If sin α + sin β = a and cos α + cos β = b, prove that \[ \cos(\alpha-\beta)=\frac{a^2+b^2-2}{2} \]
Question
If \[ \sin\alpha+\sin\beta=a \] and \[ \cos\alpha+\cos\beta=b, \] prove that \[ \cos(\alpha-\beta)=\frac{a^2+b^2-2}{2}. \]
Solution
Given,
\[ \sin\alpha+\sin\beta=a \]
and
\[ \cos\alpha+\cos\beta=b \]
Squaring both equations,
\[ (\sin\alpha+\sin\beta)^2=a^2 \]
\[ \sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta=a^2 \]
Similarly,
\[ (\cos\alpha+\cos\beta)^2=b^2 \]
\[ \cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta=b^2 \]
Adding the two equations,
\[ \sin^2\alpha+\cos^2\alpha + \sin^2\beta+\cos^2\beta \]
\[ + 2(\sin\alpha\sin\beta+\cos\alpha\cos\beta) = a^2+b^2 \]
Using
\[ \sin^2\theta+\cos^2\theta=1 \]
we get
\[ 1+1 + 2(\sin\alpha\sin\beta+\cos\alpha\cos\beta) = a^2+b^2 \]
\[ 2+2(\sin\alpha\sin\beta+\cos\alpha\cos\beta) = a^2+b^2 \]
Using the identity
\[ \cos(\alpha-\beta) = \cos\alpha\cos\beta+\sin\alpha\sin\beta \]
Therefore,
\[ 2+2\cos(\alpha-\beta) = a^2+b^2 \]
\[ 2\cos(\alpha-\beta) = a^2+b^2-2 \]
Hence,
\[ \boxed{ \cos(\alpha-\beta) = \frac{a^2+b^2-2}{2} } \]