Question

Найти cos(a)^8-sin(a)^8, если cos(a)=1/3

Answer 1

(cos a)^8 - (sin a)^8 = [(cos a)^4 - (sin a)^4]*[(cos a)^4 + (sin a)^4] =
= (cos^2 a - sin^2 a)(cos^2 a + sin^2 a)*(cos^4 a + sin^4 a) =
= (cos 2a)*1*(cos^4 a + 2sin^2 a*cos^2 a + sin^4 a - 2sin^2 a*cos^2 a) =
= cos 2a *[ (cos^2 a + sin^2 a)^2 - 0,5*4sin^2 a*cos^2 a ] =
= cos 2a *(1^2 - 1/2*(sin 2a)^2) = cos 2a *(1 - 1/2*sin^2 (2a)) =
= cos 2a - 1/2*cos 2a*sin^2 (2a)
(cos a)^8 - (sin a)^8 = cos 2a - 1/2*cos 2a*sin^2 (2a)
Теперь подставляем. Так как cos a = 1/3, то:
cos 2a = 2cos^2 a - 1 = 2*1/9 - 1 = -7/9
sin^2 (2a) = 1 - cos^2 (2a) = 1 - 49/81 = 32/81
(cos a)^8 - (sin a)^8 = -7/9 - 1/2*(-7/9)*32/81 = -7/9 + 16*7/(9*81) =
= (-7*81+16*7)/729 = -455/729