Задача 54094 вычислите : 1)...

5f42a255b5ac0b3cc606fbf7

06.10.2020 17:55:25

вычислите :
1) 2arcsin(sqrt(-3)\2)-3arctg(-sqrt(3)\3)+arccos(sqrt(3)\2)-2arcctg(-1)
2) arccos(- sqrt(2)\2)+2arcctg(-sqrt(3))+arcsin(- sqrt(3)\2)+arctg1
3) 3arcsinsqrt(3)\2-arctg(-1)+arccos(- sqrt(2)\2)+arcctg(- sqrt(3))
4) arccos(-sqrt(3)\2)+arctg(-sqrt(3))-arcsin(-1)-2arctgsqrt(3)

5f3ea7e3faf909182968ddd9

06.10.2020 18:33:35

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1) 2arcsin(√–3/2)–3arctg(–√3/3)+arccos(-√3/2)–2arcctg(–1)=2*(-π/3)-3*(-π/6)+(π-(π/6))-2*(π-(π/4))=

=(-2π/3)+(π/2)+(5π/6)-(3π/2)=[b]-5π/6[/b]


2) arccos(– √2/2)+2arcctg(–√3)+arcsin(– √3/2)+arctg1=(π-(π/4))+2*(π-(π/6))+(-π/3)+(π/4)=[b](3π/4)[/b]+(10π/6)-(π/3)+[b](π/4)[/b]=

=π+(4π/3)=[b]7π/3
[/b]

3) 3arcsin(√3/2)–arctg(–1)+arccos(– √2/2)+arcctg(– √3)=3*(π/3)-(-π/4)+(π-(π/4))+(π-(π/6))=2π+(5π/6)=17π/6


4) arccos(–√3/2)+arctg(–√3)–arcsin(–1)–2arctg√3=(π -(π/6))+(-π/3)-(-π/2)-2*(π/3)=

=(5π/6)+(-2π/6)+(π/2)-(2π/3)=

=(3π/6)+(π/2)-(2π/3)=

=π-(2π/3)=π/3