If $\cos 5 θ = a \cos θ + b \cos^{3} θ + c \cos^{5} θ + d$ , then

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b = - 20

a = 20

d = 5

c = 16

answer is B, C.

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Detailed Solution

Given

\cos 5 θ = a \cos θ + b \cos^{3} θ + c \cos^{5} θ + d

$\Rightarrow \cos 5 θ = \cos [3 θ + 2 θ]$

$= \cos 3 θ . \cos 2 θ - \sin 3 θ . \sin 2 θ$

$= [4 \cos^{3} θ - 3 \cos θ] [2 \cos^{2} θ - 1] - [3 \sin θ - 4 \sin^{3} θ] 2 \sin θ . \cos θ$

$= 8 \cos^{5} θ - 6 \cos^{3} θ - 4 \cos^{3} θ + 3 \cos θ - 2 \sin^{2} θ [3 - 4 [1 - \cos^{2} θ]] \cos θ$

$= 8 \cos^{5} θ - 10 \cos^{3} θ + 3 \cos θ - 2 [[1 - \cos^{2} θ] [3 \cos θ - 4 \cos θ + 4 \cos^{3} θ]]$

$= 8 \cos^{5} θ - 10 \cos^{3} θ + 3 \cos θ - 2 [[1 - \cos^{2} θ] [4 \cos^{3} θ - \cos θ]]$

$= 8 \cos^{5} θ - 10 \cos^{3} θ + 3 \cos θ - 8 \cos^{3} θ + 2 \cos θ + 3 \cos^{5} θ - 2 \cos^{3} θ$

$= 16 \cos^{5} θ - 20 \cos^{3} θ + 5 \cos θ$

$= 5 \cos θ - 20 \cos^{3} θ + 16 \cos^{5} θ$

$∴ a = 5, b = - 20, c = 16, d = 0$

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