计算:tan 72度，用根号表示？ 用三角函数求tan72°和tan36°，要过程不要解释和公式...

tan72\u5ea6\u7b49\u4e8e\u6839\u53f7\u4e0b5+2\u500d\u68395\u5bf9\u5417\uff1f

\u5bf9\uff0csin18\u00b0=\uff08\u221a5-1)/4,
cos18\u00b0=\u221a1-\u3010\u221a5-1)/4\u3011^2=\u221a(10+2\u221a5)/4,
tan72\u5ea6=sin72\u00b0/cos72\u00b0=cos18\u00b0/sin18\u00b0=[\u221a(10+2\u221a5)/4]/[\uff08\u221a5-1)/4]
=\u221a[(10+2\u221a5)/(6-2\u221a5)]=\u221a(5+2\u221a5)

\u5982\u56fe\u8bbe\u5b9a\u25b3ABC
\u4f5c AH\u22a5BC \u4e8e H \uff0c \u5404\u4e2a\u89d2\u53ef\u7531 \u4e09\u89d2\u5f62 \u5185\u89d2 \u548c \u539f\u7406 \u9010\u6b65\u7b97\u51fa BC=BF=AF
\u8bbe BC=BF=AF=R

\u56e0 \u25b3ABC \u76f8\u4f3c\u4e8e \u25b3CDF =\u3009 DF/R =CF /\uff08R + CF\uff09 \u2460
\u53c8 \u56e0 DC= BD = CF =\u3009 DF + CF = R \u2461
\u8bbe CF = x
\u7531 \u2460\u2461 =\u3009x² + Rx - R² = 0
\u89e3\u4e8c\u5143\u4e00\u6b21\u65b9\u7a0b \u5f97 x = \uff08\u221a5 -1\uff09/ \uff082R\uff09
\u7531 CD = x =\uff08\u221a5 -1\uff09/ \uff082R\uff09 \u548c CH = R/2 \u5f97 DH = \u221a(5-2\u221a5) R / 2

tan36\u00b0= DH / HC = \u221a(5-2\u221a5) \u2248 0.73

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\u7531 GF / HC = AF / AC =\u3009 GF = AF * HC / AC = R * (R/2) / (R+x) = R / (\u221a5 +1)
\u90a3\u4e48 AG = \u221a(5+2\u221a5) R / (\u221a5 +1)
tan72\u00b0 = AG / GF = \u221a(5+2\u221a5) \u2248 3.08

tan72°=√(5+2√5).

作一个顶角36度,底角72度的等腰三角形ABC,〈A=36度,〈B=〈C=72度,
作〈B平分线BD,交AC于D,
则三角形BDC相似于三角形ABC,
设CD=x,
设BC=1,BD=BC=AD=1,BC/AC=CD/BC,
(1+x)*x=1,
x=(√5-1）/2,
AC=CD+AD=(√5+1）/2,
作AE⊥BC,垂足E,
〈EAC=18度,
sin18°=CE/AC=（1/2）/[(√5+1）/2]=（√5-1）/4,
cos72°=sin18°=（√5-1）/4,
AE=√（AC^2-CE^2)=√(5+2√5)/2,
sin72°=AE/AC=)=√(10+2√5)/4,
tan72°=AE/CE=√(5+2√5).

tan72°=√(5+2√5).作一个顶角36度,底角72度的等腰三角形ABC,〈A=36度,〈B=〈C=72度,作〈B平分线BD,交AC于D,则三角形BDC相似于三角形ABC,设CD=x,设BC=1,BD=BC=AD=1,BC/AC=CD/BC,(1+x)*x=1,x...

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