Question

If $P,Q$ and $R$ are angles of an isosceles triangle and $\angle P=\frac{\pi }{2}$, then the value of

$\{{(\cos \frac{P}{3}-i\sin \frac{P}{3})}^3+(\cos Q+i\sin Q)(\cos R-i\sin R)+(\cos P-i\sin P)(\cos Q-i\sin Q)(\cos R-i\sin R)\}$

• A. $i$
• B. $-i$
• C. $1$
• D. $-1$

Answer 1

The correct option is B $-i$

Given that $P.Q$ and $R$ are angles of an isosceles triangle and $\angle P=\frac{\pi }{2}$
$\therefore$ $Q=R=\frac{\pi }{4}$ ($\because$ $P+Q+R={180}^o$)
Now

${(\cos \frac{P}{3}-i\sin \frac{P}{3})}^3+(\cos Q+i\sin Q)(\cos R+i\sin R)+(\cos P-i\sin P)(\cos Q-i\sin Q)(\cos R-i\sin R)$

$={\left(e^{\frac{P}{3}}\right)}^3+e^{iQ}.e^{-iR}+e^{-iP}{e}^{-iQ}.e^{-iR}(\because \cos \theta +i\sin \theta =e^{i\theta })$

$=e^{-iP}+e^{(Q-R)}+e^{-i(P+Q+R)}$

$=e^{-i\frac{\pi }{2}}+e^{i\left(0\right)}+e^{-i\pi }(\because Q=R=\frac{\pi }{4};P=\frac{\pi }{2})$

$=(-i+1)-1=-i$