Answer
$\dfrac{\sqrt{6} + 3 \sqrt{2}}{4}$
Work Step by Step
$\cos{\alpha} + \cos{\beta} = 2 \cos{\dfrac{\alpha+\beta}{2}} \cos{\dfrac{\alpha-\beta}{2}} $
$\cos{45} + \cos{15} = 2 \cos{30} \cos{15} = \dfrac{\sqrt{6} + 3 \sqrt{2}}{4}$
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