Sin2xcos2x=1/2sin4x"

"cos^2x/2-sin^2x/2=cosx

Question

Алгебра

Ленин

18 ноября, 14:53

Sin2xcos2x=1/2sin4x"

"cos^2x/2-sin^2x/2=cosx

+1

Ответы (1)

Знаете ответ на вопрос?

Венюша · Answer 1

Sin2α = 2sinα • cosα

sin2x • cos2x = (2sin2x • cos2x) / 2 = (sin (2 • 2x)) / 2 = (1/2) • sin4x

(1/4) • sin4x = (1/4) • sin4x

cos2α = (cosα) ² - (sinα) ²

(cos (x/2)) ² - (sin (x/2)) ² = cos (2 • (x/2)) = cosx

cosx = cosx

Sin2xcos2x=1/2sin4x""cos^2x/2-sin^2x/2=cosx

Sin2xcos2x=1/2sin4x"

"cos^2x/2-sin^2x/2=cosx