Trigonometry> compound angle...

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

rakes prasad , 15 Years ago

Grade

4 Answers

shreyas ramoji

90^o
There 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 = 90^o

sin (A+B) = sin (90^o+90^o) = sin 180^o = 0

Last Activity: 15 Years ago

vikas askiitian expert

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

Last Activity: 15 Years ago

Chetan Mandayam Nayakar

sin(theta)≤1, thus sin a=sin b=1, cos a=cos b=0,sin(a+b)=1*0+0*1=0

Last Activity: 15 Years ago

Arkajyoti Banerjee

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.

Last Activity: 9 Years ago