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Question Number 35223 by abdo mathsup 649 cc last updated on 16/May/18

$$whatisthevalueofcos(i+j)with{i}^2=-1and$$$$j=e^{i\frac{2\pi }{3}}?$$

Answered by sma3l2996 last updated on 17/May/18

$$cos(i+j)=\frac{e^{(i+j)i}+e^{-(i+j)i}}{2}=\frac{e^{-1+ij}+e^{1-ij}}{2}$$$$ij=i\times e^{i\frac{2\pi }{3}}=i(\frac{1}{2}-i\frac{\sqrt{3}}{2})=\frac{\sqrt{3}}{2}+i\frac{1}{2}$$$$cos(i+j)=\frac{e^{-1+\frac{\sqrt{3}}{2}+i\frac{1}{2}}+e^{1-\frac{\sqrt{3}}{2}-i\frac{1}{2}}}{2}=\frac{e^{\frac{\sqrt{3}}{2}-1}\times e^{i/2}+\frac{e^{-i/2}}{e^{\sqrt{3}/2-1}}}{2}$$$$=\frac{1}{2e^{\sqrt{3}/2-1}}(e^{\sqrt{3}-2}(cos\frac{1}{2}+isin\frac{1}{2})+cos\frac{1}{2}-isin\frac{1}{2})$$$$\frac{1}{2e^{\sqrt{3}/2-1}}((e^{\sqrt{3}-2}+1)cos\left(1/2\right)+i(e^{\sqrt{3}-2}-1)sin\left(1/2\right))$$

Commented by abdo imad last updated on 17/May/18

$$sirSma3lyourmethodiscorrectbutyouhavecommited$$$$asmallerror{e}^{i\frac{2\pi }{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}!...$$

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