Question

помогите пожалуйста решить

Answer 1

4)sin²(π/4)-2cos²(π/3)-5tg²(π/4)=(√2/2)²-2.(1/2)²-5.(1)²=1/2-1/2-5=-5

5)(5+9-1)/(1-1-1)=13/(-1)=-13

(ctg(π/4)=1, ctg(π/6)=√3, tg(π/4)=1, cos(π/3)=1/2, sin(π/6)=1/2)

6)(a³-b³-0)/(a²+ab+b²)=(a³-b³)/(a²+ab+b²)=(a-b)(a²+ab+b²)/(a²+ab+b²)=a-b

(A³-B³=(A-B)(A²+AB+B²))

(cos(π/2)=0, cos0=1, cos(π/4)=(√2)/2)

Answer 2

6)(aˇ3 -bˇ3-0)/aˇ2+ab+bˇ2)=(a-b)(aˇ2+ab+bˇ2)/(aˇ2+ab+bˇ2)=a-b (izvini ošibku...)

Answer 3

исправьте пожалуйста в самом решении

Answer 4

$4) sin^2(fracpi4) - 2cos^2(fracpi3) - 5tg^2(fracpi4) = frac12 - frac12 - 5\5) frac{5+ctg^4(fracpi6) - tg^2(fracpi4)}{ctg(fracpi4) - 4cos^2(fracpi3) - 8sin^3(fracpi6)} = frac{5+9-1}{1-1-1} = -13\6) frac{(2a sin(fracpi6))^3 - (btg(fracpi4))^3 - (2abcos(fracpi2))^2}{(acos0)^2 + 2ab(cos(fracpi3) + 2b^2cos^2(fracpi4)} = frac{a^3-b^3}{a^2 +ab + b^2} = a-b$