If cos - - sin - - -2 cos - prove that cos - - sin - - -2 sin - - Maths Q - A

Prove the given expression:Given, cos(θ)+sin(θ)=√(2)cos(θ)0ex0ex⇒(cos(θ)+sin(θ))^2=(√(2)cos(θ))^2[Squaringonbothside]0ex0ex⇒cos^2(θ)+2cos(θ)sin(θ)+sin^2(θ)=2cos ...

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Standard X

Mathematics

Relation between Trigonometric Ratios

If cos θ + si...

Question

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

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Solution

Prove the given expression:
Given,
$\cos (θ) + \sin (θ) = \sqrt{2} \cos (θ) \Rightarrow {(\cos (θ) + \sin (θ))}^{2} = {(\sqrt{2} \cos (θ))}^{2} [Squaring on both side] \Rightarrow \cos^{2} (θ) + 2 \cos (θ) \sin (θ) + \sin^{2} (θ) = 2 \cos^{2} (θ) [∵ {(a + b)}^{2} = a^{2} + 2 a b + b^{2}] \Rightarrow 2 \cos (θ) \sin (θ) = 2 \cos^{2} (θ) - \sin^{2} (θ) - \cos^{2} (θ) \Rightarrow 2 \cos (θ) \sin (θ) = \cos^{2} (θ) - \sin^{2} (θ) \Rightarrow 2 \cos (θ) \sin (θ) = (\cos (θ) - \sin (θ)) (\cos (θ) + \sin (θ)) [∵ (a + b) (a - b) = a^{2} - b^{2}] \Rightarrow 2 \cos (θ) \sin (θ) = (\cos (θ) - \sin (θ)) \sqrt{2} \cos (θ) [∵ \cos (θ) + \sin (θ) = \sqrt{2} \cos (θ)] \Rightarrow \sqrt{2} \sin (θ) = (\cos (θ) - \sin (θ)) ∴ \cos (θ) - \sin (θ) = \sqrt{2} \sin (θ)$
Hence proved, $\cos (θ) - \sin (θ) = \sqrt{2} \sin (θ)$ .

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If cosθ+sinθ=2cosθ prove that cosθ-sinθ=2sinθ

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