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laurapop42
@laurapop42
July 2022
1
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ВНИМАНИЕ!!!!! 70 БАЛЛОВ Помогите пожалуйста.... напишите в виде произведения: 1)√3-2cosA. 2)2sinA-√3. 3) √2+2cosA. 4)0,5-sinA
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sedinalana
1
√3-2cosA=2(√3/2-cosA)=2(cosπ/6-cosA)=
=-4sin((π/12-A/2)sin(π/12+A/2)
2
2sinA-√3=2(sinA-√3/2)=2(sinA-sinπ/3)=
=4sin(A/2-π/6)cos(A/2+π/6)
3
√2+2cosA=2(√2/2+cosA)=2(cosπ/4+cosA)=
=4cos(π/8+A/2)cos(π/8-A/2)
4
0,5-sinA=sinπ/6-sinA=2sin(π/12-A/2)cos(π/12+A/2)
1 votes
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Answers & Comments
√3-2cosA=2(√3/2-cosA)=2(cosπ/6-cosA)=
=-4sin((π/12-A/2)sin(π/12+A/2)
2
2sinA-√3=2(sinA-√3/2)=2(sinA-sinπ/3)=
=4sin(A/2-π/6)cos(A/2+π/6)
3
√2+2cosA=2(√2/2+cosA)=2(cosπ/4+cosA)=
=4cos(π/8+A/2)cos(π/8-A/2)
4
0,5-sinA=sinπ/6-sinA=2sin(π/12-A/2)cos(π/12+A/2)