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romikkjh
@romikkjh
August 2022
1
10
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cosA+sinA=1
A-?
решите 3 способами
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sedinalana
CosA+cos(π/2-A)=1
2cosπ/4cos(A-π/4)=1
2*√2/2cos(A-π/4)=1
cos(A-π/4)=1/√2
A-π/4=-π/4+2πk U A-π/4=π/4+2πk
A=2πk U A=π/2+2πk,k∈z
1-2sin²A/2+2sinA/2*cosA/2=1
2sinA/2*(cosA/2-sinA/2)=0
sinA/2=0⇒a/2=πk⇒a=2πk
cosA/2-sinA/2=0/cosA/2
1-tgA/2=0⇒tgA/2=1⇒A/2=π/4+πk⇒A=π/2+2πk,k∈z
sin(π/2-A)+sinA=1
2sinπ/4*cos(x-π/4)=1
2*√2/2cos(A-π/4)=1
cos(A-π/4)=1/√2
A-π/4=-π/4+2πk U A-π/4=π/4+2πk
A=2πk U A=π/2+2πk,k∈z
1 votes
Thanks 1
romikkjh
cosA+cos(π/2-A)=1
romikkjh
а откуда взялось (π/2-A)
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Answers & Comments
2cosπ/4cos(A-π/4)=1
2*√2/2cos(A-π/4)=1
cos(A-π/4)=1/√2
A-π/4=-π/4+2πk U A-π/4=π/4+2πk
A=2πk U A=π/2+2πk,k∈z
1-2sin²A/2+2sinA/2*cosA/2=1
2sinA/2*(cosA/2-sinA/2)=0
sinA/2=0⇒a/2=πk⇒a=2πk
cosA/2-sinA/2=0/cosA/2
1-tgA/2=0⇒tgA/2=1⇒A/2=π/4+πk⇒A=π/2+2πk,k∈z
sin(π/2-A)+sinA=1
2sinπ/4*cos(x-π/4)=1
2*√2/2cos(A-π/4)=1
cos(A-π/4)=1/√2
A-π/4=-π/4+2πk U A-π/4=π/4+2πk
A=2πk U A=π/2+2πk,k∈z