If the real part of the complex number 1 - cos θ + 2 i sin θ - 1 is 1 5 for θ ∈ ( 0 , π ) , then the value of the integral ∫ 0 θ sin x d x is equal to:

Question

If the real part of the complex number 1 - cos θ + 2 i sin θ - 1 is 1 5 for θ ∈ ( 0 , π ) , then the value of the integral ∫ 0 θ sin x d x is equal to:, 1, 2, - 1, 0

MARKS App · Accepted Answer

Let z = 1 1 - cos θ + 2 i sin θ ⇒ z = 1 - cos θ - 2 i sin θ 1 - cos θ + 2 i sin θ 1 - cos θ - 2 i sin θ ⇒ z = 2 sin 2 θ 2 - 2 i sin θ ( 1 - cos θ ) 2 + 4 sin 2 θ Put, sin θ = 2 sin θ 2 cos θ 2 , 1 - cos θ = 2 sin 2 θ 2 ⇒ z = 2 sin 2 θ 2 - 4 i sin θ 2 cos θ 2 4 sin 4 θ 2 + 16 sin 2 θ 2 cos 2 θ 2 Hence, R e z = 2 sin 2 θ 2 4 sin 4 θ 2 + 16 sin 2 θ 2 cos 2 θ 2 ⇒ 1 2 sin 2 θ 2 + 8 cos 2 θ 2 = 1 5 ⇒ 1 2 sin 2 θ 2 + 8 cos 2 θ 2 = 1 5 ⇒ sin 2 θ 2 + 4 cos 2 θ 2 = 5 2 ⇒ 1 - cos 2 θ 2 + 4 cos 2 θ 2 = 5 2 ⇒ 3 cos 2 θ 2 = 3 2 ⇒ cos 2 θ 2 = 1 2 ⇒ θ 2 = n π ± π 4 , n ∈ Z ⇒ θ = 2 n π ± π 2 For θ ∈ ( 0 , π ) , θ = π 2 . Hence, ∴ ∫ 0 θ sin θ d θ = ∫ 0 π 2 sin θ d θ = - cos θ 0 π 2 = - 0 - 1 = 1

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