If \tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha} where \theta and \alpha (\alpha\neq\frac{\pi}{4}) are acute angles, then what is \sqrt{2}\sin \theta equal to?

Explanation

Construct a right triangle where perpendicular is \sin \alpha - \cos \alpha and base is \sin \alpha + \cos \alpha. The hypotenuse is \sqrt{(\sin \alpha - \cos \alpha)^2 + (\sin \alpha + \cos \alpha)^2} = \sqrt{2}. Thus, \sin \theta = \frac{\sin \alpha - \cos \alpha}{\sqrt{2}}, so \sqrt{2} \sin \theta = \sin \alpha - \cos \alpha.

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