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

[MP PET 1986]

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Solution

The correct option is D

As given $\frac{s i n A + s i n B}{c o s A + c o s B}$ = $\frac{C}{D}$

$\Rightarrow$ $\frac{2 s i n \frac{A + B}{2} . c o s \frac{A - B}{2}}{2 c o s \frac{A + B}{2} . c o s \frac{A - B}{2}}$ = $\frac{C}{D}$ $\Rightarrow$ tan $\frac{A + B}{2}$ = $\frac{C}{D}$

Thus, sin(A + B) = $\frac{2 t a n \frac{A + B}{2}}{1 + t a n^{2} \frac{A + B}{2}}$

= $\frac{2 \frac{C}{D}}{1 + \frac{C^{2}}{D^{2}}}$ = $\frac{2 C D}{(C^{2} + D^{2})}$ .

Suggest Corrections

Similar questions

sin A + sin B = C

cos A + cos B = D

sin (A + B) =

sin A + sin B + sin C = 0

cos A + cos B + cos C = 0

cos (A + B) + cos (B + C) + cos (C + A) =

\frac{sin A - sin B}{cos A + cos B} + \frac{cos A - cos B}{sin A + sin B}

sin A = sin B

cos A = cos B

sin \frac{A - B}{2} =

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