The expression [(1+sin (π/8)+i cos (π/8)/1+ sin (π)8-i cos (π)8]/8) =

Question

The expression [(1+sin (π/8)+i cos (π/8)/1+ sin (π)8-i cos (π)8]/8) = (A) 1 (B) -1 (C) i (D) -i

Tardigrade Admin · Accepted Answer

Let sin(π/8) +i cos (π/8)=z ⇒ [(1+sin (π/8)+i cos (π/8)/1+ sin (π)8-i cos (π)8]/8) =((1+z/1+(1)z))8 =z8 =(sin (π/8) +i cos (π/8))8 =(cos ((π/2)-(π/8))+i sin ((π/2)-(π/8)))8 =(cos (3π/8) +i sin (3π/8))8 = cos 3π =-1

Q. The expression $[\frac{1 + s in \frac{π}{8} + i cos \frac{π}{8}}{1 + s in \frac{π}{8} - i cos \frac{π}{8}}]^{^{^{8}}} =$

$1$

$- 1$

$i$

$- i$

Q. The expression [1+sin8π​−icos8π​1+sin8π​+icos8π​​]8=

1

−1

i

−i

Let sin8π​+icos8π​=z ⇒[1+sin8π​−icos8π​1+sin8π​+icos8π​​]8 =(1+z1​1+z​)8 =z8 =(sin8π​+icos8π​)8 =(cos(2π​−8π​)+isin(2π​−8π​))8 =(cos83π​+isin83π​)8 =cos3π=−1

Q. The expression $[\frac{1 + s in \frac{π}{8} + i cos \frac{π}{8}}{1 + s in \frac{π}{8} - i cos \frac{π}{8}}]^{^{^{8}}} =$

$1$

$- 1$

$i$

$- i$