Arkajyoti Banerjee
Last Activity: 8 Years ago
It is evident that the answer is 0. But I worked out a trigonometric solution:
SinA+SinB=2
SinA+SinB=1+1
SinA-1=1-SinB
Cos(90-A)-1=1-Cos(90-B)
Cos(90-A)-1=2Sin^2(45 - B/2)
-2Sin^2(45 – A/2)=2Sin^2(45 – B/2)
-Sin^2(45 – A/2)=Sin^2(45 – B/2)
Taking absolute values on both the sides,
Sin^2(45 – A/2)=Sin^2(45 – B/2) [SinX cannot be imaginary, so Sin^2(x) cannot be negative]
Sin^2(45 – A/2)-Sin^2(45 – B/2)=0
[Sin(45 – A/2)+Sin(45 – B/2)][Sin(45 – A/2)-Sin(45 – B/2)]=0
Hence, we have 2 cases to solve now:
Case 1:
Sin(45 – A/2)+Sin(45 – B/2)=0
Sin(45 – A/2)=-Sin(45 – B/2)
Sin(45 – A/2)=Sin(B/2 – 45)
Taking principal values, we have:
45 – A/2 = B/2 – 45
90-A=B-90
A+B=90 => Sin(A+B)=Sin(180) => Sin(A+B) = 0.
Case 2:
Sin(45 – A/2)-Sin(45 – B/2)=0
Sin(45 – A/2)=Sin(45 – B/2)
Taking principal values, we have:
45 – A/2 = 45 – B/2
A/2 = B/2
A=B
Let A=B=x, then in the original equation:
SinA+SinB=2
Sinx+Sinx=2
2Sinx=2
Sinx=1
x=90
A=B=90, which proves the other solutions too, that there is only one such possibility.
Once again, Sin(A+B)=Sin(90+90)=Sin180=0.