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 Exercise 2, Question 4

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$\\cos\\theta = \\dfrac{e^{i\\theta} + e^{-i\\theta}}{2i}$

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$\\sin\\theta = \\dfrac{e^{i\\theta} - e^{-i\\theta}}{2i}$

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Euler’s formula turns to be very useful in evaluating integrals since products of exponentials are easier to integrate than products of trigonometric functions.
Obtain the following relations by solving the equations

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$e^{i\\theta} = \\cos{\\theta} +i\\sin{\\theta}$ and $e^{-i\\theta} = \\cos{\\theta} - i\\sin{\\theta}$  

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for $\\sin{\\theta}$ and $\\cos\\theta$

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