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lil5
@lil5
August 2022
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Помогите,пожалуйста,с задачами.Необходимо полное решение.
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IUV
Verified answer
(cos(pi/6)-3*tg(pi)+tg(5*pi/3))*ctg(5*pi/4)=-корень(3)/2= -0,8660254
2*tg(1,5)*cctg(1,5)-sin^2(2)-cos^2(2)=2*1-1=1
cos(-7pi/12)*sin(3*pi/8)*ctg(-pi/6)=(-)*(+)*(-1)=(+) > 0
sin(2)*cos(-1)*ctg(3)=(+)*(+)*(-)=(-) < 0
a=cos(3,5) < 0
b=sin(3,5) < 0
pi < 3,5 < 5pi/4 => cos(3,5) < sin(3,5)
3 votes
Thanks 4
IUV
1/a/a
(cos(pi/6)-3*tg(pi)+tg(5*pi/3))*ctg(5*pi/4)=(корень(3)/2-3*0-корень(3))*(1)=(-корень(3)/2)=-корень(3)/2= -0,8660254
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Answers & Comments
Verified answer
(cos(pi/6)-3*tg(pi)+tg(5*pi/3))*ctg(5*pi/4)=-корень(3)/2= -0,86602542*tg(1,5)*cctg(1,5)-sin^2(2)-cos^2(2)=2*1-1=1
cos(-7pi/12)*sin(3*pi/8)*ctg(-pi/6)=(-)*(+)*(-1)=(+) > 0
sin(2)*cos(-1)*ctg(3)=(+)*(+)*(-)=(-) < 0
a=cos(3,5) < 0
b=sin(3,5) < 0
pi < 3,5 < 5pi/4 => cos(3,5) < sin(3,5)
(cos(pi/6)-3*tg(pi)+tg(5*pi/3))*ctg(5*pi/4)=(корень(3)/2-3*0-корень(3))*(1)=(-корень(3)/2)=-корень(3)/2= -0,8660254