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Lifa
@Lifa
July 2022
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Помогите прошу срочно 30 баллов
1) Вычислите: √32-√128sin^2(2085°)
2) Вычислите: cos^2(8430°)
3) Вычислите: sina, cosa, tga, ctga, если a=67Π/3.
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DariosI
Verified answer
1) √32-√128sin^2(2085°)=√32-√128sin²(1800+285°)=√32-√128sin²(5*360+180+105°)=√32-√128sin²(5*2π+π+105°)=√32-√128sin²(π/2+15)=√32-√128cos²(15)=√32-√128(√(1+cos30)/2)²=√32-√128*(1+√3/2)/2=√32-√128/2-√(128*3)/4=√32-√32-√24=-√24=-2√6
2) cos^2(8430°)=cos²(8280+150)=cos²(23*2π+π+60)=cos²(π+60)=(-cos60)²=1/4
3) Вычислите: sina, cosa, tga, ctga, если a=67Π/3.
67π/3=22π+π/3=11*2π+π/3
sin(11*2π+π/3)=sin(π/3)=√3/2
cos(
(11*2π+π/3)=cos(π/3)=1/2
tg(67Π/3)=√3/2:1/2=√3
ctg(
67Π/3)=1/2:
√3/2=1/√3=√3/3
2 votes
Thanks 1
Lifa
Помоги пожалуйста здесь тоже спасибо огрлмноеhttp://znanija.com/task/17121859
Lifa
http://znanija.com/task/17122023
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Answers & Comments
Verified answer
1) √32-√128sin^2(2085°)=√32-√128sin²(1800+285°)=√32-√128sin²(5*360+180+105°)=√32-√128sin²(5*2π+π+105°)=√32-√128sin²(π/2+15)=√32-√128cos²(15)=√32-√128(√(1+cos30)/2)²=√32-√128*(1+√3/2)/2=√32-√128/2-√(128*3)/4=√32-√32-√24=-√24=-2√62) cos^2(8430°)=cos²(8280+150)=cos²(23*2π+π+60)=cos²(π+60)=(-cos60)²=1/4
3) Вычислите: sina, cosa, tga, ctga, если a=67Π/3.
67π/3=22π+π/3=11*2π+π/3
sin(11*2π+π/3)=sin(π/3)=√3/2
cos((11*2π+π/3)=cos(π/3)=1/2
tg(67Π/3)=√3/2:1/2=√3
ctg(67Π/3)=1/2:√3/2=1/√3=√3/3