If $\sin \theta + \cos \theta = \sqrt{2} \cos \theta$, then $\cos \theta - \sin \theta =$
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There is a famous identity: if $a \cos\theta + b \sin\theta = c$, then $b \cos\theta - a \sin\theta = \pm \sqrt{a^2+b^2-c^2}$. Here, $1^2 + 1^2 - (\sqrt{2}\cos\theta)^2 = 2 - 2\cos^2\theta = 2\sin^2\theta$, giving $\sqrt{2}\sin\theta$.