Question

if cos theta + sin theta =√2 cos theta then prove that cos that - sin theta = √2 sin theta ​

Answer 1

Given:-

• $\cos \theta + \sin\theta = \sqrt{2} \cos\theta$

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Need to prove:-

• $\cos \theta - \sin\theta = \sqrt{2} \sin \theta$

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Solution:-

$\cos\theta + \sin \theta = \sqrt{2} \cos\theta$

Squaring on both sides we get,

$\implies {( \cos \theta + \sin \theta ) }^{2} = {( \sqrt{2} \cos \theta ) }^{2}$

$\implies { { \cos }^{2} \theta } + { \sin }^{2} \theta + 2 \sin \theta \cos\theta = 2 { \cos }^{2} \theta$

$\implies 2 { \cos }^{2} \theta - {cos}^{2} \theta - {sin}^{2} \theta = 2sin \theta cos \theta$

$\implies {cos}^{2} \theta - {sin}^{2} \theta = 2sin \theta \: cos \theta$

$\implies (cos \theta + sin \theta)(cos \theta - sin \theta) = 2sin \theta \: cos \theta$

We know,

$\cos \theta + \sin\theta = \sqrt{2} \cos\theta$

Therefore,

$\implies ( \sqrt{2} \cos\theta)(cos \theta - sin \theta) = 2sin \theta \: cos \theta$

$\implies (cos \theta - sin \theta) = 2sin \theta \: cos \theta \div \sqrt{2}cos \theta$

$\red{\boxed{\bold{\large{\implies cos \theta - sin \theta = \sqrt{2}sin \theta}}}}$ (Hence proved!)

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Knowledge cell:-

Trigonometric Identities:-

• $\sin^2\theta+\cos^2\theta=1$

• $\sin^2\theta= 1-\cos^2\theta$

• $\cos^2\theta=1-\sin^2\theta$

• $1+\cot^2\theta={cosec}^2 \theta$

• ${cosec}^2 \theta-\cot^2\theta =1$

• ${cosec}^2 \, \theta= 1+\cot^2\theta$

• $\sec^2\theta=1+\tan^2\theta$

• $\sec^2\theta-\tan^2\theta=1$

• $\tan^2\theta=\sec^2\theta-1$

Answer 2

Corrected Question :-

$\sf \: if \: cosθ + sinθ = \sqrt{2} cosθ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\\sf then \: prove \: that \: \sf \: cosθ - sinθ = \sqrt{2} sinθ$

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Given :-

• $\sf{cosθ + sinθ = \sqrt{2}cosθ }$

To prove :-

• $\sf \: cosθ - sinθ = \sqrt{2} sinθ$

Solution :-

$\sf{we \: know \: that} \\ {\sf \: cosθ + sinθ = \sqrt{2} cosθ}$

$\sf \: squaring \: both \: sides$

$\sf \: {(cosθ + sinθ)}^{2} = { (\sqrt{2}cosθ) }^{2}$

$\sf \: {cos}^{2} θ + {sin}^{2}θ + 2sinθcosθ = 2 {cos}^{2} θ$

$\sf2 {cos}^{2}θ - {cos}^{2} θ - {sin}^{2} θ = 2sinθcosθ$

$\sf {cos}^{2} θ - {sin}^{2} θ = 2sinθcosθ$

$\sf(cosθ + sinθ)(cosθ - sinθ) = 2sinθcosθ$

$\sf \: since \: cosθ + sinθ = \sqrt{2} cosθ$

$\sf(\sqrt{2} cosθ ) (cosθ - sinθ) = 2cosθsinθ$

$\sf \: cosθ - sinθ = \frac{2cosθsinθ}{ \sqrt{2}cosθ }$

$\sf {\red{ \star{cosθ - sinθ = \sqrt{2}sin θ \star}}}$

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${\red{ \sf{ hence \: proved}}}$

${ \rm{ \color{cyan}{❤||ur \: shona||❤}}}$