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`        if sin a + sin b=2 , then find the value of sin (a+b). `
7 years ago

shreyas ramoji
19 Points
```										90oThere is only 1 possibility for sin A +sin B to be equal to 2.
max value of sinθ is 1...
so sin A = 1 and sin B = 1.....
therefore A = B = 90o
sin (A+B) = sin (90o+90o) = sin 180o = 0

```
7 years ago
510 Points
```										sina +sinb =2 is possible only when sina as well as sinb is in its maximum value....
sina=1,sinb=1
a=b=90
sina+b=sin180=0
```
7 years ago
Chetan Mandayam Nayakar
312 Points
```										sin(theta)≤1, thus sin a=sin b=1, cos a=cos b=0,sin(a+b)=1*0+0*1=0
```
7 years ago
Arkajyoti Banerjee
26 Points
```										It is evident that the answer is 0. But I worked out a trigonometric solution:SinA+SinB=2SinA+SinB=1+1SinA-1=1-SinBCos(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)]=0Hence, we have 2 cases to solve now:Case 1:Sin(45 – A/2)+Sin(45 – B/2)=0Sin(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 – 4590-A=B-90A+B=90 => Sin(A+B)=Sin(180) => Sin(A+B) = 0.Case 2:Sin(45 – A/2)-Sin(45 – B/2)=0Sin(45 – A/2)=Sin(45 – B/2)Taking principal values, we have:45 – A/2 = 45 – B/2A/2 = B/2A=BLet A=B=x, then in the original equation:SinA+SinB=2Sinx+Sinx=22Sinx=2Sinx=1x=90A=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.
```
one year ago
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