Question

$ {{(\sin \theta +i\cos \theta )}^{n}} $ is equal to

• A. $ {{i}^{n}}(i\sin \theta +\cos \theta ) $
• B. $ (\sin n\theta +i\cos n\theta ) $
• C. $ \cos n\left( \frac{\pi }{2}-\theta \right)+i\sin n\left( \frac{\pi }{2}-\theta \right) $
• D. $ \cos \theta +i\sin n\theta $

Answer 1

Answer: (C)

$ (\sin \theta +i\cos \theta )hn $
$ ={{\left[ \cos \left( \frac{\pi }{2}-\theta \right)+i\sin \left( \frac{\pi }{2}-\theta \right) \right]}^{n}} $
$ =\cos n\left( \frac{\pi }{2}-\theta \right)+i\sin n\left( \frac{\pi }{2}-\theta \right) $