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If $\sin \theta+\cos \theta=\sqrt{2} \cos \theta$, then what is $(\cos \theta-\sin \theta)$ equal
to ?
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to ?
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Verified Answer
The correct answer is:
$\sqrt{2} \sin \theta$
$\sin \theta+\cos \theta=\sqrt{2} \cos \theta$
Let $\cos \theta-\sin \theta=\mathrm{P}$
$(1)^{2}+(2)^{2} \Rightarrow \sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta+\cos ^{2} \theta+\sin ^{2} \theta$
$-2 \sin \theta \cos \theta=2 \cos ^{2} \theta+\mathrm{p}^{2}$
$\Rightarrow 2=2 \cos ^{2} \theta+\mathrm{p}^{2}$
$\Rightarrow \mathrm{p}^{2}=2\left(1-\cos ^{2} \theta\right)=2 \sin ^{2} \theta$
$\Rightarrow p=\sqrt{2} \sin \theta$
Let $\cos \theta-\sin \theta=\mathrm{P}$
$(1)^{2}+(2)^{2} \Rightarrow \sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta+\cos ^{2} \theta+\sin ^{2} \theta$
$-2 \sin \theta \cos \theta=2 \cos ^{2} \theta+\mathrm{p}^{2}$
$\Rightarrow 2=2 \cos ^{2} \theta+\mathrm{p}^{2}$
$\Rightarrow \mathrm{p}^{2}=2\left(1-\cos ^{2} \theta\right)=2 \sin ^{2} \theta$
$\Rightarrow p=\sqrt{2} \sin \theta$
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