高中数学题，三角函数，谁会进来做一下，来个过程
\u9ad8\u4e2d\u6570\u5b66\u4e09\u89d2\u51fd\u6570\u6d4b\u8bd5\u9898\u7b2c\u4e00\u9898\uff1a\uff081\uff09 f(x)=\uff08\u6839\u53f73/2\uff09sin2x-(cosx)^2-1/2
=\uff08\u6839\u53f73/2\uff09-(1+cos2x)/2-1/2 \uff08\u7531cos2x=2\uff08cosx\uff09^2-1\u5f97\uff09
= (\u6839\u53f73/2)sin2x-(1/2)cos2x-1
= sin2x\u00b7cos(\u03c0/6)-cos2x\u00b7sin(\u03c0/6)-1
= sin(2x-\u03c0/6)-1 \uff08\u6b63\u5f26\u548c\u5dee\u516c\u5f0f\uff09
\u56e0\u4e3a-1\u300a sin(2x-\u03c0/6)\u300a1\uff0c\u6240\u4ee5f(x)\u7684\u6700\u5c0f\u503c\u4e3a-1-1=-2
\u6700\u5c0f\u6b63\u5468\u671fT=2\u03c0/2x=\u03c0
(2)\u56e0\u4e3a\u5411\u91cfm\u4e0en\u5171\u7ebf\uff0c\u6545\u67092sinA-sinB=0\uff0c
\uff08\u82e5\u5411\u91cfa=\uff08m\uff0cn\uff09\u4e0e\u5411\u91cfb=\uff08p\uff0cq\uff09\u5171\u7ebf\uff0c\u5219\u6709np-mq=0\uff09
\u6240\u4ee52a=b\uff08\u6b63\u5f26\u5b9a\u7406\uff09
\u53c8f(C)=sin(2C-\u03c0/6)-1=0\uff0c\u53ef\u89e3\u5f97C=\u03c0/3
\u4f59\u5f26\u5b9a\u7406c^2=a^2+b^2-2abcosC\uff0c\u5c06c=\u6839\u53f73\uff0cC=\u03c0/3\u53ca2a=b\u5e26\u5165\u53ef\u89e3\u5f97
a=1\uff0cb=2
\u7b2c\u4e8c\u9898\uff1a(1)f(x)=\uff082\u500d\u6839\u53f73\uff09sin(x/3)\u00b7cos(x/3)-2(sin(x/3))^2
= \u6839\u53f73\u00b7sin(2x/3)+cos(2x/3)-1
=2\u3016\uff08\u6839\u53f73/2\uff09sin(2x/3)+\uff081/2)cos(2x/3)\u3017-1
=2\u3016cos(\u03c0/6)\u00b7sin(2x/3)+sin(\u03c0/6)\u00b7cos(2x/3)\u3017-1
=2sin\u3016(2x/3)+\u03c0/6\u3017-1
\u56e0\u4e3a0\u300ax\u300a\u03c0\uff0c\u6240\u4ee5\u03c0/6\u300a(2x/3)+\u03c0/6\u300a5\u03c0/6\uff0c\u8bbet=(2x/3)+\u03c0/6\uff0c
\u90a3\u4e48\u03c0/6\u300at\u300a5\u03c0/6\uff0c\u753b\u56fe\u53ef\u77e51/2\u300asint\u300a1\uff0c\u56e0\u6b640\u300a2sin\u3016(2x/3)+\u03c0/6\u3017-1\u300a1
\u6545f(x)\u7684\u503c\u57df\u4e3a\u30100\uff0c1\u3011
(2)f(C)=2sin\u3016(2C/3)+\u03c0/6\u3017-1\uff0c\u53ef\u89e3\u5f97C=\u03c0/2\uff0c
\u7531b^2=ac\u77e5\uff08sinB\uff09^2=sinAsinC=sinAsin(\u03c0/2)\uff0c\u6240\u4ee5sinA=(sinB)^2
\u53c8sinA=sin\u3016\u03c0-(C+B)\u3017=sin\uff08\u03c0/2+B)=cosB\uff0c\u6240\u4ee5
cosB=(sinB)^2=1-(cosB)^2\uff0c\u5373(cosB)^2+cosB-1=0\uff0c
\u5f97cosB=\u4e8c\u5206\u4e4b\u6839\u53f75\u51cf1\uff0c\u6545sinA=cosB=\u4e8c\u5206\u4e4b\u6839\u53f75\u51cf1
\uff08\u3016\uff08\u6839\u53f75\uff09-1\u3017/2)
\u89d2a\u7684\u6b63\u5f26\u7ebf\u548c\u4f59\u5f26\u7ebf\u662f\u65b9\u5411\u76f8\u53cd\uff0c\u957f\u5ea6\u76f8\u7b49\u7684\u6709\u5411\u7ebf\u6bb5,\u5373sin\uff08x\uff09=-cos\uff08x\uff09
\u6545sin(x)+cos(x)=0;\u89e3\u5f97x=-\u03c0/4\u62163\u03c0/4\uff0c\u5373\u5728\u7b2c\u4e8c\u3001\u7b2c\u56db\u8c61\u9650\u7684\u89d2\u5e73\u5206\u7ebf\u4e0a\u3002
（1）
sin20°·cos100°-cos170°·sin70°
= sin20°·cos(90°+10°)-cos(180°-10°)·sin(90°-20°)
= -sin20°cos10°+cos10°sin20°
= 0
（2）tan（α+四分之派）=3 求α
即[tanα+tan(π/4)]/[1-tanαtan(π/4)]=3
即 (tanα+1)/(1-tanα)=3
∴ tanα+1=3-3tanα
∴ 4tanα=2
∴ tanα=1/2
∴ α不是特殊角
∴ α=kπ+arctan(1/2),k∈Z
(1)
cos100°=cos(180°-80°)=-cos80°=-sin10°
cos170°=cos(180°-10°)=-cos10°
sin20°·cos100°-cos170°·sin70°=-sin20°sin10°-(-cos10°)sin70°
=-sin20°sin10°+cos10°cos20°=cos(20°+10°)=cos30°=√3/2
(2)
tan（α+π/4）=3
(tanα+tanπ/4)/(1-tanα*tanπ/4)=3
即
1+tanα=3*(1-tanα)
tanα=1/2
α=kπ+arctan(1/2) k为任意整数
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解：
（1）
根据sin(a-b)=sinacosb-cosasinb得
原式=-sin20°cos80°＋cos10°sin70°
=cos10°sin70°-sin10°cos70°
=sin(70°-10°)
=sin60°
=√3/2
（2）
tan(α+π/4)=3
(tanα+tanπ/4)/(1-tanαtanπ/4)=3
(tanα+1)/(1-tanα)=3
tanα+1=3-3tanα
4tanα=2
tanα=1/2
α=arctan(1/2)
（1）、sin20°·cos100°-cos170°·sin70°
=sin20°*(-sin10°)-(-cos10°)*cos20°
=cos(10°+20°)
=v3/2；
（2）、tan(a+π/4)
=(tana+tanπ/4)/(1-tanatanπ/4)
=(tana+1)/(1-tana)=3，
——》tana=1/2，
——》a=kπ+arctan1/2，k∈Z。
利用sin与cos之间的关系加减pi/2
然后利用和差化积公式
转化成sin/cos乘开再利用积化和差
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