Given that $\displaystyle z = cos A + j sin A$ , show that
zē + z + 2 + (1/z) + (1/(zē)) = $\displaystyle 2 cos A ( 2 cos A + 1 )$
can someone help me prove?? urgent
Are you familiar with deMoivre's theorem? Then:
$\displaystyle z^2 = \text{cis} (2A) = \cos (2A) + j \sin (2A)$.
$\displaystyle \frac{1}{z^2} = z^{-2} = \text{cis} (-2A) = \cos (-2A) + j \sin (-2A) = cos (2A) - j \sin (2A)$.
$\displaystyle \frac{1}{z} = z^{-1} = \text{cis} (-A) = \cos (-A) + j \sin (-A) = cos A - j \sin A$.
So the left hand side of the identity is equal to $\displaystyle 2 \cos (2A) + 2 \cos A + 2$.
From here you should be able to substitute from the appropriate double angle formula and simplify to get $\displaystyle 2 \cos A ( 2 \cos A + 1 )$ .....