Question

If
$\cos( \alpha ) - \sin( \alpha ) = \sqrt{2} \sin( \alpha )$
then ptove that
$\cos( \alpha ) + \sin( \alpha ) = \sqrt{2} \cos( \alpha )$

Plz help with explained answer​

Answer 1

Answer:

cosa - sina = $\sqrt{2}sina$  -(1)

Squaring both sides

$(cosa - sina)^2 = 2sin^2a\\cos^2a + sin^2a - 2sinacosa = 2sin^2a\\1 - 2sinacosa = 2sin^2a\\1 - 2sin^2a = 2sinacosa\\1 - sin^2a - sin^2a = 2sinacosa\\cos^2a - sin^2a = 2sinacosa\\(cosa - sina)(cosa + sina) = 2sinacosa - Using (a-b)(a+b) = a^2 + b^2\\\\Now, from (1)\\\sqrt{2}sina (cosa + sina) = {2sinacosa} \\cosa + sina = \frac{2sinacosa}{\sqrt{2}sina }\\ cosa + sina = \sqrt{2}cosa$

Answer 2

Step-by-step explanation:

Let alpha = θ for a moment .

cos θ - sin θ = √2 sin θ

(cos θ)^2 + (sin θ)^2 -2sinθcosθ = 2(sinθ)^2

1 - 2sinθcosθ = 2(sinθ)^2

1 - 2(sinθ)^2 = 2sinθcosθ

(cosθ)^2 - (sinθ)^2 = 2sinθcosθ

(cosθ + sinθ)(cosθ - sinθ) = 2sinθcosθ

(cosθ + sinθ)√2sin θ = 2sinθcosθ

cosθ + sinθ = √2cosθ