${sin}^{6} θ + {cos}^{6} θ = 1 - k {sin}^{2} 2 θ$

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Solution

${sin}^{6} θ + {cos}^{6} θ = ({sin}^{2} θ)^{3} + ({cos}^{2} θ)^{3}$ $= ({sin}^{2} θ + {cos}^{2} θ) {{sin}^{4} θ - {sin}^{2} θ {cos}^{2} θ + {cos}^{4} θ}$ [ $∵$ $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ ] $= 1 \times {({sin}^{4} θ + {cos}^{4} θ) - {sin}^{2} θ {cos}^{2} θ}$ $= {({sin}^{2} θ + {cos}^{2} θ)^{2} - 2 {sin}^{2} θ {cos}^{2} θ - {sin}^{2} θ {cos}^{2} θ}$ [ $∵ a^{4} + b^{4} = (a^{2} + b^{2})^{2} - 2 a^{2} b^{2}$ ] $= 1 - 3 {sin}^{2} θ {cos}^{2} θ = 1 - \frac{3}{4} \times 4 {sin}^{2} θ {cos}^{2} θ = 1 - \frac{3}{4} {sin}^{2} 2 θ$ Therefore, $k = \frac{3}{4}$

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