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Question Number 151559 by peter frank last updated on 21/Aug/21

$$\mathrm{prove}\mathrm{that}$$$${(1+\cos \theta +\mathrm{isin}\theta )}^{\mathrm{n}}$$$$+{(1+\cos \theta -\mathrm{isin}\theta )}^{\mathrm{n}}=2^{\mathrm{n}+1}\cos \frac{\theta }{2}\cos \frac{\mathrm{n}\theta }{2}$$

Answered by Olaf_Thorendsen last updated on 22/Aug/21

$${\mathrm{S}}_n={(1+\cos \theta +i\sin \theta )}^n+{(1+\cos \theta -i\sin \theta )}^n$$$${\mathrm{S}}_n={(1+e^{i\theta })}^n+{(1+e^{-i\theta })}^n$$$${\mathrm{S}}_n={(1+e^{i\theta })}^n+\mathrm{conj}\left({(1+e^{i\theta })}^n\right)$$$${\mathrm{S}}_n=2\mathrm{Re}{(1+e^{i\theta })}^n$$$${\mathrm{S}}_n=2\mathrm{Re}\left(e^{i\frac{n\theta }{2}}{(e^{i\frac{\theta }{2}}+e^{-i\frac{\theta }{2}})}^n\right)$$$${\mathrm{S}}_n=2\mathrm{Re}\left(e^{i\frac{n\theta }{2}}{2}^n{\cos }^n\left(\frac{\theta }{2}\right)\right)$$$${\mathrm{S}}_n=2^{n+1}\cos \left(\frac{n\theta }{2}\right){\cos }^n\left(\frac{\theta }{2}\right)$$

Commented by peter frank last updated on 22/Aug/21

$$\mathrm{thank}\mathrm{you}$$

Answered by peter frank last updated on 22/Aug/21

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