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Waqi
@Waqi
August 2022
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армения20171
Решение во вложении......
0 votes
Thanks 1
kirichekov
Verified answer
A)cos11π/2=cos(4π+3π/2)=cos3π/2=0
tg765°=tg(4*180°+45°)=tg45°=1
sin630°=sin(2*360°-90°)=sin(-90°)=-1
b)4arcsin(-√3/2)+arctg(-1)+arccos0=
4•(-π/3)+(-π/4)+π/2=(-16π-3π+6π)/12=
-13π/12
1 votes
Thanks 1
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Answers & Comments
Verified answer
A)cos11π/2=cos(4π+3π/2)=cos3π/2=0tg765°=tg(4*180°+45°)=tg45°=1
sin630°=sin(2*360°-90°)=sin(-90°)=-1
b)4arcsin(-√3/2)+arctg(-1)+arccos0=
4•(-π/3)+(-π/4)+π/2=(-16π-3π+6π)/12=
-13π/12