Дано: sin α + cos α = 0,5
Вычислить: sin⁵α + cos⁵α
I. sin α + cos α = 1/2 возвести обе части в квадрат
sin²α + 2sinα cosα + cos²α = 1/4
1 + sin (2α) = 1/4
sin (2α) = -3/4
II.
sin⁵α + cos⁵α =
=(sinα + cosα)(sin⁴α - sin³α*cosα + sin²α*cos²α - sinα*cos³α + cos⁴α) =
=0,5((sin⁴α + 2sin²α*cos²α + cos⁴α) - sin²α*cos²α - (sin³α*cosα+sinα*cos³α))=
=0,5( (sin²α + cos²α)² - sin²α*cos²α - sinα*cosα (sin²α + cos²α) )=
=1= =1=
=0,5 (1 - (1/4)(2sinα*cosα)² - (1/2)*(2sinα*cosα) ) =
=0,5 (1 - (1/4) sin²(2α) - (1/2) sin(2α)) =
![= \frac{1}{2} (1 - \frac{1}{4}*(- \frac{3}{4} )^2- \frac{1}{2} *(- \frac{3}{4} ) )= \\ \\ = \frac{1}{2} (1 - \frac{1}{4}* \frac{9}{16} + \frac{3}{8} )= \\ \\ = \frac{1}{2} (1 - \frac{9}{64} + \frac{24}{64} )= \frac{1}{2} * \frac{79}{64} = \frac{79}{128}](https://tex.z-dn.net/?f=%3D+%5Cfrac%7B1%7D%7B2%7D+%281+-++%5Cfrac%7B1%7D%7B4%7D%2A%28-+%5Cfrac%7B3%7D%7B4%7D+%29%5E2-+%5Cfrac%7B1%7D%7B2%7D+%2A%28-+%5Cfrac%7B3%7D%7B4%7D+%29+%29%3D+%5C%5C++%5C%5C+%3D+%5Cfrac%7B1%7D%7B2%7D+%281+-+%5Cfrac%7B1%7D%7B4%7D%2A+%5Cfrac%7B9%7D%7B16%7D+%2B+%5Cfrac%7B3%7D%7B8%7D++%29%3D+%5C%5C++%5C%5C+%3D+%5Cfrac%7B1%7D%7B2%7D+%281+-++%5Cfrac%7B9%7D%7B64%7D+%2B+%5Cfrac%7B24%7D%7B64%7D+%29%3D+%5Cfrac%7B1%7D%7B2%7D+%2A+%5Cfrac%7B79%7D%7B64%7D+%3D+%5Cfrac%7B79%7D%7B128%7D+)
sin⁵α + cos⁵α =
![\frac{79}{128}](https://tex.z-dn.net/?f=+%5Cfrac%7B79%7D%7B128%7D+)
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Использованы формулы
sin²α + cos²α = 1
2 sinα cosα = sin (2α)
a⁵ + b⁵ = (a + b)(a⁴ - a³b + a²b² - ab³ + b⁴)