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Alicia15074
@Alicia15074
July 2022
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Cosα=0,6 α∈(3π/2;2π) ⇒ sinα= -0,8 ( sin²α+cos²α=1)
tg(π/3+α)= [tg(π/3)+tgα]/(1-tg(π/3)·tgα) =[√3+sinα/cosα]/[1-(√3)·(sinα/cosα)]=
=[(√3)·cosα+sinα]/[1·cosα-(√3)·(sinα)]=[0,6·(√3)-0,8]/[0,6+0,8·(√3)]=
=[2(3√3-4)]/[2(3+4√3)]=[(3√3-4)·(3-4√3)]/[(3+4√3)·(3-4√3)]=
=[(3√3-4)·(3-4√3)]/(9-48)=[9√3-12-36+16√3]/(-39)=[25√3-48]/(-39)=
=[48-25√3]/39
tg(π/3)=√3
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Answers & Comments
Verified answer
Cosα=0,6 α∈(3π/2;2π) ⇒ sinα= -0,8 ( sin²α+cos²α=1)tg(π/3+α)= [tg(π/3)+tgα]/(1-tg(π/3)·tgα) =[√3+sinα/cosα]/[1-(√3)·(sinα/cosα)]=
=[(√3)·cosα+sinα]/[1·cosα-(√3)·(sinα)]=[0,6·(√3)-0,8]/[0,6+0,8·(√3)]=
=[2(3√3-4)]/[2(3+4√3)]=[(3√3-4)·(3-4√3)]/[(3+4√3)·(3-4√3)]=
=[(3√3-4)·(3-4√3)]/(9-48)=[9√3-12-36+16√3]/(-39)=[25√3-48]/(-39)=
=[48-25√3]/39
tg(π/3)=√3