Question

Trigonometrie cls 9-a

Answer 1

$sin^6a+cos^6a=1-3sin^2a*cos^2a\\\\sin^6a+(cos^2a)^3=1-3sin^2a*cos^2a\\\\sin^6a+(1-sin^2a)^3=1-3sin^2a*cos^2a\\\\sin^6a+1-3*1^2*sin^2a+3*1*(sin^2a)^2-(sin^2a)^3=1-3sin^2a*cos^2a\\\\sin^6a+1-3*sin^2a+3*sin^4a-sin^6a=1-3sin^2a*cos^2a\\\\1-3*sin^2a+2sin^4a=1-3sin^2a*cos^2a\\\\1-3*sin^2a(1-sin^2a)=1-3sin^2a*cos^2a\\\\1-3sin^2a*cos^2a=1-3sin^2a*cos^2a$

$Formule\ folosite\\\\sin^2x+cos^2x\ =>\ cos^2x=1-sin^2x\\(a-b)^3=a^3-3a^2b+3ab^2-b^3$

Answer 2

Salut, raspunsul lui Delacey este bun ,dar vreau sa iti arat o varianta mai directa.

$\sin^6 a+\cos^6a=1-3\cos^2a\sin^2a(\text{asta avem de demonstrat})\\\sin^6 a+\cos^6 a = (\sin^2a)^3 +(\cos^2a)^3=\\=(\sin^2a+\cos^2a)(\sin^4a-\sin^2a\cdot\cos^2a+\cos^4 a)=\\= \sin^4 a+2\sin^2a\cdot\cos^2a+\cos^4a-3\sin^2a\cdot\cos^2a=\\=(\sin^2a+\cos^2a)^2-3\sin^2 a\cdot \cos^2a = 1-3\sin^2a\cdot\cos^2a~~Q.E.D.\\\bold{Formule~ folosite:}\\\bullet~ a^3+b^3=(a+b)(a^2-ab+b^2)\\\bullet~ a^2+2ab+b^2=(a+b)^2\\\bullet \sin^2a+\cos^2a=1$