Question

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

• A. $70^{\circ }$
• B. $-110^{\circ }$
• C. $110^{\circ }$
• D. $-70^{\circ }$

Answer 1

Answer: (B)

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