$A=\frac{5\pi}{{\arccos\frac{1}{2}}-i\cdot\arcsin\frac{1}{2}}$ ise $A$'yı bulunuz.

Question

$A=12+6i$

Answer 1

$\begin{align*} & arc\cos \dfrac {1} {2}=\alpha \Rightarrow \cos \alpha =\dfrac {1} {2}\Rightarrow \alpha =\dfrac {\pi } {3}\\ & \arcsin \dfrac {1} {2}=\beta \Rightarrow \sin \beta =\dfrac {1} {2}=>\beta =\dfrac {\pi } {6}\end{align*}$

 $\begin{align*} & \dfrac {5\pi } {\dfrac {\pi } {3}-i\dfrac {\pi } {6}}=\dfrac {5\pi } {\dfrac {\pi } {6}\left( 2-i\right) }=\dfrac {30} {2-i}=\dfrac {60+30i} {5}\\ & =12+6i\end{align*}$