Question

if cos theta + sin theta = root2 cos Theta, show that cos theta - sin theta = root2 sin theta​

Answer 1

$\textsf{We\:have :-}$

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

$\textsf{Hence\:Shown.}$

Answer 2

Answer:

Given,

cosθ+sinθ=

2

cosθ

squaring on both the sides, we get,

cos

2

θ+sin

2

θ+2sinθcosθ=2cos

2

θ

cos

2

θ−sin

2

θ=2cosθsinθ

(cosθ+sinθ)(cosθ−sinθ)=2cosθsinθ

2

cosθ(cosθ−sinθ)=2cosθsinθ [ Given cosθ+sinθ=

2

cosθ]

∴cosθ−sinθ=

2

sinθ [henceproved