task/29474948 Доказать тождество sin(3arctg√3-arccos0)=1
sin(3arctg√3-arccos0)= sin(3*π/3 - π /2) = sin(π - π /2) = sin(π /2) = 1
* * * arcsina , arccosa ( |a| ≤ 1 ) такие углы ,что * * *
1. - π/2 ≤ arcsina ≤ π/2 ; 2. sin (arcsina) = a
2. 0 ≤ arcsina ≤ π ; 2. cos (arccos) = a
1)<span>f(x)=3x⁴*(6x-5)
f ' (x) = 12x</span>³*(6x-5)+3x⁴ *6 = 12x³*(6x-5)+18x<span>⁴
2)</span><span>f(x)=(5x-1)sin x/2
f '(x)=5 *sin x/2 -(5x-1)cos x/2 * 1/2
3)</span><span>f(x)=(5x-1)²(6x+2)
f '(x) = 2*5*(5x-1) *(6x+2) +6* (5x-1)</span>² =10*(5x-1) *(6x+2)+6* (5x-1)<span>²</span>
2x+1/3 - 4x-2/4= 1 -2x-1/6=1 -2x=1/6 x=-1/12