First slide

Trigonometric equations

Question

If $\sin^{2} A = x$ then $\sin A \sin 2 A \sin 3 A \sin 4 A$ is a polynomial in x, the sum of whose coefficients is

Moderate

A

0
B

40
C

168
D

336

Solution

$W e h a v e$ $\sin A \sin 2 A \sin 3 A \sin 4 A$ $= \sin A (2 \sin A \cos A) (3 \sin A - 4 \sin^{3} A) \times 2 \sin 2 A \cos 2 A$
$= 2 \sin^{2} A \cos A \times \sin A (3 - 4 \sin^{2} A) \times 2 \times 2 \sin A \cos A (1 - 2 \sin^{2} A)$
$= 8 \sin^{4} A \cos^{2} A (3 - 4 \sin^{2} A) (1 - 2 \sin^{2} A)$
$= 8 x^{2} (1 - x) (3 - 4 x) (1 - 2 x)$
$= 24 x^{2} - 104 x^{3} + 144 x^{4} - 64 x^{5}$
$∴$ The sum of coefficients = 0

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