If sin α + sin β = a and cos α + cos β = b, prove that \[ \sin(\alpha+\beta)=\frac{2ab}{a^2+b^2} \]
Question
If \[ \sin\alpha+\sin\beta=a \] and \[ \cos\alpha+\cos\beta=b, \] prove that \[ \sin(\alpha+\beta)=\frac{2ab}{a^2+b^2}. \]
Solution
Given,
\[ \sin\alpha+\sin\beta=a \]
and
\[ \cos\alpha+\cos\beta=b \]
Using sum-to-product identities,
\[ \sin\alpha+\sin\beta = 2\sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \]
Therefore,
\[ a = 2\sin\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \]
Also,
\[ \cos\alpha+\cos\beta = 2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \]
Hence,
\[ b = 2\cos\frac{\alpha+\beta}{2} \cos\frac{\alpha-\beta}{2} \]
Now multiply \(a\) and \(b\):
\[ ab = 4 \sin\frac{\alpha+\beta}{2} \cos\frac{\alpha+\beta}{2} \cos^2\frac{\alpha-\beta}{2} \]
Using
\[ 2\sin A\cos A=\sin2A \]
\[ ab = 2\sin(\alpha+\beta) \cos^2\frac{\alpha-\beta}{2} \]
Now find \[ a^2+b^2 \]
\[ a^2 = 4\sin^2\frac{\alpha+\beta}{2} \cos^2\frac{\alpha-\beta}{2} \]
\[ b^2 = 4\cos^2\frac{\alpha+\beta}{2} \cos^2\frac{\alpha-\beta}{2} \]
Adding,
\[ a^2+b^2 = 4\cos^2\frac{\alpha-\beta}{2} \left( \sin^2\frac{\alpha+\beta}{2} + \cos^2\frac{\alpha+\beta}{2} \right) \]
\[ = 4\cos^2\frac{\alpha-\beta}{2} \]
Therefore,
\[ \frac{2ab}{a^2+b^2} = \frac{ 2\left[ 2\sin(\alpha+\beta)\cos^2\frac{\alpha-\beta}{2} \right] } { 4\cos^2\frac{\alpha-\beta}{2} } \]
\[ = \sin(\alpha+\beta) \]
Hence proved.
Final Answer
\[ \boxed{ \sin(\alpha+\beta) = \frac{2ab}{a^2+b^2} } \]