sin-6-cos-6-1-sin-4-sin-2-

Question Number 146866 by mathdanisur last updated on 16/Jul/21

\frac{{s i n}^{6} α + {c o s}^{6} α - 1}{{s i n}^{4} α - {s i n}^{2} α} = ?

Answered by Olaf_Thorendsen last updated on 16/Jul/21

a = \sin^{6} α + \cos^{6} α

a = {(\sin^{2} α + \cos^{2} α)}^{3} - {3 \sin}^{2} α \cos^{2} α (\sin^{2} α + \cos^{2} α)

a = 1 - {3 \sin}^{2} α \cos^{2} α = 1 - \frac{3}{4} \sin^{2} 2 α

b = \sin^{4} α - \cos^{4} α

b = (\sin^{2} α - \cos^{2} α) (\sin^{2} α + \cos^{2} α)

b = - \cos^{2} 2 α

X = \frac{\sin^{6} α + \cos^{6} α - 1}{\sin^{4} α - \cos^{4} α}

X = \frac{a - 1}{b}

X = \frac{1 - \frac{3}{4} \sin^{2} 2 α - 1}{- {2 \cos}^{2} 2 α}

X = \frac{3}{8} \tan^{2} 2 α

Commented by mathdanisur last updated on 23/Jul/21

S i r, b = {s i n}^{4} α - {s i n}^{2} α

Commented by mathdanisur last updated on 16/Jul/21

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