If sinα + cosα =m then for m2≤ 2, sin6α + cos6α =

sin α+cos α=M sin6α+cos6α=(sin2α+cos2α)3-3sin2α cos2α (sin2α+cos2α) =1-3sin2α cos2α =1-34(4sin2α cos2α) =1-34sin22α =1-34(M2-1)2 =4-3(m2-1)24

If sinα + cosα =m then for m2≤ 2, sin6α + cos6α =

sin α+cos α=M sin6α+cos6α=(sin2α+cos2α)3-3sin2α cos2α (sin2α+cos2α) =1-3sin2α cos2α =1-34(4sin2α cos2α) =1-34sin22α =1-34(M2-1)2 =4-3(m2-1)24

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$I f \sin α + \cos α = m t h e n f o r m^{2} \leq 2, \sin^{6} α + \cos^{6} α =$

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$\frac{4 - 3 {(m^{2} - 1)}^{2}}{4}$

$\frac{4 - 3 {(m^{2} + 1)}^{2}}{4}$

$\frac{4 + 3 {(m^{2} + 1)}^{2}}{4}$

$\frac{4 + 3 {(m^{2} - 1)}^{2}}{4}$

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Detailed Solution

$\sin α + \cos α = M \sin^{6} α + \cos^{6} α = {(\sin^{2} α + \cos^{2} α)}^{3} - 3 \sin^{2} α \cos^{2} α (\sin^{2} α + \cos^{2} α) = 1 - 3 \sin^{2} α \cos^{2} α = 1 - \frac{3}{4} (4 \sin^{2} α \cos^{2} α) = 1 - \frac{3}{4} \sin^{2} 2 α = 1 - \frac{3}{4} {(M^{2} - 1)}^{2} = \frac{4 - 3 {(m^{2} - 1)}^{2}}{4}$