# Cos X+CosY=1/3,SinX+SinY=1/4 then Sin (X+Y))=Please send andere for this question .This question is from trignomdtri

Samyak Jain
333 Points
5 years ago
cosX + cosY = 1/3  $\dpi{100} \Rightarrow$  2 cos(X+Y)/2 cos(X–Y)/2 = 1/3         ...(1)
sinX + sinY  =  1/4  $\dpi{100} \Rightarrow$  2 sin(X+Y)/2 cos(X–Y)/2 = 1/4         ...(2)
Divide (2) by (1)
[2 sin(X+Y)/2 cos(X–Y)/2] /[2 cos(X+Y)/2 cos(X–Y)/2] = (1/4)/(1/3)
tan(X + Y)/2  =  3/4.
We know that sin2x = 2 tanx / (1 + tan2x)
$\dpi{100} \therefore$ sin(X + Y) = 2 tan(X + Y)/2 / (1 + tan2(X + Y)/2) = 2.(3/4) / (1 + 9/16)
= (3/2)(25/16)
sin(X + Y) = 24/25.
SANDEEP
14 Points
3 years ago
cosX + cosY = 1/3 \Rightarrow 2 cos(X+Y)/2 cos(X–Y)/2 = 1/3 ...(1)sinX + sinY = 1/4 \Rightarrow 2 sin(X+Y)/2 cos(X–Y)/2 = 1/4 ...(2)Divide (2) by (1)[2 sin(X+Y)/2 cos(X–Y)/2] /[2 cos(X+Y)/2 cos(X–Y)/2] = (1/4)/(1/3)tan(X + Y)/2 = 3/4.We know that sin2x = 2 tanx / (1 + tan2x)\therefore sin(X + Y) = 2 tan(X + Y)/2 / (1 + tan2(X + Y)/2) = 2.(3/4) / (1 + 9/16) = (3/2)(25/16) sin(X + Y) = 24/25.