A) \[\sin 4\,\theta +i\cos 4\,\theta \]
B) \[\sin 4\,\theta -i\sin 2\theta \]
C) 0
D) none of these
Correct Answer: D
Solution :
\[{{(\sin \theta +i\cos \theta )}^{4}}\] \[={{\left[ \cos \left( \frac{\pi }{2}-\theta \right)+i\sin \left( \frac{\pi }{2}-\theta \right) \right]}^{4}}\] Using De-Moivre theorem \[=\cos (2\pi -4\theta )+i\,\sin (2\pi -4\theta )\] \[=\cos 4\theta -i\sin 4\theta \] Note: The De-moivre theorem is only applicable, when the real part is in cos form and the imaginary part is in sin form.
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