Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.

1.	Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
Answer» Given: sin θ +2 cos θ = 1 Squaring on both sides, (sin θ +2 cos θ)² = 1 ⇒ sin² θ + 4 cos² θ + 4sin θcos θ = 1 Since, sin² θ = 1 – cos² θ and cos² θ = 1 – sin² θ ⇒ (1 – cos² θ) + 4(1 – sin² θ) + 4sin θcos θ = 1 ⇒ 1 – cos² θ + 4 – 4 sin² θ + 4sin θcos θ = 1 ⇒ – 4 sin² θ – cos² θ + 4sin θcos θ = – 4 ⇒ 4 sin² θ + cos² θ – 4sin θcos θ = 4 We know that, a²+ b² – 2ab = ( a – b)² So, we get, (2sin θ – cos θ)² = 4 ⇒ 2sin θ – cos θ = 2 Hence proved.

Answer»

Given: sin θ +2 cos θ = 1

Squaring on both sides,

(sin θ +2 cos θ)² = 1

⇒ sin² θ + 4 cos² θ + 4sin θcos θ = 1

Since, sin² θ = 1 – cos² θ and cos² θ = 1 – sin² θ

⇒ (1 – cos² θ) + 4(1 – sin² θ) + 4sin θcos θ = 1

⇒ 1 – cos² θ + 4 – 4 sin² θ + 4sin θcos θ = 1

⇒ – 4 sin² θ – cos² θ + 4sin θcos θ = – 4

⇒ 4 sin² θ + cos² θ – 4sin θcos θ = 4

We know that,

a²+ b² – 2ab = ( a – b)²

So, we get,

(2sin θ – cos θ)² = 4

⇒ 2sin θ – cos θ = 2

Hence proved.

Discussion