Prove that:

$\displaystyle

\left[ \frac{\cos (A)+i \sin (A)}{\sin(A)+i\cos(A)} \right]^4 = \cos(8A)+i\sin(8A)

$

Multiply top and bottom inside the bracket on the left hand side by the

complex conjugate of what is on the bottom:

$\displaystyle

\left[ \frac{\cos (A)+i \sin (A)}{\sin(A)+i\cos(A)} \right]^4 =$$\displaystyle \left[ \frac{(\cos (A)+i \sin (A))(\sin(A)-i\cos(A))}{(\sin(A)+i\cos(A))(\sin(A)-i\cos(A))} \right]^4$

$\displaystyle

=\left[ \frac{2\sin(A)\cos (A)-i (2 (\cos(A))^2-1)}{1} \right]^4

$

$\displaystyle

=\left[ \sin(2A)-i \cos(2A) \right]^4

$

multiply top and bottom inside the bracket by $\displaystyle i$:

$\displaystyle

=\left[ \frac{\cos(2A)+i \sin(2A)}{i} \right]^4

$

$\displaystyle

=\frac{(\cos(2A)+i \sin(2A))^4}{i^4}

$

Then DeMoivre's theorem gives:

$\displaystyle

=\frac{\cos(8A)+i \sin(8A)}{i^4}=\cos(8A)+i \sin(8A)

$

RonL