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

$4) sin^2(\frac\pi4) - 2cos^2(\frac\pi3) - 5tg^2(\frac\pi4) = \frac12 - \frac12 - 5\\5) \frac{5+ctg^4(\frac\pi6) - tg^2(\frac\pi4)}{ctg(\frac\pi4) - 4cos^2(\frac\pi3) - 8sin^3(\frac\pi6)} = \frac{5+9-1}{1-1-1} = -13\\6) \frac{(2a sin(\frac\pi6))^3 - (btg(\frac\pi4))^3 - (2abcos(\frac\pi2))^2}{(acos0)^2 + 2ab(cos(\frac\pi3) + 2b^2cos^2(\frac\pi4)} = \frac{a^3-b^3}{a^2 +ab + b^2} = a-b$