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Complex Numbers

If cosθ+sinθ=...

Question

If $\cos (θ) + \sin (θ) = \sqrt{2} \cos (θ)$ , then prove that $\sin (θ) - \cos (θ) = \sqrt{2} \sin (θ)$

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Solution

Prove the given expression
Given,
$\cos (θ) + \sin (θ) = \sqrt{2} \cos (θ) \Rightarrow {(\cos (θ) + \sin (θ))}^{2} = {(\sqrt{2} \cos (θ))}^{2} \Rightarrow \cos^{2} (θ) + 2 \cos (θ) \sin (θ) + \sin^{2} (θ) = 2 \cos^{2} (θ) [∵ {(a + b)}^{2} = a^{2} + 2 a b + b^{2}] \Rightarrow 2 \cos (θ) \sin (θ) + \sin^{2} (θ) = 2 \cos^{2} (θ) - \cos^{2} (θ) \Rightarrow 2 \cos (θ) \sin (θ) + \sin^{2} (θ) = \cos^{2} (θ) \Rightarrow \sin^{2} (θ) = \cos^{2} (θ) - 2 \cos (θ) \sin (θ) \Rightarrow \sin^{2} (θ) + \sin^{2} (θ) = \cos^{2} (θ) + \sin^{2} (θ) - 2 \cos (θ) \sin (θ) [adding \sin^{2} (θ) on both side] \Rightarrow 2 \sin^{2} (θ) = {(\sin (θ) - \cos (θ))}^{2} [∵ a^{2} - 2 ab + b^{2} = {(a - b)}^{2}] \Rightarrow \sqrt{2 \sin^{2} (θ)} = \sqrt{{(\sin (θ) - \cos (θ))}^{2}} [Taking root on both side] \Rightarrow \sqrt{2} \sin (θ) = (\sin (θ) - \cos (θ))$
Hence Proved, $\sin (θ) - \cos (θ) = \sqrt{2} \sin (θ)$

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