Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 128805 by bramlexs22 last updated on 10/Jan/21

$$\mathrm{Prove}\mathrm{that}{\mathrm{e}}^{\mathrm{i}\theta }+1={\mathrm{e}}^{-\frac{\mathrm{i}\theta }{2}}+{\mathrm{e}}^{\frac{\mathrm{i}\theta }{2}}?$$

Commented by mr W last updated on 10/Jan/21

$$LHS=(1+\cos \theta )+i\sin \theta$$$$RHS=2\cos \frac{\theta }{2}$$$$LHS\ne RHS!$$

Answered by benjo_mathlover last updated on 10/Jan/21

$${\mathrm{e}}^{i\theta }+1=\cos \theta +i\sin \theta +1$$$${\mathrm{e}}^{-\frac{i\theta }{2}}=\cos \frac{\theta }{2}-i\sin \frac{\theta }{2}$$$${\mathrm{e}}^{\frac{i\theta }{2}}=\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}$$$${\mathrm{e}}^{-\frac{i\theta }{2}}+{\mathrm{e}}^{\frac{i\theta }{2}}=2\cos \frac{\theta }{2}\ne 1+\cos \theta +i\sin \theta$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com