Question

If $a=\cos 2\alpha +i\sin 2\alpha ,b=\cos 2\beta +i\sin 2\beta ,c=\cos 2\gamma +i\sin 2\gamma$ and $d=\cos 2\delta +i\sin 2\delta$, then $\sqrt{abcd}+\frac{1}{\sqrt{abcd}}=$

• A. $\sqrt{2}\cos (\alpha +\beta +\gamma +\delta )$
• B. $2\cos (\alpha +\beta +\gamma +\delta )$
• C. $\cos (\alpha +\beta +\gamma +\delta )$
• D. None of these

Answer 1

The correct option is B $2\cos (\alpha +\beta +\gamma +\delta )$

We have $a=\cos 2\alpha +i\sin 2\alpha ,b=\cos 2\beta +i\sin 2\beta ,c=\cos 2\gamma +i\sin 2\gamma ,d=\cos 2\delta +i\sin 2\delta$

Euler form of $a=\cos 2\alpha +i\sin 2\alpha =e^{2i\alpha },b=\cos 2\beta +i\sin 2\beta =e^{2i\beta },c=\cos 2\gamma +i\sin 2\gamma =e^{2i\gamma },d=\cos 2\delta +i\sin 2\delta =e^{2i\delta }$

Now,$abcd=e^{2i\alpha }\times e^{2i\beta }\times e^{2i\gamma }\times e^{2i\delta }=e^{2i(\alpha +\beta +\gamma +\delta )}$

Now,$\sqrt{abcd}=\sqrt{e^{2i(\alpha +\beta +\gamma +\delta )}}$

$={\left(e^{2i(\alpha +\beta +\gamma +\delta )}\right)}^{\frac{1}{2}}$ since $\sqrt[m]{m{a}^n}=a^{\frac{n}{m}}$

$=e^{i(\alpha +\beta +\gamma +\delta )}$

We have $\sqrt{abcd}+\frac{1}{\sqrt{abcd}}=e^{i(\alpha +\beta +\gamma +\delta )}+\frac{1}{e^{i(\alpha +\beta +\gamma +\delta )}}$

$=e^{i(\alpha +\beta +\gamma +\delta )}+e^{-i(\alpha +\beta +\gamma +\delta )}$

$=\cos (\alpha +\beta +\gamma +\delta )+i\cos (\alpha +\beta +\gamma +\delta )+\cos (\alpha +\beta +\gamma +\delta )-i\cos (\alpha +\beta +\gamma +\delta )$

$=2\cos (\alpha +\beta +\gamma +\delta )$