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Question

# If $\mathrm{sin}\mathit{}x+\mathrm{cos}\mathit{}x=m$, then prove that ${\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x=\frac{4-3{\left({m}^{2}-1\right)}^{2}}{4}$, where ${m}^{2}\le 2$

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Solution

## $\mathrm{sin}x+\mathrm{cos}x=m\left(\mathrm{Given}\right)\phantom{\rule{0ex}{0ex}}\mathrm{To}\mathrm{prove}:{\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x=\frac{4-3\left({m}^{2}-1{\right)}^{2}}{4},\mathrm{where}{m}^{2}\le 2\phantom{\rule{0ex}{0ex}}\mathrm{Proof}:\phantom{\rule{0ex}{0ex}}\mathrm{LHS}:\phantom{\rule{0ex}{0ex}}{\mathrm{sin}}^{6}x+{\mathrm{cos}}^{6}x\phantom{\rule{0ex}{0ex}}={\left({\mathrm{sin}}^{2}x\right)}^{3}+{\left({\mathrm{cos}}^{2}x\right)}^{3}\phantom{\rule{0ex}{0ex}}={\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)}^{3}-3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x\left({\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x\right)\phantom{\rule{0ex}{0ex}}=1-3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{RHS}:\phantom{\rule{0ex}{0ex}}\frac{4-3\left({m}^{2}-1{\right)}^{2}}{4}\phantom{\rule{0ex}{0ex}}=\frac{4-3{\left[{\left(\mathrm{sin}x+\mathrm{cos}x\right)}^{2}-1\right]}^{2}}{4}\phantom{\rule{0ex}{0ex}}=\frac{4-3{\left[{\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x+2\mathrm{sin}x\mathrm{cos}x-1\right]}^{2}}{4}\phantom{\rule{0ex}{0ex}}=\frac{4-3{\left[{\mathrm{sin}}^{2}x-\left(1-{\mathrm{cos}}^{2}x\right)+2\mathrm{sin}x\mathrm{cos}x\right]}^{2}}{4}\phantom{\rule{0ex}{0ex}}=\frac{4-3×4{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}{4}\phantom{\rule{0ex}{0ex}}=1-3{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x\phantom{\rule{0ex}{0ex}}\mathrm{LHS}=\mathrm{RHS}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}\mathrm{proved}$

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