Discussion Forum



1)

If   $a=\cos 2\alpha+i\sin2\alpha,b=\cos 2\beta+i\sin2\beta,c=\cos2\gamma+i\sin2\gamma$   and   $d=\cos2\delta+i\sin2\delta$

 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:

Option B

Explanation:

 we have,

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

$\sqrt{abcd}=\cos(\alpha+\beta+\gamma+\delta)+i\sin(\alpha+\beta+\gamma+\delta)$  ..(i)

$\therefore$   $\frac{1}{\sqrt{abcd}}=\cos(\alpha+\beta+\gamma+\delta)-i\sin(\alpha+\beta+\gamma+\delta)$   ....(ii)

Adding (i) and (ii) , we obtain

$\sqrt{abcd}+\frac{1}{\sqrt{abcd}}=2\cos(\alpha+\beta+\gamma+\delta)$