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Emurgaa
@Emurgaa
August 2022
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1)Докажите тождество (sina+sin3a)/(cosa+cos3a)=tg2a;
2)Докажите тождество (sina+sin4a)/(cos2a-cos4a)=ctga;
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hlopushinairina
1) (sinα+sin3α)/(cosα+cos3α)=tg2α;
(sinα+sin3α)/(cosα+cos3α)=
=2sin((α+3α)/2)cos((α-3α)/2)/2cos((α+3α)/2)cos((α-3α)/2)=
=sin2α/cos2α=tg2α;
tg2α=g2α.
2) (sinα+sin4α)/(cos2α-cos4α)=ctgα;
(sinα+sin4α)/(cos2α-cos4α)=
=2sin((α+4α)/2)cos((α-4α)/2)/(-2)·sin((2α-4α)/2)sin((2α+4α)/2)=
=-sin(5α/2)cos(-3α/2)/sin(-α)sin3α≠ctgα
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Answers & Comments
(sinα+sin3α)/(cosα+cos3α)=
=2sin((α+3α)/2)cos((α-3α)/2)/2cos((α+3α)/2)cos((α-3α)/2)=
=sin2α/cos2α=tg2α;
tg2α=g2α.
2) (sinα+sin4α)/(cos2α-cos4α)=ctgα;
(sinα+sin4α)/(cos2α-cos4α)=
=2sin((α+4α)/2)cos((α-4α)/2)/(-2)·sin((2α-4α)/2)sin((2α+4α)/2)=
=-sin(5α/2)cos(-3α/2)/sin(-α)sin3α≠ctgα