tan2x=？？不要万能公式的那个，要用sin和cos表示的。。。 给定tanα/2的值，能求出两个sin和cos值。为什么使用...
\u7528\u4e09\u89d2\u51fd\u6570\u6c42tan72\u00b0\u548ctan36\u00b0\uff0c\u8981\u8fc7\u7a0b\u4e0d\u8981\u89e3\u91ca\u548c\u516c\u5f0f\u7684
\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
\u7ed9\u51fatana/2,\u80fd\u6c42\u51fa\u4e24\u7ec4sina/2,cosa/2,\u4f46\u662f\u53ea\u80fd\u5f97\u51fa\u4e00\u7ec4sina,cosa.\u7528\u4e07\u80fd\u516c\u5f0f\u76f4\u63a5\u6c42\u51fa\u7684\u662fsina,cosa!\u6240\u4ee5\u5e76\u4e0d\u77db\u76fe\uff0c\u4f8b\u5982\uff0ctana/2=1/2,\u90a3\u4e48sina/2=sqrt(5)/5,cosa/2=2sqrt(5)/5;sina/2=-sqrt(5)/5,cosa/2=-2sqrt(5)/5,\u4f46\u662fsina=4/5,cosa=.../
tan2x没有直接导出的公式，只能用sin2x和cos2x来求。sin2x=2sinxcosx,cos2x=cosx平方-sinx平方。用sin2x和cos2x一比就出来了tan2x=sin2x/cos2x
=(2sinxcosx)/((cosx)^2-(sinx)^2)
=2tgx/(1-(tgx)^2)
2sinxcosx/（2cos²x-1）
三八
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绛旓細tan2x=4/3 sin2x=4/5,cos2x=3/5 sin(2x+pie/4)=7sqt(2)/10
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