Question

for the value of A and B given below sin (A+B)=sin (A)+ sin (B)?

Answer 1

Answer:

A = 180° &  B=0°

Step-by-step explanation:

On the LHS we have, sin (A+B)

∴  LHS

= sin (A+B)

= sinA cosB + sinB cosA

RHS = sin A + sin B

Equating LHS & RHS,

sinA cosB + sinB cosA  = sin A + sin B

Comparing, we conclude,

cos(B) = 1, ∴ B=0° & ,

cos(A) = 1, ∴ A = 180°

Answer 2

Answer:

(i) A+B=360, 720..........

(II) B=0

Step-by-step explanation:

Formula used:

$SinA=2\:sin\frac{A}{2}\:cos\frac{A}{2}$

$sinC+sinD=2\:sin(\frac{C+D}{2})\:cos(\frac{C-D}{2})$

Given:

$sin(A+B)=sinA+sinB$

$2\:sin(\frac{A+B}{2})\:cos(\frac{A+B}{2})=2\:sin(\frac{A+B}{})\:cos(\frac{A-B}{2})$

$2\:sin(\frac{A+B}{2})[cos(\frac{A+B}{2})-cos(\frac{A-B}{2})=0$

$sin(\frac{A+B}{2})[cos(\frac{A+B}{2})-cos(\frac{A-B}{2})]=0$

This implies

case(i)

$sin(\frac{A+B}{2})=0$

$\frac{A+B}{2}=n\:\pi$

$A+B=n\:2\pi$ where n∈ Z

Therefore

A+B=360, 720, 1080...........

case(ii):

$cos(\frac{A+B}{2})-cos(\frac{A-B}{2})=0$

$cos(\frac{A+B}{2})=cos(\frac{A-B}{2})$

$\frac{A+B}{2}=\frac{A-B}{2}$

$A+B=A-B$

This implies B=0