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Article #14010Re: a simple trigonometric problem
From: H'Osamu
Date: Sat, 13 Jan 1996 00:00
Date: Sat, 13 Jan 1996 00:00
31 lines
1759 bytes
1759 bytes
>I have a simple problem about trigonometric function of angle. >Could anyone tell me why tan(theta) = (3^1/2)/3, is not 1/(3^1/2). >If anyone can help me to solve it, please sell the e-mail to directly. >I am looking forward someone feed back. > >k.k. A=3^1/2 $@$r(J 3 $@$N@5$NJ?J}:,$H$9$l$P(J$@!$(J (3^1/2)/3 = A/(A*A) = (A/A)/A = 1/A = 1/(3^1/2), $@$H$J$C$F0l7oMnCe$G$9$M!%(J $@!t(J $@?t3X$NEz0F$G$O!$J,Jl(J$@$rM-M}2=$7$F$*$+$J$$$H8:E@$5$l$k$+$bCN$l$^$;$s!%(J $@!t(J $@$J$^$8M-M}2=$J$I$9$k$H!$$+$($C$F8+1I$($,0-$/$J$C$F(J$@$7$^$&>l9g$,B?$$(J $@!t(J $@$G$9$1$l$I$b!%(J $@>e5-$G$O0l1~!$(J3^1/2 = 3^(1/2) = sqrt(3) $@$H$7$F07$$$^$7$?!%$7$+$7!$(J $@QQ>h7W;;$,;MB'7W;;$KM%@h$9$k?tCM7W;;$N@$3&$G$O!$(J3^1/2 $@$H(J 3^(1/2) $@$H$r(J $@6hJL$9$kI,MW$,$"$j$^$9!%(JFORTRAN $@$N$h$&$K!$@0?t7W;;$H<B?t7W;;$H$r6hJL$9$k(J $@>l9g$b$"$j$^$9!%$=$3$X$5$i$K!$L58B>.?t$rM-8B7e$GBG$A@Z$C$?(J$@$3$H$K$h$k(J $@8m:9$,Mm$^$j$^$9!%(J $@!D$F$J$o$1$G!$(J >Could anyone tell me why tan(theta) = (3^1/2)/3, is not 1/(3^1/2). $@$H$$$&$3$H$,5/$3$C$F$b!$JL$KIT;W5D$G$O$J$$$N$G$9$M!%(J -- HOSHINO.Osamu@EIC.ERI.TODAY TEL: +81-3-3812-2111 Ex.5774 Internet: o3m@eri.u-tokyo.ac.jp WWW: http://www.eri.u-tokyo.ac.jp/~o3m/
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X-Google-Language: Japanese,JIS X-Google-Thread: f4956,9e8f56e8937629f0,start X-Google-Attributes: gidf4956,public From: o3m@eri.u-tokyo.ac.jp (H'Osamu) Subject: Re: a simple trigonometric problem Date: 1996/01/13 Message-ID: <O3M.96Jan14012931@abe-sun.eri.u-tokyo.ac.jp> X-Deja-AN: 135044766 references: <4d85jq$7ff@news.asiaonline.net> organization: Earthquake Research Institute, University of Tokyo, Tokyo, Japan. reply-to: o3m@eri.u-tokyo.ac.jp newsgroups: fj.sci.math