∑sina(a的取值范围在0到90度
\u6c42sinAcosA+sinA\u7684\u53d6\u503c\u8303\u56f4\uff08A\u57280\u523090\u5ea6\u4e4b\u95f4)\u662f0<a<\u03c0/2\u5417\uff1f\u5047\u8bbe\u5982\u6b64.
sinacoa+sina=sina(1+cosa)=2sinacos^2(a/2)=4sin(a/2)[cos(a/2)]^3
=4{[1-cos^2(a/2)]cos^2(a/2)cos^2(a/2)cos^2(a/2)}^(1/2)
=(4/\u221a3){[3-3cos^2(a/2)]cos^2(a/2)cos^2(a/2)cos^2(a/2)}^(1/2)
<=(4/\u221a3\uff09[\uff083/4)^4]^(1/2)
=(3/4)\u221a3
\u5f53\u4e14\u4ec5\u5f533-3cos^2(a/2)=cos^2(a/2)\u5373cos(a/2)=\u221a3/2,a=\u03c0/3\u65f6\u7b49\u53f7\u6210\u7acb\u3002
\u6240\u4ee50<a<\u03c0/2\u65f60<sinacosa+sina<=(3/4)\u221a3
\u53e6\u89e3\uff1a\u8bbey=f(a)=sinacosa+sina(0<a<\u03c0/2)
y'=cos^2a-sin^2a-cosa=2cos^2a-cosa-1=2(cosa-1/4)^2-9/8
y'=0\u7684\u6839\u662fa=\u03c0/3
y''=-sina(1+4cosa)<0
\u6240\u4ee5f(\u03c0/3)=(3/4)\u221a3\u662f\u552f\u4e00\u7684\u6781\u5927\u503c\uff0c\u4e5f\u662f\u6700\u5927\u503c
lim(a\u21920\uff09f(a)=0,lim(a\u2192\u03c0/2)f(a)=1
\u6bd4\u8f83(3/4)\u221a3,0,1\u5f970<y<=(3/4)\u221a3
\u7531\u9898:cosA+sinA=tanA
\u5219\u6839\u636e\u5747\u503c\u4e0d\u7b49\u5f0f tanA>=2*\u6839\u53f7\u4e0bsinAsinA=\u6839\u53f7\u4e0b2*sin2A
\u56e0\u4e3a0<2A<180 \u5219tanA\u53ea\u9700\u5927\u4e8e\u7b49\u4e8e\u6839\u53f72
\u5219 arctan\u6839\u53f72=<A<90 \u65e2\u53ef
sina的积分为-cosa，所以∑sina=-cos90-（-cos0）=1
将x轴等分成n份 每个分点求sin值 利用积化和差求和 （乘以sin（π/4n）再除以它） 最后全消掉了 剩两项 令n趋向正无穷求极限 就行了
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