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khokhlovakater
@khokhlovakater
June 2021
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СРОЧНО!АЛГЕБРА 99БДАЮ
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sedinalana
2
sin²2a+2sin2acos2a+cos²2a-sin²2a-cos²2a=2sin2acos2a=sin4a
2
sina=√(1-cos²a)=√(1-225/289)=√(64/289)=8/17
tga=sina/cosa=8/17:15/17=8/17*17/15=8/15
ctga=1/tga=1:8/15=15/8
4
(cosφcos2x+sinφsin2x-sinφsin2x)/(sinφsin2x+cosφcos2x-sinφsin2x)=
=cosφcos2x/cosφcos2x=1
5
ctga=1/2
tga=1:ctga=1:1/2=2
tg2a=2tga/(1-tgΑa)=2*2/(1-4)=-4/3
tg(π/4+2a)=(tgπ/4+tg2a)/(1-tg2a*tgπ/4)=(1-4/3):(1+4/3*1)=
=-1/3:7/3=-1/3*3/7=-1/7
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Answers & Comments
sin²2a+2sin2acos2a+cos²2a-sin²2a-cos²2a=2sin2acos2a=sin4a
2
sina=√(1-cos²a)=√(1-225/289)=√(64/289)=8/17
tga=sina/cosa=8/17:15/17=8/17*17/15=8/15
ctga=1/tga=1:8/15=15/8
4
(cosφcos2x+sinφsin2x-sinφsin2x)/(sinφsin2x+cosφcos2x-sinφsin2x)=
=cosφcos2x/cosφcos2x=1
5
ctga=1/2
tga=1:ctga=1:1/2=2
tg2a=2tga/(1-tgΑa)=2*2/(1-4)=-4/3
tg(π/4+2a)=(tgπ/4+tg2a)/(1-tg2a*tgπ/4)=(1-4/3):(1+4/3*1)=
=-1/3:7/3=-1/3*3/7=-1/7