Sin(-11π/6).cos(19π/6).tg(5π/4)=
=sin(12π/6-11π/6).cos(3π+π/6).tg(π+π/4)=
=sin(π/6).cos(2π+π+π/6).tg(π/4)=
=1/2.cos(7π/6).1=1/2.(-cos(π/6)).1=
=1/2.(-√3/2).1=-√3/4
.........................
(Sin^2(b)-cos^2(b))/sin2b=(-cos2b)/sin2b=-ctg2b
1. а) (√3-1)^2=3-2*√3*1+1=4-2√3
б)(√2)^2/ <span>√2-1=2/√2-1=2*(√2-1)/(√2-1)*(√2-1)=2√2-2/2-√2-√2+1=2√2-2/-2√2+1
2. <span>(4-b)^2 - b*(b+10)=16-8b+b^2-b^2-10b=16-18b
3. </span></span>3х+10>-5*1 7-4х>3
x>-5 x<1
Ответ: (-5;1)