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Prove that ${\mathrm{sin}}^{6}\frac{\theta }{2}+{\mathrm{cos}}^{6}\frac{\theta }{2}=\frac{1}{4}\left(1+3{\mathrm{cos}}^{2}\theta \right)$
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grcalia1
${\mathrm{sin}}^{6}\frac{\theta }{2}+{\mathrm{cos}}^{6}\frac{\theta }{2}=\cdots =1-3\left(1-{\mathrm{cos}}^{2}\frac{\theta }{2}\right){\mathrm{cos}}^{2}\frac{\theta }{2}$
is correct. From here, note that
$\begin{array}{rl}{\mathrm{sin}}^{6}\frac{\theta }{2}+{\mathrm{cos}}^{6}\frac{\theta }{2}& =1-3\left(1-{\mathrm{cos}}^{2}\frac{\theta }{2}\right){\mathrm{cos}}^{2}\frac{\theta }{2}\\ & =1-3\left(1-\frac{\mathrm{cos}\theta +1}{2}\right)\cdot \frac{\mathrm{cos}\theta +1}{2}\\ & =1-3\cdot \frac{1-\mathrm{cos}\theta }{2}\cdot \frac{1+\mathrm{cos}\theta }{2}\\ & =\frac{4-3\left(1-{\mathrm{cos}}^{2}\theta \right)}{4}\\ & =\frac{1+3{\mathrm{cos}}^{2}\theta }{4}\end{array}$

Zion Wheeler
You may begin factoring:
$\begin{array}{rl}{\mathrm{sin}}^{6}\frac{\theta }{2}+{\mathrm{cos}}^{6}\frac{\theta }{2}& =\left({\mathrm{sin}}^{2}\frac{\theta }{2}+{\mathrm{cos}}^{2}\frac{\theta }{2}\right)\left({\mathrm{sin}}^{4}\frac{\theta }{2}-{\mathrm{sin}}^{2}\frac{\theta }{2}{\mathrm{cos}}^{2}\frac{\theta }{2}+{\mathrm{cos}}^{4}\frac{\theta }{2}\right)\\ & =\left({\mathrm{cos}}^{2}\frac{\theta }{2}-{\mathrm{sin}}^{2}\frac{\theta }{2}{\right)}^{2}+{\mathrm{sin}}^{2}\frac{\theta }{2}{\mathrm{cos}}^{2}\frac{\theta }{2}={\mathrm{cos}}^{2}\theta +\frac{1}{4}{\mathrm{sin}}^{2}\theta \\ & =\frac{1}{4}\left(3{\mathrm{cos}}^{2}\theta +1\right).\end{array}$
You also may linearise:
${\mathrm{cos}}^{2}\theta +\frac{1}{4}{\mathrm{sin}}^{2}\theta =\frac{1}{2}\left(1+\mathrm{cos}2\theta \right)+\frac{1}{8}\left(1-\mathrm{cos}2\theta \right)=\frac{1}{8}\left(3\mathrm{cos}2\theta +5\right).$

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