Question

Как выразить tg 22 градусов 30 минут? ctg 22 градусов 30 минут?

Answer 1

tg (22°30') > 0 ;  сtg (22°30') > 0 - первая четверть

$tg (22textdegree 30')=sqrt{dfrac{1-cos (45textdegree)}{1+cos (45textdegree)}}=sqrtdfrac{1-frac{sqrt2}2}{1+frac{sqrt2}2}}=\\\=sqrt{dfrac{2-sqrt2}{2+sqrt2}}=sqrt{dfrac{(2-sqrt2)^2}{(2+sqrt2)cdot (2-sqrt2)}}=\\\=sqrt{dfrac{4-4sqrt2+2}{2}}boldsymbol{=sqrt{3-2sqrt2}}$

$ctg (22textdegree 30')=sqrt{dfrac{1+cos (45textdegree)}{1-cos (45textdegree)}}=sqrtdfrac{1+frac{sqrt2}2}{1-frac{sqrt2}2}}=\\\=sqrt{dfrac{2+sqrt2}{2-sqrt2}}=sqrt{dfrac{(2+sqrt2)^2}{(2-sqrt2)cdot (2+sqrt2)}}=\\\=sqrt{dfrac{4+4sqrt2+2}{2}}boldsymbol{=sqrt{3+2sqrt2}}$

Проверка :

$tg (22textdegree 30')cdot ctg (22textdegree 30')=\\=sqrt{3-2sqrt2}cdot sqrt{3+2sqrt2}=3^2-(2sqrt 2)^2=1$

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Использованы формулы

$tg^2dfrac {alpha }2=dfrac{1-cos alpha }{1+ cos alpha }\\ctg^2dfrac {alpha }2=dfrac{1+cos alpha }{1- cos alpha }\\a^2-b^2=(a-b)(a+b)$