Задача 12013 ...

SOVA

12 марта 2016 г. в 00:00

Вычислить
sin⁴(π/16)+sin⁴(3π/16)+sin⁴(5π/16)+sin⁴(7π/16)

sin²α=(1–cos2α)/2
sin⁴α=(1–cos2α)²/4

sin⁴(π/16)+sin⁴(3π/16)+sin⁴(5π/16)+sin⁴(7π/16)=

=((1–cos(π/8))²+(1–cos(3π/8))²+(1–cos(5π/8))²+(1–cos(7π/8))²)/4=

=(1+1+1+1–2cos(π/8)–2cos(3π/8)–2cos(5π/8)–2cos(7π/8)+cos²(π/8)+cos²(3π/8)+cos²(5π/8)+cos²(7π/8))/4=

=1–(cos(π/8)+cos(3π/8)+cos(5π/8)+cos(7π/8))/2 + (cos²(π/8)+cos²(3π/8)+cos²(5π/8)+cos²(7π/8))/4=

= так как cos²α=(1+cos2α)/2=

=1–(cos(π/8)+cos(3π/8)+cos(5π/8)+cos(7π/8))/2 + (1+cos(π/4)+1+cos(3π/4)+1+cos(5π/4)+1+cos(7π/4))/8=

= так как cos(7π/8)=–cos(π/8); cos(5π/8)=–cos(3π/8), то
(cos(π/8)+cos(3π/8)+cos(5π/8)+cos(7π/8))=0
cos(7π/4)=–cos(π/4); cos(5π/4)=–cos(3π/4), тоответ
=1+(4/8)=3/2

Ответ: 3/2