If sin &#x2061;<!-- ⁡ --> x + sin 2 </msup> &#x2061;<!-- ⁡ --> x

If $\mathrm{sin}x+{\mathrm{sin}}^{2}x=1$ then find the value of ${\mathrm{cos}}^{8}x+2{\mathrm{cos}}^{6}x+{\mathrm{cos}}^{4}x$
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bgu999dq
$\mathrm{sin}x+{\mathrm{sin}}^{2}x=1$
$\mathrm{sin}x=1-{\mathrm{sin}}^{2}x$
$\mathrm{sin}x={\mathrm{cos}}^{2}x$
Now
${\mathrm{cos}}^{8}x+2{\mathrm{cos}}^{6}x+{\mathrm{cos}}^{4}x$
$={\mathrm{sin}}^{4}x+2{\mathrm{sin}}^{3}x+{\mathrm{sin}}^{2}x$
$={\mathrm{sin}}^{4}x+{\mathrm{sin}}^{3}x+{\mathrm{sin}}^{3}x+{\mathrm{sin}}^{2}x$
$={\mathrm{sin}}^{3}x\left(\mathrm{sin}x+1\right)+{\mathrm{sin}}^{2}x\left(\mathrm{sin}x+1\right)$
$\left(\mathrm{sin}x+1\right)\left({\mathrm{sin}}^{3}x+{\mathrm{sin}}^{2}x\right)\phantom{\rule{0ex}{0ex}}=\left(\mathrm{sin}x+1\right)\left(\mathrm{sin}x+1\right){\mathrm{sin}}^{2}x\phantom{\rule{0ex}{0ex}}=\left({\mathrm{sin}}^{2}x+\mathrm{sin}x\right)\left({\mathrm{sin}}^{2}x+\mathrm{sin}x\right)\phantom{\rule{0ex}{0ex}}=1$
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Angel Malone
Even shorter:
You know $\mathrm{sin}x={\mathrm{cos}}^{2}x$. Then
${\mathrm{cos}}^{8}x+2{\mathrm{cos}}^{6}x+{\mathrm{cos}}^{4}x={\mathrm{sin}}^{4}x+2{\mathrm{sin}}^{3}x+{\mathrm{sin}}^{2}x=\left({\mathrm{sin}}^{2}x+\mathrm{sin}x{\right)}^{2}={1}^{2}=1$