Question

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

• A. $0$
• B. $40$
• C. $168$
• D. $336$

Answer 1

Answer: (A)

$0$

$\sin A\sin 2A\sin 3A\sin 4A$

$=\sin A\sin 2A\left(2\sin 2A\cos 2A\right)\sin 3A..........[\because \sin 2x=2\sin x\cos x]$

$=2\sin A\left(\sin 2A{)}^2\cos 2A\right[3\sin A-4{\sin }^{3}A]...............[\because \sin 3x=3\sin x-4{\sin }^{3}x]$

$=8{\sin }^{3}A{\cos }^{2}A[1-2{\sin }^{2}A][3\sin A-4{\sin }^{3}A]$

$=(8{\sin }^{3}A-16{\sin }^{5}A)(1-{\sin }^{2}A)(3\sin A-4{\sin }^{3}A)$

$=(8{\sin }^{3}A-16{\sin }^{5}A)(3\sin A-4{\sin }^{3}A-3{\sin }^{3}A+4{\sin }^{5}A)$

$=(24{\sin }^{4}A-32{\sin }^{6}A-24{\sin }^{6}A+32{\sin }^{8}A-48{\sin }^{6}A+64{\sin }^{8}A+48{\sin }^{8}A-64{\sin }^{10}A)$

$=(24{\sin }^{4}A-104{\sin }^{6}A+144{\sin }^{8}A-64{\sin }^{10}A)$

$=(24x^2-104x^3+144x^4-64x^5)$

Sum of coefficients $=24-104+144-64$

$=168-168=0$