Question

If we express $z=\frac{{(\cos 2\theta -i\sin 2\theta )}^4{(\cos 4\theta +i\sin 4\theta )}^{-5}}{{(\cos 3\theta +i\sin 3\theta )}^{-2}{(\cos 3\theta -i\sin 3\theta )}^{-9}}$ in the form of $x+iy$, we get

• A. $\cos 49\theta -i\sin 49\theta$
• B. $\cos 23\theta -i\sin 23\theta$
• C. $\cos 49\theta +i\sin 49\theta$
• D. $\cos 21\theta +i\sin 21\theta$

Answer 1

The correct option is A $\cos 49\theta -i\sin 49\theta$

The given complex numbers is $z=\frac{{(\cos 2\theta -i\sin 2\theta )}^4{(\cos 4\theta +i\sin 4\theta )}^{-5}}{{(\cos 3\theta +i\sin 3\theta )}^{-2}{(\cos 3\theta -i\sin 3\theta )}^{-9}}$

The given number may be written as $z=\frac{{\left[{(\cos \theta +i\sin \theta )}^{-2}\right]}^4{\left[{(\cos \theta +i\sin \theta )}^4\right]}^{-5}}{{\left[{(\cos \theta +i\sin \theta )}^3\right]}^{-2}{\left[{(\cos \theta +i\sin \theta )}^4\right]}^{-9}}=\frac{{(\cos \theta +i\sin \theta )}^{-8}{(\cos \theta +i\sin \theta )}^{-20}}{{(\cos \theta +i\sin \theta )}^{-6}{(\cos \theta +i\sin \theta )}^{27}}$

$={(\cos \theta +i\sin \theta )}^{-8-20+6-27}={(\cos \theta +i\sin \theta )}^{-49}=\cos 49\theta -i\sin 49\theta$