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Prove formula relating trig functions of real numbers to the complex exponential

  1. Prove that for \theta \in \mathbb{R} we have the following formulas,

        \[ \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2}, \qquad \sin  \theta = \frac{e^{i \theta} - e^{-i\theta}}{2i}. \]

  2. Using part (a) prove that

        \[ \cos^2 \theta = \frac{1}{2}(1 + \cos (2 \theta)), \qquad \sin^2 \theta = \frac{1}{2} (1 - \cos (2 \theta)). \]


  1. Proof. We compute, using the definition of the complex exponential, e^{i \theta} = \cos \theta + i \sin \theta:

        \begin{align*}  \cos \theta &= \frac{1}{2} (2 \cos \theta) \\  &= \frac{1}{2} \left( \cos \theta + i \sin \theta + \cos (\theta) - i \sin (\theta) ) \\  &= \frac{1}{2} \left( \cos \theta + i \sin \theta + \cos (-\theta) + i \sin (-\theta)) \\  &= \frac{e^{i \theta} + e^{-i \theta}}{2}.  \end{align*}

    (Where in the second to last line we used that cosine is an even function and sine is odd, i.e., \cos \theta = \cos (-\theta) and \sin \theta = -\sin \theta.)
    For the second formula we compute similarly,

        \begin{align*}  \sin \theta &= \frac{1}{2i} (2i \sin \theta) \\  &= \frac{1}{2i} \left( \cos \theta + i \sin \theta - \cos (-\theta) - i \sin (-\theta)) \\  &= \frac{e^{i \theta} - e^{-i \theta}}{2i}. \qquad \blacksquare \end{align*}

  2. Proof. We can compute these directly using the expressions we obtained in part (a),

        \begin{align*}  \cos^2 \theta &= \left( \frac{e^{i \theta} + e^{-i \theta}}{2} \right)^2 \\  &= \frac{e^{2i \theta} + e^{-2i \theta} + 2}{4} \\  &= \frac{ \cos (2 \theta) + 2}{2} \\  &= \frac{1}{2} (1 + \cos (2 \theta)). \\ \sin^2 \theta &= \left( \frac{e^{i \theta} - e^{-i \theta}}{2i} \right)^2 \\  &= \frac{e^{2 i \theta} + e^{-2i \theta} - 2}{-4} \\  &= \frac{2 - \cos (2 \theta)}{2} \\  &= \frac{1}{2} (1 - \cos (2 \theta)). \qquad \blacksquare \end{align*}

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