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logangeor
@logangeor
August 2022
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помогите решить формулы приведения
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sedinalana
1
(cos²a-sin²a):(ctg²a-tg²a)=(cos²a-sin²a):(cos²a/sin²a-sin²a/cos²a)=
=(cos²a-sin²a)*sin²acos²a/[(cos²a-sin²a)(cos²a+sin²a)]=sin²acos²a=1/4*sin²2a
2
[(1+ctg²a)*sin²a]/[(1+tg²a)*(cos²a-1)=[1/sin²a *sin²a]/(1/cos²a*(-sin²a)]=
=1:(-tg²a)=-ctg²a
3
sina=√(1-cos²a)=√(1-0,64)=√0,36=0,6
tga=sina/cosa=-0,6/0,8=-3/4
ctga=1:tga=-4/3
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Answers & Comments
(cos²a-sin²a):(ctg²a-tg²a)=(cos²a-sin²a):(cos²a/sin²a-sin²a/cos²a)=
=(cos²a-sin²a)*sin²acos²a/[(cos²a-sin²a)(cos²a+sin²a)]=sin²acos²a=1/4*sin²2a
2
[(1+ctg²a)*sin²a]/[(1+tg²a)*(cos²a-1)=[1/sin²a *sin²a]/(1/cos²a*(-sin²a)]=
=1:(-tg²a)=-ctg²a
3
sina=√(1-cos²a)=√(1-0,64)=√0,36=0,6
tga=sina/cosa=-0,6/0,8=-3/4
ctga=1:tga=-4/3