The expression cos 3 -sin 3 -2 sin 2 -3sin-cos- is positive for all - R inA- 2 n -3 -4- 2 n -4- n - IB- 2 n -2- 2 n -6- n - I- 2 n -3- 2 n -3- n - ID- 2 n -4- 2 n -3 -4- n - I

The correct options are A (2nπ 3π4,2nπ+π4), n∈ I B (2nπ π2,2nπ+π6), n∈ I(cos 3θ +sin 3 θ )+(2sin 2θ 3)(sinθ cosθ ) =4(cosθ sinθ)(cos ^2 θ +cosθsinθ +sin ^2 ...

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Byju's Answer

Standard XII

Mathematics

Cos(A+B)Cos(A-B)

The expressio...

Question

The expression $(cos 3 θ + sin 3 θ) + (2 sin 2 θ - 3) (sin θ - cos θ)$ is positive for all $θ \in R$ in

(2 n π - \frac{3 π}{4}, 2 n π + \frac{π}{4}), n \in I

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(2 n π - \frac{π}{2}, 2 n π + \frac{π}{6}), n \in I

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(2 n π - \frac{π}{3}, 2 n π + \frac{π}{3}), n \in I

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(2 n π - \frac{π}{4}, 2 n π + \frac{3 π}{4}), n \in I

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Solution

The correct options are
A $(2 n π - \frac{3 π}{4}, 2 n π + \frac{π}{4}), n \in I$
B $(2 n π - \frac{π}{2}, 2 n π + \frac{π}{6}), n \in I$
$(cos 3 θ + sin 3 θ) + (2 sin 2 θ - 3) (sin θ - cos θ)$
$= 4 (cos θ - sin θ) ({cos}^{2} θ + cos θ sin θ + {sin}^{2} θ - sin θ cos θ)$
$= - 4 \sqrt{2} sin (θ - \frac{π}{4}) > 0$
$\Rightarrow sin (θ - \frac{π}{4})$ is negative.
$\Rightarrow (2 n - 1) π < θ - \frac{π}{4} < 2 n π, n \in I$

Suggest Corrections

The expression (cos3θ+sin3θ)+(2sin2θ−3)(sinθ−cosθ) is positive for all θ∈R in

The correct options are A (2nπ−3π4,2nπ+π4), n∈I B (2nπ−π2,2nπ+π6), n∈I(cos3θ+sin3θ)+(2sin2θ−3)(sinθ−cosθ) =4(cosθ−sinθ)(cos2θ+cosθsinθ+sin2θ−sinθcosθ) =−4√2sin(θ−π4)>0 ⇒sin(θ−π4) is negative. ⇒(2n−1)π<θ−π4<2nπ, n∈I

The expression $(cos 3 θ + sin 3 θ) + (2 sin 2 θ - 3) (sin θ - cos θ)$ is positive for all $θ \in R$ in