If $s i n^{2} A = x$ , then $sin A sin 2 A sin 3 A sin 4 A = a x^{2} + b x^{3} + c x^{4} + d x^{5}$ , where $12 a - 10 b + 8 c - 11 d$ is equal to

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Solution

Given ${sin}^{2} A = x$ ,

Also given $sin A sin 2 A sin 3 A sin 4 A = a x^{2} + b x^{3} + c x^{4} + d x^{5}$ .....(1)
Consider $sin A sin 2 A sin 3 A sin 4 A$

$= 2 {sin}^{2} A cos A (3 sin A - 4 {sin}^{3} A) (4 sin A cos A cos 2 A)$

$= 8 {sin}^{4} A {cos}^{2} A (3 - 4 {sin}^{2} A) (1 - 2 {sin}^{2} A)$

$= 8 {sin}^{4} A (1 - {sin}^{2} A) (3 - 4 {sin}^{2} A) (1 - 2 {sin}^{2} A)$

$= 8 x^{2} (1 - x) (3 - 4 x) (1 - 2 x)$

$= 24 x^{2} - 104 x^{3} + 144 x^{4} - 64 x^{5}$
Put this value in (1), we get

$24 x^{2} - 104 x^{3} + 144 x^{4} - 64 x^{5} = a x^{2} + b x^{3} + c x^{4} + d x^{5}$

On comparing we get

$a = 24, b = - 104, c = 144, d = - 64$

So, $12 a - 10 b + 8 c - 11 d = (12 \times 24) + (10 \times 104) + (8 \times 144) + (11 \times 64) = 3184$

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