三角函数证明题
\u4e09\u89d2\u51fd\u6570\u8bc1\u660e\u9898\u73b0\u5bf9\uff081+sinA/2\uff09/ \uff082cosA/2\uff09\u5316\u7b80
\u4ee4tanA/4=t
\u5219\u7531\u4e07\u80fd\u516c\u5f0f\uff1a
\uff081+sinA/2\uff09/ \uff082cosA/2\uff09
=[1+2t/\uff081+t^2)]/[(1-t^2)/(2+t^2]
=(1+t)^2/(2-2t^2)
=(1+t)/(2-2t)
=2*{\uff08tan\u03c0/4+t)/(1-tan\u03c0/4*t)
=2*tan(\u03c0/4+A/2\uff09=1/2*cot(\u03c0/2-A/2\uff09
\u6545\u539f\u4e0d\u7b49\u5f0f\u5373\u8bc1\u660e\uff1aa/R>=2*tan(\u03c0/2-A/2\uff09
\u8fde\u63a5\uff1aBO,CO,DO
\u5219\u6709DO\u22a5BC DO=R
\u4ee4\u2220BOD=\u03b1 \u2220COD=\u03b2
\u03b1\u03b2\u5747\u4e3a\u9510\u89d2
\u7531\u4e8eBO\u5e73\u5206\u2220EOD
CO\u5e73\u5206\u2220FOD
OE\u22a5EA
OF\u22a5FA
\u6545\u6709\u03b1+\u03b2=\uff08\u03c0-A\uff09/2
\u540c\u65f6a=BD+DC=Rtan\u03b1+Rtan\u03b2=R\uff08tan\u03b1+tan\u03b2\uff09
\u6545a/R=tan\u03b1+tan\u03b2
\u73b0\u8bc1\u660e\uff1a2tan\u3010\uff08\u03b1+\u03b2\uff09/2\u3011<=tan\u03b1+tan\u03b2
\u770b\u8fd9\u91cc\u5427\uff1ahttp://zhidao.baidu.com/question/61537776.html?si=1
\u6545a/R=tan\u03b1+tan\u03b2>=2tan\u3010(\u03b1+\u03b2\uff09/2\u3011=2tan(\u03c0/2-A/2\uff09
\u539f\u547d\u9898\u5f97\u8bc1
\u6211\u5df2\u7ecf\u6700\u5927\u7a0b\u5ea6\u7684\u5199\u5f97\u8be6\u7ec6\u4e86\uff0c\u8fd8\u6709\u4e0d\u61c2\u5f97\u8bdd\u95ee\u6211\u54af
\u4ee5\u4e0a
\u30101\u3011\u2235x+y=90º,\u4e140\uff1cx\uff1c90º,\u22340\uff1cy\uff1c90º,\u4e14y=90º-x.\u2234tany=tan(90º-x)=cotx=1/tanx.\u663e\u7136\uff0ctanx\uff1e0\u4e14tany=1/tanx\uff1e0.\u30102\u3011\u7531\u201c\u57fa\u672c\u4e0d\u7b49\u5f0f\u201d\u53ef\u77e5\uff0ctanx+tany=tanx+(1/tanx)\u22652.\u7b49\u53f7\u4ec5\u5f53x=y=45º\u65f6\u53d6\u5f97\u3002\u2234tanx+tany\u22652
sin3a=sin(a+2a)=sinacos2a+cosasin2a=sina(1-2sin²a)+cosa*2sinacosa
=sina-2sin³a+2sinacos²a
=sina-2sin³a+2sina(1-sin²a)
=sina-2sin³a+2sina-2sin³a
=3sina-4sin³a
cos3a=cos(2a+a)=cos2acosa-sin2asina
=(2cos²a-1)cosa-2sinacosa*sina
=2cos³a-cosa-2sin²acosa
=2cos³a-cosa-2(1-cos²a)cosa
=2cos³a-cosa-2cosa+2cos³a
=4cos³a-3cosa
绛旓細锛坰in伪+cos伪锛²-锛坰in伪-cos伪锛²=1+2sin伪cos伪-1+2sin伪cos伪 =4sin伪cos伪 sin伪cos伪*tan²伪=sin³伪/cos伪=sin伪(1-cos²伪)/cos伪 =tan伪-sin伪cos伪 鍒欏師鍛介鎴愮珛銆
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