The root of equation 1 – cos θ = sin θ.sin \(\frac{θ}{2}\) is (a

1.	The root of equation 1 – cos θ = sin θ.sin \(\frac{θ}{2}\) is (a) k π, k∈I (b) 2kπ, k∈I (c) k. \(\frac{π}{2}\), k ∈ I(d) None of these
Answer» Answer : (b) 2kπ, k∈I 1 – cos θ = sin θ . sin θ/2 ⇒ 2 sin² θ/2 = 2 sin θ/2 . cos θ/2 . sin θ/2 ⇒ 2 sin² θ/2 (1 – cos θ/2) = 0 ⇒ 2 sin² θ/2 = 0 or 1 – cos θ/2 = 0 ⇒ sin θ/2 = 0 or 2 sin² θ/4 = 0 (Using, 1 – cos 2θ = 2 sin² θ) ⇒ \(\frac{θ}{2}\) = kπ or \(\frac{θ}{2}\) = kπ , where k∈I. ⇒ θ = 2kπ or θ = 4kπ, k∈I. ⇒ θ = 2kπ, k∈I

The root of equation 1 – cos θ = sin θ.sin \(\frac{θ}{2}\) is (a) k π, k∈I (b) 2kπ, k∈I (c) k. \(\frac{π}{2}\), k ∈ I(d) None of these

Answer»

Answer : (b) 2kπ, k∈I

1 – cos θ = sin θ . sin θ/2

⇒ 2 sin² θ/2 = 2 sin θ/2 . cos θ/2 . sin θ/2

⇒ 2 sin² θ/2 (1 – cos θ/2) = 0

⇒ 2 sin² θ/2 = 0 or 1 – cos θ/2 = 0 ⇒ sin θ/2 = 0 or 2 sin² θ/4 = 0

(Using, 1 – cos 2θ = 2 sin² θ)

⇒ \(\frac{θ}{2}\) = kπ or \(\frac{θ}{2}\) = kπ , where k∈I.

⇒ θ = 2kπ or θ = 4kπ, k∈I.

⇒ θ = 2kπ, k∈I