DE-MOVIRE’S THEOREM (REVISION)

\[1.\ Prove\ that:\ [\frac{cos⁡\ θ + i sin⁡\ θ} {sin⁡\ θ – i cos\ θ}]^4\ = 1\ \hspace{18cm}\]
\[\color {black}{Solution:}\ Question\ No.\ 8\ Solution\ \hspace{2cm}\]

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\[2:\ Simplify\ using\ DeMoivre’s\ theorem:\ \frac{(cos⁡\ 2θ – i sin⁡\ 2θ)^7\ (cos⁡\ 3θ + i sin⁡\ 3θ)^{-5}} {(cos⁡\ 4θ + i sin⁡\ 4θ)^2\ (cos⁡\ 5θ – i sin⁡\ 5θ)^{-6}}\ \hspace{10cm}\]
\[\color {black}{Solution:}\ \frac{(cos⁡\ 2θ – i sin⁡\ 2θ)^7\ (cos⁡\ 3θ + i sin⁡\ 3θ)^{-5}} {(cos⁡\ 4θ + i sin⁡\ 4θ)^2\ (cos⁡\ 5θ – i sin⁡\ 5θ)^{-6}}\ \hspace{18cm}\]
\[= \frac{(cos⁡\ θ + i sin⁡\ θ)^{7 \times -2}\ (cos⁡\ θ + i sin⁡\ θ)^{3 \times -5}} {(cos⁡\ θ + i sin⁡\ θ)^{4 \times 2}\ (cos⁡\ θ + i sin⁡\ θ)^{-5 \times -6 }}\ \hspace{8cm}\]
\[= \frac{(cos⁡\ θ + i sin⁡\ θ)^{-14}\ (cos⁡\ θ + i sin⁡\ θ)^{-15}} {(cos⁡\ θ + i sin⁡\ θ)^{8}\ (cos⁡\ θ + i sin⁡\ θ)^{30}}\ \hspace{8cm}\]
\[= (cos\ θ + i sin⁡\ θ )^{-\ 14\ -\ 15\ -\ 8\ -\ 30}\ \hspace{10cm}\]
\[= (cos\ θ + i sin⁡\ θ )^{-67}\ \hspace{10cm}\]
\[= cos\ 67\ θ\ -\ i sin⁡\ 67\ θ\ \hspace{10cm}\]