求题——高中三角函数
\u9ad8\u4e2d\u4e09\u89d2\u51fd\u6570\u7ec3\u4e60\u9898y=sin2x+3sinxcosx+4cos2x
=sin2x+(3/2)sin2x+4cos2x
=(5/2)sin2x+4cos2x
\u6700\u5927\u503c\u4e3a\u221a[(5/2)^2+4^2]=\u221a89/2
\u6700\u5c0f\u503c\u4e3a-\u221a89/2
f(x)=2sinxf(x)=2sinxcos^2\u03c6/2+cosxsin\u03c6-sinx
=2cos\u03c6/2(sinxcos\u03c6/2+sin\u03c6/2cosx)-sinx
\u4ee3\u5165X=\u03c0
\u5f97 -2sin\u03c6/2*cos\u03c6/2=-2sin\u03c6
\u6240\u4ee5\u03c6=\u03c0/2\u65f6\u6709\u6700\u5c0f\u503c-2
\uff082\uff09\u7531\u4e00\u5f97\u4ee3\u5165\u5f97cosA=\u221a3/2\u6240\u4ee5A\u7b49\u4e8e30\u5ea6\uff08\u56e0\u4e3ab>a,\u6240\u4ee5\u6392\u9664A=150)
\u7136\u540e\u6b63\u4f59\u5f26\u5b9a\u7406\u56e0\u8be5\u4f1a\u4e86\u5427
sinBsin(180-A-B)=(1+cosA)/2
2sinBsin(A+B)=1+cosA
2sinB(sinAcosB+cosAsinB)=1+cosA
sin2BsinA+2cosAsin²B-cosA-1=0
sin2BsinA+cosA(2sin²B-1)=1
sin2BsinA-cosAcos2B=1
cos2BcosA-sin2BsinA=-1
cos(2B+A)=-1
因为A，B是三角形内角
2B+A=180
因为A+B+C=180
所以B=C
三角形ABC是等腰三角形
计算cos20°-cos40°+cos60°+cos100°
=cos20°-cos40°-cos80°+1/2
=cos20°-cos40°-cos(20°+60°)+1/2
=cos20°-cos40°+sin20°sin60°-cos20°cos60°+1/2
=cos20°-cos40°-1/2cos20°+√3/2sin20°+1/2
=1/2cos20°+√3/2sin20°-cos40°+1/2
=cos(60°-20°)-cos40°+1/2
=cos40°-cos40°+1/2
=1/2
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