If $f(x)=\sin x \cdot \sin 2 x \cdot \sin 3 x$ and $f^{\prime \prime}(x)=a(\sin b x)+c(\sin d x)+e(\sin k…
If $f(x)=\sin x \cdot \sin 2 x \cdot \sin 3 x$ and $f^{\prime \prime}(x)=a(\sin b x)+c(\sin d x)+e(\sin k x)$, then the value of $(a+c+e)-(b+d+k)$ equals
8
-8
16
12
Solution
$
\text { } \begin{aligned}
f(x) & =\sin x \sin 2 x \cdot \sin 3 x \\
& =\frac{(2 \sin x \sin 2 x) \sin 3 x}{2} \\
& =\frac{1}{2}\{\cos x-\cos 3 x\} \sin 3 x \\
& =\frac{1}{2}[\cos x \sin 3 x-\cos 3 x \sin 3 x] \\
& =\frac{2}{4} \cos x \sin 3 x-\frac{2}{4} \cos 3 x \sin 3 x \\
& =\frac{1}{4}[\sin 4 x+\sin 2 x]-\frac{1}{4} \sin 6 x \\
f^{\prime}(x) & =\cos 4 x+\frac{1}{2} \cos 2 x-\frac{3}{2} \cos 6 x \\
f^{\prime \prime}(x) & =-4 \sin 4 x-\sin 2 x+9 \sin 6 x \\
\therefore \quad a & =-4, b=4, c=-1, d=2, e=9, k=6 \\
(a & +c+e)-(b+d+k)=4-12=-8
\end{aligned}
$