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kirrito228
@kirrito228
August 2022
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arctg(x)=p/24 если х=√6-√3+√2-2
и
arctg(x)=p/12 если х=(√3-1)/(√3+1)
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армения20171
1)х=√6-√3/(√2-2)=√3(√2-1)-√2(√2-1)=
(√2-1)(√3-√2)
tgπ/24=(√2-1)(√3-√2)
2)x=(√3-1)/(√3+1)=(√3-1)²/(3-1)=(3-2√3+1)/2
=(4-2√3)/2=2-√3
arctg(2-√3)=π/12
tgπ/12=2-√3
(1-cosπ/6)/sinπ/6=(1-√3/2)/(1/2)=
(2-√3):1/2=2-√3
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Answers & Comments
(√2-1)(√3-√2)
tgπ/24=(√2-1)(√3-√2)
2)x=(√3-1)/(√3+1)=(√3-1)²/(3-1)=(3-2√3+1)/2
=(4-2√3)/2=2-√3
arctg(2-√3)=π/12
tgπ/12=2-√3
(1-cosπ/6)/sinπ/6=(1-√3/2)/(1/2)=
(2-√3):1/2=2-√3