if cos A + cos B = (√3 + 1)/ 2 and sin A + sin B = ( √3+ 1)/2 then value of sin( A+B).?

　　　cosA + cosB = (√3 + 1)/2

　　　sinA + sinB = (√3 + 1)/2

　　　cosA + cosB = sinA + sinB

　　　cosA - sinA = sinB - cosB

　　　(cosA - sinA)² = (sinB - cosB)²

　　　　　　sinAcosA = sinBcosB

　　　　　　　　sin2A = sin2B

　　　　　2A = 2B , 2A = π- 2B 　　[ 2B = π - 2A ]

　　　　　　A = B , A = (π/2) - B 　　[ B = (π/2) - A ]

(1) A = B

cosA = (√3 + 1)/4

sinA = (√3 + 1)/4 ... unsuitable

(2) B = (π/2) - A →　A + B = π/2

　　　then　sin(A + B) = sin(π/2) = 1

That means:

cosA = rt3/2

and

cosB = 1/2

This means that

sinA = 1/2

and

sinB = rt3/2

So, we can then use basic trig to calculate A and B.

A = 30

B = 60

sin(A+B) = sin(90) = 1

Since the sin and cos of A and B are equal, the angles are complimentary (add to 90 degrees), therefore the sin(A+B) where A+B=pi/2 is 1.