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

The correct option is B ${-110}^{\circ }$

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