( + i , ) ^4 ( + i , ) ^5 = - Vidyadip

MCQ

$\frac{{{{(\cos \alpha + i\,\sin \alpha )}^4}}}{{{{(\sin \beta + i\,\cos \beta )}^5}}} = $

A
$\cos (4\alpha + 5\beta ) + i\,\sin (4\alpha + 5\beta )$
B
$\cos (4\alpha + 5\beta ) - i\,\sin (4\alpha + 5\beta )$
✓
$\sin (4\alpha + 5\beta ) - i\cos (4\alpha + 5\beta )$
D
None of these

✓

Answer

Correct option: C.

$\sin (4\alpha + 5\beta ) - i\cos (4\alpha + 5\beta )$

c
(c) $\frac{{{{(\cos \alpha + i\sin \alpha )}^4}}}{{{{(\sin \beta + i\cos \beta )}^5}}}$ $ = \frac{{\cos 4\alpha + i\sin 4\alpha }}{{{i^5}{{(\cos \beta - i\sin \beta )}^5}}}$
= $ - i\,(\cos 4\alpha + i\sin 4\alpha )\,{(\cos \beta - i\sin \beta )^{ - 5}}$
= $ - i\,[\cos 4\alpha + i\sin 4\alpha ]\,\,[\cos 5\beta + i\sin 5\beta ]$
= $ - i\,[\cos (4\alpha + 5\beta ) + i\sin (4\alpha + 5\beta )]$
= $\sin (4\alpha + 5\beta ) - i\,\cos (4\alpha + 5\beta )$.

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