<s>[INST] 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 $d$.
 [/INST] Interesting answers .Let me add another, just for filing a somewhat more "theoretical" one, using Chebyshev polynomials $U_n(x)$ of the second kind defined by
$$U_0(x)=1\\U_1(x)=2x\\U_{n+1}(x)=2x\space U_n(x)-U_{n-1}(x)$$
and for which one has the formula
$$U_n(\cos x)=\frac{\sin (n+1)x}{\sin x}\qquad (*)$$ 
We use the data (if you don’t want to calculate it)
$$\begin{cases}U_6(x)=64x^6-80x^4+24x^2-1\\U_4(x)=16x^4-12x^2+1\\U_2(x)=4x^2-1\\U_0(x)=1\end{cases}$$
Therefore, using $(*)$ and simplifying  the given identity,we have
$$(\sin x)^6=aU_6(\cos x)+bU_4(\cos x)+cU_2(\cos x)+dU_0(\cos x)$$
It follows, putting  $\cos x= t$,
$$(1-t^2)^3=64at^6+(-80a+16b)t^4+(24a-12b+4c)t^2+(-a-c+b+d)$$
Hence the easy system
 $$\begin{cases}64a=-1\\-80a+16b=3\\24a-12b+4c=-3\\-a+b-c+d=1\end{cases}$$
which is resolved line by line giving in succession 
$$(a,b,c,d)=\left(\frac {-1}{64},\frac {7}{64},\frac{-21}{64},\frac {35}{64}\right)$$
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