Question
If $\cos\theta+\sin\theta)=\sqrt{2}\sin\theta,$ show that $\sin\theta-\cos\theta=\sqrt{2}\cos\theta.$
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Answer
Given:$\cos\theta+\sin\theta=\sqrt{2}\sin\theta$
We have $(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2=2\big(\sin^2\theta+\cos^2\theta\big)$
$\Rightarrow\big(\sqrt{2}\sin\theta\big)^2+(\sin\theta-\cos\theta)^2=2$
$\Rightarrow2\sin^2\theta+(\sin\theta-\cos\theta)^2=2$
$\Rightarrow(\sin\theta-\cos\theta)^2=2-2\sin^2\theta$
$\Rightarrow(\sin\theta-\cos\theta)^2=2\big(1-\sin^2\theta\big)$
$\Rightarrow(\sin\theta-\cos\theta)^2=2\cos^2\theta$
$\Rightarrow(\sin\theta-\cos\theta)=\sqrt{2}\cos\theta$
Hence proved.
We have $(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2=2\big(\sin^2\theta+\cos^2\theta\big)$
$\Rightarrow\big(\sqrt{2}\sin\theta\big)^2+(\sin\theta-\cos\theta)^2=2$
$\Rightarrow2\sin^2\theta+(\sin\theta-\cos\theta)^2=2$
$\Rightarrow(\sin\theta-\cos\theta)^2=2-2\sin^2\theta$
$\Rightarrow(\sin\theta-\cos\theta)^2=2\big(1-\sin^2\theta\big)$
$\Rightarrow(\sin\theta-\cos\theta)^2=2\cos^2\theta$
$\Rightarrow(\sin\theta-\cos\theta)=\sqrt{2}\cos\theta$
Hence proved.
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