The solution of 3sin- - sin 3-1 - cos- - 3cos- - cos 3-1 - sin- - 4-2cos - - -4

The correct option is B nπ sin3θ = #160; 3sinθ 4sin^3θ cos3θ = 4cos^3θ 3cosθ 4sin^3θ1+cosθ + 4cos^3θ1 sinθ 4sinθ ( 1 cos^2θ)1+cosθ ...

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Byju's Answer

Standard XII

Mathematics

Sin(A+B)Sin(A-B)

The solution ...

Question

The solution of
$\frac{3 sin θ - sin 3 θ}{1 + cos θ} + \frac{3 cos θ + cos 3 θ}{1 - sin θ} = 4 \sqrt{2} cos (θ + \frac{π}{4})$

n π

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n π + \frac{π}{12}

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n π \pm \frac{π}{2}

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2 n π

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Solution

The correct option is B $n π$
$sin 3 θ = 3 sin θ - 4 {sin}^{3} θ$
$cos 3 θ = 4 {cos}^{3} θ - 3 cos θ$
$⟹ \frac{4 {sin}^{3} θ}{1 + cos θ} + \frac{4 {cos}^{3} θ}{1 - sin θ}$
$⟹ \frac{4 sin θ (1 - {cos}^{2} θ)}{1 + cos θ} + \frac{4 cos θ (1 - {sin}^{2} θ)}{1 - sin θ}$
$⟹ 4 sin θ (1 - cos θ) + 4 cos θ (1 + sin θ)$
$⟹ 4 (sin θ + cos θ)$
$⟹ 4 \sqrt{2} (cos (θ - \frac{Π}{4}))$
Equating with RHS, we get
$θ = n π$

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The solution of 3sinθ−sin3θ1+cosθ+3cosθ+cos3θ1−sinθ=4√2cos(θ+π4)

The correct option is B nπsin3θ=3sinθ−4sin3θcos3θ=4cos3θ−3cosθ⟹4sin3θ1+cosθ+4cos3θ1−sinθ⟹4sinθ(1−cos2θ)1+cosθ+4cosθ(1−sin2θ)1−sinθ ⟹4sinθ(1−cosθ)+4cosθ(1+sinθ)⟹4(sinθ+cosθ)⟹4√2(cos(θ−Π4))Equating with RHS, we get θ=nπ

The solution of
$\frac{3 sin θ - sin 3 θ}{1 + cos θ} + \frac{3 cos θ + cos 3 θ}{1 - sin θ} = 4 \sqrt{2} cos (θ + \frac{π}{4})$