If \[ \cos\alpha+\cos\beta=0 \] and \[ \sin\alpha+\sin\beta=0, \] prove that \[ \cos2\alpha+\cos2\beta=-2\cos(\alpha+\beta) \]
Question
If \[ \cos\alpha+\cos\beta=0 \] and \[ \sin\alpha+\sin\beta=0, \] prove that \[ \cos2\alpha+\cos2\beta=-2\cos(\alpha+\beta). \]
Solution
Given,
\[ \cos\alpha+\cos\beta=0 \]
and
\[ \sin\alpha+\sin\beta=0 \]
Squaring both equations,
\[ (\cos\alpha+\cos\beta)^2=0 \]
\[ \cos^2\alpha+\cos^2\beta+2\cos\alpha\cos\beta=0 \]
Similarly,
\[ (\sin\alpha+\sin\beta)^2=0 \]
\[ \sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta=0 \]
Adding the two equations,
\[ \sin^2\alpha+\cos^2\alpha + \sin^2\beta+\cos^2\beta \]
\[ + 2(\sin\alpha\sin\beta+\cos\alpha\cos\beta)=0 \]
Using
\[ \sin^2\theta+\cos^2\theta=1 \]
we get
\[ 1+1+2\cos(\alpha-\beta)=0 \]
\[ 2+2\cos(\alpha-\beta)=0 \]
\[ \cos(\alpha-\beta)=-1 \]
Now use the identity
\[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \]
Taking \[ C=2\alpha \quad \text{and} \quad D=2\beta, \]
\[ \cos2\alpha+\cos2\beta = 2\cos(\alpha+\beta)\cos(\alpha-\beta) \]
Substitute \[ \cos(\alpha-\beta)=-1 \]
\[ \cos2\alpha+\cos2\beta = 2\cos(\alpha+\beta)(-1) \]
\[ \boxed{ \cos2\alpha+\cos2\beta = -2\cos(\alpha+\beta) } \]
Hence proved.