Question

If a = cos 2a + i sin 2a, b = cos 2b + i sin 2b,

c = cos 2g + i sin 2g and d = cos 2d + i sin 2d, then $\sqrt{abcd} + \frac{1}{\sqrt{abcd}}$=

• A. $\sqrt{2}$cos (a + b + g + d)
• B. 2 cos (a + b + g + d)
• C. cos (a + b + g + d)
• D. None of these

Answer 1

Answer: (B)

2 cos (a + b + g + d)

Answer 2

Sol. We have,

abcd = cos(2a + 2b + 2g + 2d) + i sin (2a + 2b + 2g + 2d)

\ $\sqrt{abcd}$

= [cos (2a + 2b + 2g + 2d) + i sin (2a + 2b + 2g + 2d)]^1/2

or $\sqrt{abcd}$= cos (a + b + g + d) + i sin (a + b + g + d)] … (1)

[DeMoiver’s Theorem]

\ $\frac{1}{\sqrt{abcd}}$ = cos (a + b + g + d) – i sin (a + b + g + d)… (2)

Adding (1) and (2), we obtain

$\sqrt{abcd}$+$\frac{1}{\sqrt{abcd}}$ = 2 cos (a + b + g + d)