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Q. The principal amplitude of $\left(\sin\, 40^{\circ} + i\, \cos\, 40^{\circ}\right)^{5}$ is

WBJEEWBJEE 2008Complex Numbers and Quadratic Equations

Solution:

$\left(\sin 40^{\circ}+i \cos 40^{\circ}\right)^{5}$
$=i^{5}\left(\cos 40^{\circ}-i \sin 40^{\circ}\right)^{5}$
$=i\left(\cos 200^{\circ}-i \sin 200^{\circ}\right)$
$=i\left[\cos \left(180^{\circ}+20^{\circ}\right)-i \sin \left(180^{\circ}+20^{\circ}\right)\right]$
$=i\left(-\cos 20^{\circ}+i \sin 20^{\circ}\right)$
$=-i \cos 20^{\circ}-\sin 20^{\circ}$
$=\cos \left(-110^{\circ}\right)+i \sin \left(-110^{\circ}\right)$
$\therefore \quad$ Principal amplitude $=-110^{\circ}$