Question

$\frac{(\cos 2\theta -i\sin 2\theta {)}^4(\cos 4\theta +i\sin 4\theta {)}^{-5}}{(\cos 3\theta +i\sin 3\theta {)}^{-2}(\cos 3\theta -i\sin \theta {)}^{-9}}=\cos 7k\theta -i\sin 7k\theta$. Find the value of $k$.

Answer 1

$\frac{(\cos 2\theta -i\sin 2\theta {)}^4(\cos 4\theta +i\sin 4\theta {)}^{-5}}{(\cos 3\theta +i\sin 3\theta {)}^{-2}(\cos 3\theta -i\sin \theta {)}^{-9}}$

$=\frac{{\left[{(\cos \theta +i\sin \theta )}^{-2}\right]}^4{\left[\right(\cos \theta +i\sin \theta {)}^4]}^{-5}}{{\left[\right(\cos \theta +i\sin \theta {)}^3]}^{-2}{\left[\right(\cos \theta +i\sin \theta {)}^{-3}]}^{-9}}$

$=\frac{(\cos \theta +i\sin \theta {)}^{-8}(\cos \theta +i\sin \theta {)}^{-20}}{(\cos \theta +i\sin \theta {)}^{-6}(\cos \theta +i\sin \theta {)}^{27}}$

$=(\cos \theta +i\sin \theta {)}^{-8-20+6-27}$

$=(\cos \theta +i\sin \theta {)}^{-49}$

$=\cos 49\theta -i\sin 49\theta$

$\therefore ,7k=49\Rightarrow k=7$