Question

If (cos theta + sin theta) = root2 cos theta, then prove that cos theta - sin theta = root2 sin theta

Answer 1

Let θ = x
cos x  + sin x =  √2 cos x
squaring on both side, we get......
cos2x + sin2x + 2cosxsinx = 2cos2x
2sinxcosx = 2cos2x - cos2x - sin2x
2sinxcosx = cos2x - sin2x
2sinxcosx = (cosx+sinx) (cosx - sinx)
2sinxcosx = (root2 cosx) (cosx - sinx)
2sinxcosx/root2 cosx = cosx - sinx
 √2 sinx = cosx - sinx

Answer 2

Question :-

→ If cos θ + sin θ = √2cos θ , then prove that cos θ - sin θ = √2sin θ .

Answer :-

We have,

→ cos θ + sin θ = √2cos θ .

[ Squaring both side, we get ] .

⇒ ( cos θ + sin θ )² = 2cos²θ .

⇒ cos²θ + sin²θ + 2cosθsinθ = 2cos² .

⇒ sin²θ + 2cosθsinθ = 2cos²θ - cos²θ .

⇒ sin²θ + 2cosθsinθ = cos²θ .

⇒ cos²θ - 2cosθsinθ = sin²θ .

[ Adding sin²θ both side, we get ] .

⇒ cos²θ - 2cosθsinθ + sin²θ = sin²θ + sin²θ .

⇒ ( cos θ - sin θ )² = 2sin²θ .

⇒ cos θ - sin θ = √( 2sin²θ ) .

∴ cos θ - sin θ = √2sin θ .

Hence, it is proved .