cos3θ⋅sin4θ=1k(cos7θ−cos5θ−3cosθ+psinθ), then k and p are

x=cosθ+isinθ⇒1x=cosθ−isinθ2cosθ=x+1x⇒2isinθ=x−1x⇒(x+1x)3(x−1x)4=(23cos3θ)(24(−1)4sin4θ)⇒27cos3θ.sin4θ=((x+1x)(x−1x))3(x−1x)⇒27cos3θsin4θ=(x2−1x2)3(x−1x)⇒27cos3θsin4θ=(x6−1x6−3.x2.1x2(x2−1x2))(x−1x)⇒27cos3θsin4θ=(x6(x−1x)−1x6(x−1x)−3x2(x−1x)+3x2(x−1x))⇒27.cos3θsin4θ=x7−x5−1x5+1x7−3x3+3x+3x−3x3⇒27.cos3θsin4θ=(x7+1x7)−(x5+1x5)−3(x3+1x3)+3(x+1x)⇒27.cos3θsin4θ=2cos7θ−2cos5θ−6cos3θ+6cosθ⇒cos3θsin4θ=164(cos7θ−cos5θ−3cos3θ+3cosθ)

cos3θ⋅sin4θ=1k(cos7θ−cos5θ−3cosθ+psinθ), then k and p are

x=cosθ+isinθ⇒1x=cosθ−isinθ2cosθ=x+1x⇒2isinθ=x−1x⇒(x+1x)3(x−1x)4=(23cos3θ)(24(−1)4sin4θ)⇒27cos3θ.sin4θ=((x+1x)(x−1x))3(x−1x)⇒27cos3θsin4θ=(x2−1x2)3(x−1x)⇒27cos3θsin4θ=(x6−1x6−3.x2.1x2(x2−1x2))(x−1x)⇒27cos3θsin4θ=(x6(x−1x)−1x6(x−1x)−3x2(x−1x)+3x2(x−1x))⇒27.cos3θsin4θ=x7−x5−1x5+1x7−3x3+3x+3x−3x3⇒27.cos3θsin4θ=(x7+1x7)−(x5+1x5)−3(x3+1x3)+3(x+1x)⇒27.cos3θsin4θ=2cos7θ−2cos5θ−6cos3θ+6cosθ⇒cos3θsin4θ=164(cos7θ−cos5θ−3cos3θ+3cosθ)

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$\cos^{3} θ \cdot \sin^{4} θ = \frac{1}{k} (\cos 7 θ - \cos 5 θ - 3 \cos θ + p \sin θ)$ , then k and p are

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$k = 62$

$k = 64$

$p = \frac{1}{3}$

$p = 3$

answer is B, C.

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Detailed Solution

$x = \cos θ + i \sin θ \Rightarrow \frac{1}{x} = \cos θ - i \sin θ$

$2 \cos θ = x + \frac{1}{x} \Rightarrow 2 i \sin θ = x - \frac{1}{x}$

$\Rightarrow {(x + \frac{1}{x})}^{3} {(x - \frac{1}{x})}^{4} = (2^{3} \cos^{3} θ) (2^{4} {(- 1)}^{4} \sin^{4} θ)$

$\Rightarrow 2^{7} \cos^{3} θ . \sin^{4} θ = {((x + \frac{1}{x}) (x - \frac{1}{x}))}^{3} (x - \frac{1}{x})$

$\Rightarrow 2^{7} \cos^{3} θ \sin^{4} θ = {(x^{2} - \frac{1}{x^{2}})}^{3} (x - \frac{1}{x})$

$\Rightarrow 2^{7} \cos^{3} θ \sin^{4} θ = (x^{6} - \frac{1}{x^{6}} - 3. x^{2} . \frac{1}{x^{2}} (x^{2} - \frac{1}{x^{2}})) (x - \frac{1}{x})$

$\Rightarrow 2^{7} \cos^{3} θ \sin^{4} θ = (x^{6} (x - \frac{1}{x}) - \frac{1}{x^{6}} (x - \frac{1}{x}) - 3 x^{2} (x - \frac{1}{x}) + \frac{3}{x^{2}} (x - \frac{1}{x}))$

$\Rightarrow 2^{7} . \cos^{3} θ \sin^{4} θ = x^{7} - x^{5} - \frac{1}{x^{5}} + \frac{1}{x^{7}} - 3 x^{3} + 3 x + \frac{3}{x} - \frac{3}{x^{3}}$

$\Rightarrow 2^{7} . \cos^{3} θ \sin^{4} θ = (x^{7} + \frac{1}{x^{7}}) - (x^{5} + \frac{1}{x^{5}}) - 3 (x^{3} + \frac{1}{x^{3}}) + 3 (x + \frac{1}{x})$