A) \[-2\sin (\alpha +\beta )\]
B) \[2\cos (\alpha +\beta )\]
C) \[2\sin (\alpha -\beta )\]
D) \[-2\cos (\alpha +\beta )\]
E) \[-2\cos (\alpha -\beta )\]
Correct Answer: D
Solution :
Given, \[cos\text{ }\alpha -cos\text{ }\beta =0\] and \[sin\text{ }\alpha +sin\text{ }\beta =0\] Let \[\alpha =cos\text{ }\alpha +i\text{ }sin\text{ }\alpha \] and \[b=cos\beta +i\text{ }sin\beta \] \[\Rightarrow \] \[a+b=0\] \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}+2ab=0\] \[\Rightarrow \] \[{{(cos\text{ }\alpha +i\text{ }sin\text{ }\alpha )}^{2}}\] \[+{{(\cos \beta +i\sin \beta )}^{2}}+2{{e}^{-i\alpha }}{{e}^{i\beta }}=0\] \[\Rightarrow \]\[(\cos 2\alpha +\cos 2\beta )+i(\sin 2\alpha +\sin 2\beta )\] \[=-2[\cos (\alpha +\beta )+i\sin (\alpha +\beta )]\] On equating real part on both sides, we get \[\cos 2\alpha +\cos 2\beta =-2\cos (\alpha +\beta )\]
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