Question
If $$\cos \alpha = \frac{1}{2}\left( {x + \frac{1}{x}} \right),\cos \beta = \frac{1}{2}\left( {y + \frac{1}{y}} \right)$$ then $$\cos \left( {\alpha - \beta } \right)$$ is equal to
A.
$$\frac{x}{y} + \frac{y}{x}$$
B.
$${xy + \frac{1}{{xy}}}$$
C.
$$\frac{1}{2}\left( {\frac{x}{y} + \frac{y}{x}} \right)$$
D.
None of these
Answer :
$$\frac{1}{2}\left( {\frac{x}{y} + \frac{y}{x}} \right)$$
Solution :
$$\eqalign{
& \cos \alpha = \frac{1}{2}\left( {x + \frac{1}{x}} \right) \cr
& \Rightarrow \,\,x = \cos \alpha \pm i\sin \alpha . \cr
& {\text{Similarly, }}y = \cos \beta \pm i\sin \beta . \cr
& \therefore \,\,\frac{x}{y} = \cos \left( {\alpha - \beta } \right) \pm i\sin \left( {\alpha - \beta } \right), \cr
& xy = \cos \left( {\alpha + \beta } \right) \pm i\sin \left( {\alpha + \beta } \right) \cr
& {\text{and, }}\frac{y}{x} = \cos \left( {\alpha - \beta } \right) \mp i\sin \left( {\alpha - \beta } \right). \cr} $$