Question

If $x+iy=\frac{1}{1+Cos\theta +iSin\theta }$, then show that $4x^2-1=0$

Answer 1

$x+iy=\frac{1}{1+cos\theta +isin\theta }$
$=\frac{1}{2co{s}^2\frac{\theta }{2}+i2sin\frac{\theta }{2}.cos\frac{\theta }{2}}$

$=\frac{1}{2cos\frac{\theta }{2}}\times \left(\frac{1}{cos\frac{\theta }{2}+isin\frac{\theta }{2}}\right)$

$=0.5sec\left(\frac{\theta }{2}\right).\frac{1}{e^{i\frac{\theta }{2}}}$

$=0.5sec\left(\frac{\theta }{2}\right)e^{-i\frac{\theta }{2}}$

$=0.5sec\left(\frac{\theta }{2}\right)(cos\frac{\theta }{2}-isin\frac{\theta }{2})$

$=0.5-i0.5tan\frac{\theta }{2}$
Hence
$x+iy=0.5-i0.5tan\frac{\theta }{2}$
Therefore $x=0.5=\frac{1}{2}$
Or
$2x=1$
Or
$4x^2=1$ or
$4x^2-1=0$.
Hence proved.