Question

solution of this question: SinA+SinB=1/4 & CosA+CosB=1/4 than Cos(A+B)=

Answer 1

\[\sin A+\sin B=\frac{ 1 }{ 4 } \]
\[2\sin \frac{ A+B }{ 2 }\cos \frac{ A-B }{ 2 }=\frac{ 1 }{ 4 } ...(1)\]
\[\cos A+\cos B=\frac{ 1 }{ 4 }\]
\[2\cos \frac{ A+B }{ 2 } \cos \frac{ A-B }{ 2 }=\frac{ 1 }{ 4 } ...(2)\]
divide (1)by (2)
\[\tan \frac{ A+B }{ 2 }=1=\tan \frac{ \pi }{ 4 }\]
\[A+B=\frac{ \pi }{ 2 }\]
\[\cos \left( A+B \right)=0\]