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Question Number 75983 by mathocean1 last updated on 21/Dec/19

$$\mathrm{show}\mathrm{that}$$$${\cos }^6\mathrm{x}+{\sin }^6\mathrm{x}=\frac{1}{8}(5+3\cos 4\mathrm{x})$$

Commented by MJS last updated on 21/Dec/19

$$\mathrm{answer}\mathrm{is}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{other}$$$$\mathrm{question}$$

Commented by mathmax by abdo last updated on 22/Dec/19

$${cos}^{6}x+{sin}^{6}x={\left({cos}^{2}x\right)}^3+{\left({sin}^{2}x\right)}^3$$$$=({cos}^{2}x+{sin}^{2}x)({cos}^{4}x-{cos}^{2}x{sin}^{2}x+{sin}^{4}x)$$$$={cos}^{4}x+{sin}^{4}x-{cos}^{2}x{sin}^{2}x={({cos}^{2}x+{sin}^{2}x)}^2-3{cos}^{2}x{sin}^{2}x$$$$=1-\frac{3}{4}{sin}^2\left(2x\right)=1-\frac{3}{4}\left(\frac{1-cos\left(4x\right)}{2}\right)=1-\frac{3}{8}+\frac{3}{8}cos\left(4x\right)$$$$=\frac{5}{8}+\frac{3}{8}cos\left(4x\right)sotheresultisproved.$$

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