mistergoneo7

2022-07-15

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

thatuglygirlyu

Evaluate the equation for different $x$:

(after normalization of the coefficient of $d$)
Elimination by the combination $\left(1\right)+3×\left(3\right)+2×\left(4\right)$ yields
$6d=\frac{210}{64}.$
The evaluation for $x=\frac{\pi }{4}$ was not even necessary.

Jorden Pace

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

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