If-sin-10-x-cos-10-x-11-36-find-sin-12-x-cos-12-x-

Question Number 86878 by jagoll last updated on 01/Apr/20

If \sin^{10} (x) + \cos^{10} (x) = \frac{11}{36}

find \sin^{12} (x) + \cos^{12} (x) = ?

Commented by Prithwish Sen 1 last updated on 01/Apr/20

I got

\sin^{2} {xcos}^{2} x = \frac{5}{6} or \frac{1}{6}

now

\sin^{12} x + \cos^{12} x = 1 - {6 \sin}^{2} {xcos}^{2} x + {9 \sin}^{4} {xcos}^{4} x

- {2 \sin}^{6} {xcos}^{6} x

Answered by mr W last updated on 01/Apr/20

l e t s = \sin^{2} x, c = \cos^{2} x

\sin^{10} x + \cos^{10} x = \frac{11}{36}

\Rightarrow s^{5} + c^{5} = \frac{11}{36}

s + c = 1

{(s + c)}^{5} = 1

s^{5} + 5 s^{4} c + 10 s^{3} c^{2} + 10 s^{2} c^{3} + 5 {s c}^{4} + c^{5} = 1

s^{5} + c^{5} + 5 s c (s^{3} + c^{3}) + 10 s^{2} c^{2} (s + c) = 1

s^{5} + c^{5} + 5 s c (s + c) [{(s + c)}^{2} - 3 s c] + 10 {(s c)}^{2} (s + c) = 1

l e t λ = s c

\frac{11}{36} + 5 λ [1 - 3 λ] + 10 λ^{2} = 1

λ^{2} - λ + \frac{5}{36} = 0

(λ - \frac{1}{6}) (λ - \frac{5}{6}) = 0

\Rightarrow λ = \frac{1}{6}, λ = \frac{5}{6} > \frac{1}{4} \Rightarrow n o t s u i t a b l e

λ = {(\sin x \cos x)}^{2} = {(\frac{\sin 2 x}{2})}^{2} ⩽ \frac{1}{4}

\Rightarrow λ = \frac{1}{6}

{(s + c)}^{6} = 1

s^{6} + 6 s^{5} c + 15 s^{4} c^{2} + 20 s^{3} c^{3} + 15 s^{2} c^{4} + 6 {s c}^{5} + c^{6} = 1

s^{6} + c^{6} + 6 s c (s^{4} + c^{4}) + 15 {(s c)}^{2} (s^{2} + c^{2}) + 20 {(s c)}^{3} = 1

s^{6} + c^{6} + 6 s c [{({(s + c)}^{2} - 2 s c)}^{2} - 2 s^{2} c^{2}] + 15 {(s c)}^{2} [{(s + c)}^{2} - 2 s c] + 20 {(s c)}^{3} = 1

s^{6} + c^{6} + 6 λ [{(1 - 2 λ)}^{2} - 2 λ^{2}] + 15 λ^{2} [1 - 2 λ] + 20 λ^{3} = 1

s^{6} + c^{6} + 6 λ - 9 λ^{2} + 2 λ^{3} = 1

s^{6} + c^{6} = 1 - λ (2 λ^{2} - 9 λ + 6)

s^{6} + c^{6} = 1 - \frac{1}{6} (\frac{1}{18} - \frac{3}{2} + 6) = \frac{13}{54}

\Rightarrow \sin^{12} x + \cos^{12} x = \frac{13}{54}

Answered by MJS last updated on 01/Apr/20

\sin x = \pm \frac{\tan x}{\sqrt{1 + \tan^{2} x}}

\cos x = \pm \frac{1}{\sqrt{1 + \tan^{2} x}}

let \tan x = t

\sin^{10} x + \cos^{10} x = \frac{t^{8} - t^{6} + t^{4} - t^{2} + 1}{{(t^{2} + 1)}^{4}}

\sin^{12} x + \cos^{12} x = \frac{t^{12} + 1}{{(t^{2} + 1)}^{6}}

\frac{t^{8} - t^{6} + t^{4} - t^{2} + 1}{{(t^{2} + 1)}^{4}} = \frac{11}{36}

t^{8} - \frac{16}{5} t^{6} - \frac{6}{5} t^{4} - \frac{16}{5} t^{2} + 1 = 0

(t^{4} - 4 t^{2} + 1) (t^{4} + \frac{4}{5} t^{2} + 1) = 0

\Rightarrow t^{2} = 2 \pm \sqrt{3}

\sin^{12} x + \cos^{12} x = \frac{t^{12} + 1}{{(t^{2} + 1)}^{6}} = \frac{13}{54}

Commented by Prithwish Sen 1 last updated on 01/Apr/20

Excellent . Thank you sir .

Commented by jagoll last updated on 01/Apr/20

waww \dots . super easy

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