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chieftec31
@chieftec31
August 2022
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sin3a*sin2a-cos3a*cos2a-cos(3п/2-a)
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IrinaKorolevskih
Наверно так
3*2-3*2-3/2=-1,5
1 votes
Thanks 2
DAAAAAGY
((sinα-sin3α)/sin2α)·((cosα-cos3α)/cos2α=
=[2·sin((α-3α)/2)·cos((α+3α)/2)/sin2α]·[-2sin(-α)·sin2α/cos2α]=
=[-2·sinα·cos2α/sin2α]·[2sinα·sin2α/cos2α]=
=(-4)·sin²α·(1/2·sin4α)/(1/2·sin4α)=
=-4sin²α;
1 votes
Thanks 2
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Answers & Comments
3*2-3*2-3/2=-1,5
=[2·sin((α-3α)/2)·cos((α+3α)/2)/sin2α]·[-2sin(-α)·sin2α/cos2α]=
=[-2·sinα·cos2α/sin2α]·[2sinα·sin2α/cos2α]=
=(-4)·sin²α·(1/2·sin4α)/(1/2·sin4α)=
=-4sin²α;