Question

if sinA+sinB=C,cosA+cosB=D,then value of sin(A+B)=
​

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{sinA+sinB=C}$

$\mathsf{cosA+cosB=D}$

$\underline{\textsf{To find:}}$

$\textsf{The value of}$

$\mathsf{sin(A+B)}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{sinA+sinB=C}$

$\implies\mathsf{2\,sin(\dfrac{A+B}{2})\,cos(\dfrac{A-B}{2})=C}$......(1)

$\mathsf{cosA+cosB=D}$

$\implies\mathsf{2\,cos(\dfrac{A+B}{2})\,cos(\dfrac{A-B}{2})=D}$......(2)

$\textsf{Divide (1) by (2)}$

$\mathsf{\dfrac{2\,sin(\dfrac{A+B}{2})\,cos(\dfrac{A-B}{2})}{2\,cos(\dfrac{A+B}{2})\,cos(\dfrac{A-B}{2})}=\dfrac{C}{D}}$

$\mathsf{\dfrac{sin(\dfrac{A+B}{2})}{cos(\dfrac{A+B}{2})}=\dfrac{C}{D}}$

$\mathsf{tan(\dfrac{A+B}{2})=\dfrac{C}{D}}$

$\textsf{We know that, the identity}$

$\boxed{\mathsf{sinA=\dfrac{2\,tan\frac{A}{2}}{1+tan^2\frac{A}{2}}}}$

$\textsf{Now}$

$\mathsf{sin(A+B)=\dfrac{2\,tan(\dfrac{A+B}{2})}{1+tan^2(\dfrac{A+B}{2})}}$

$\mathsf{sin(A+B)=\dfrac{2\,\dfrac{C}{D}}{1+(\dfrac{C}{D})^2}}$

$\mathsf{sin(A+B)=\dfrac{\dfrac{2C}{D}}{1+\dfrac{C^2}{D^2}}}$

$\mathsf{sin(A+B)=\dfrac{\dfrac{2C}{D}}{\dfrac{D^2+C^2}{D^2}}}$

$\mathsf{sin(A+B)=\dfrac{2C}{D}{\times}\dfrac{D^2}{D^2+C^2}}$

$\mathsf{sin(A+B)=\dfrac{2C}{D}{\times}\dfrac{D^2}{D^2+C^2}}$

$\implies\boxed{\mathsf{sin(A+B)=\dfrac{2CD}{D^2+C^2}}}$

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