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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)
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