A) \[2\cos 3(\beta -\alpha )\]
B) \[2\cos 3(\beta +\alpha )\]
C) \[2\sin 3(\beta -\alpha )\]
D) \[2\sin 3(\beta +\alpha )\]
E) \[\sin 3(\beta -\alpha )\]
Correct Answer: C
Solution :
\[\because \]\[x+\frac{1}{x}=2\sin \alpha \] \[\therefore \] \[x=\frac{2\sin \alpha \pm \sqrt{4{{\sin }^{2}}\alpha -4}}{2}\] \[\Rightarrow \] \[x=\sin \alpha +i\cos \alpha \] Similarly, \[y=cos\,\beta \pm \text{ }i\text{ }sin\text{ }\beta \] \[\therefore \] \[xy=(\sin \alpha \pm i\cos \alpha )(\cos \beta \pm i\sin \beta \] \[=\sin (\beta -\alpha )\pm i\cos (\beta -\alpha )\] and\[\frac{1}{xy}=\sin (\beta -\alpha )\mp i\cos (\beta -\alpha )\] \[\therefore \]\[{{(xy)}^{3}}+\frac{1}{{{(xy)}^{3}}}=\sin 3(\beta -\alpha )\pm i\cos 3(\beta -\alpha )\] \[+\sin 3(\beta -\alpha )\mp i\cos 3(\beta -\alpha )\] \[=2\sin 3(\beta -\alpha )\]
You need to login to perform this action.
You will be redirected in
3 sec
