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If $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$, then prove that $\cos \theta – \sin \theta = \sqrt{2} \sin \theta$.
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Solution:
$\cos \theta + \sin \theta = \sqrt{2} \cos \theta$
$\sin \theta = (\sqrt{2} – 1) \cos \theta$
$= \frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1} \cos \theta$ (1.5 Mark)
$\Rightarrow (\sqrt{2} + 1) \sin \theta = \cos \theta$ (1 Mark)
$\Rightarrow \sqrt{2} \sin \theta = \cos \theta – \sin \theta$ (0.5 Mark)
$\cos \theta + \sin \theta = \sqrt{2} \cos \theta$
$\sin \theta = (\sqrt{2} – 1) \cos \theta$
$= \frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1} \cos \theta$ (1.5 Mark)
$\Rightarrow (\sqrt{2} + 1) \sin \theta = \cos \theta$ (1 Mark)
$\Rightarrow \sqrt{2} \sin \theta = \cos \theta – \sin \theta$ (0.5 Mark)