Find $d$ when $(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$ There exist constants $a$, $b$, $c$, and $d$ such that
$(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$
for all angles $x$. Find $d$.

$$\begin{align} & \sin^7x  \\ & 
=\sin^6x\cdot \sin x \\ &
=\frac{1}{16}\cdot 16\sin^6x\cdot \sin x \\  &
=\frac{1}{16}\cdot (4\sin^3x)^2 \cdot \sin x \\  &
=\frac{1}{16}\cdot (3\sin x-\sin 3x)^2 \cdot \sin x \\ &
=\frac{1}{16}\cdot (9\sin^2 x-6\sin x\sin 3x+\sin^2 3x) \cdot \sin x \\ &
=\frac{1}{16}\cdot [\frac{9}{2}(1-\cos 2x)-3(\cos 2x-\cos 4x)+\frac{1}{2}(1-\cos 6x)] \cdot \sin x \\ &
=\frac{9}{32}(\sin x-\sin x\cdot \cos 2x)-\frac{3}{16}(\sin x\cdot \cos 2x-\sin x\cdot \cos 4x)+\frac{1}{32}(\sin x-\sin x\cdot \cos 6x) \\ &
=\frac{9}{32}[\sin x-\frac{1}{2}(\sin 3x- \sin x)]-\frac{3}{32}[(\sin 3x- \sin x)-(\sin 5x - \sin 3x)]+\frac{1}{32}[\sin x-\frac{1}{2}(\sin 7x-\sin 5x)] \end{align}$$
The coefficient of $\sin x$ in the above expression is $$\color{red}{\frac{9}{32}+\frac{9}{64}+\frac{3}{32}+\frac{1}{32}}=\color{blue}{\frac{35}{64}}$$
