Euler's formula to show cost=(e^(it) +e^(-it))/2?

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hmm how do I show that cost=(e^(it) +e^(-it))/2 by the use of eulers formula? e^(it)=cost +isint ?? ty for help!

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1 Answer

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We have $e^{it} = \cos(t) + i \sin(t)$. Replacing $t$ by $-t$, we get $e^{-it} = \cos(t) - i \sin(t)$. Adding both, we get $$e^{it} + e^{-it} = 2 \cos(t) \implies \cos(t) = \dfrac{e^{it} + e^{-it}}2$$

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