求解三角函数
\u6c42\u89e3\uff01\uff01\u4e09\u89d2\u51fd\u6570~1 sin\u03b1=3/5 cos\u03b1=-4/5 sin\u65b9\u03b1/cos\u65b9\u03b1=9/16
2 sin \u03b1+cos \u03b1=1/5 \uff08 sin \u03b1+ cos \u03b1\uff09\u5e73\u65b9=1+2 sin \u03b1*cos \u03b1 \u6240\u4ee52 sin \u03b1*cos \u03b1=24/25
\u6839\u636e\u500d\u89d2\u516c\u5f0fsin 2\u03b1=2 sin \u03b1*cos \u03b1=24/25 \u6709\u56e0\u4e3a\u03c0/2\u5c0f\u4e8e\u7b49\u4e8e\u03b1\u5c0f\u4e8e\u7b49\u4e8e3\u03c0/4 cos2\u03b1=-7/25
3 sin\u03b1=3/5 (0,\u03c0/2) \u6240\u4ee5cos \u03b1=4/5 \u6240\u4ee5 cos(\u03b1+\u03c0/4)=cos \u03b1 *cos\uff08\u03c0/4\uff09-sin \u03b1*sin\uff08\u03c0/4\uff09=\u6839\u53f72 /10
\u6839\u53f72\u500d\u7684cos(\u03b1+\u03c0/4)=1/5
\u516c\u5f0f\uff1aasinx+bcosx=\u221a(a²+b²)*[sin(x+m)]
\u5176\u4e2d\u7684m, \u6709tanm=b/a
\u4f8b\u9898\u4e2d\u7684\u221a(a²+b²)=\u221a2
tanm=1, m= \u03c0/4
原式可以写为[cosx+sinx]/[cosx-sinx]=2011,
整理可得tanx=2010/2012
tan(x+225°)=tan(180+45+x)=tan(x+45)
tan(x+45)=[tanx+tan45]/[1-tanx*tan45]=(1+2010/2012)/(1-2010/2012)=2011
原式={cos(x)+sin(x)}/{cos(x)-sin(x)}=2011
tan(x+225)=sin(x+225)/cos(x+225)={sinxcos225-cosxsin225}/{cosxcos225+sinxsin225}={sinx-cosx}/{cosx+sinx}= -1/2011
化简的(cosx+sinx)/(cosx-sinx)=2011
所以 (1+tanx)/(1-tanx)=2011
有
tan(x+225°)=tan(x+45°)=2011
(cosx+sinx)/(cosx-sin(-x))=2011
tan(x+225)=tan(x-45)
接下来
你自己根据公式再算吧
原式可化为:
(cosx+sinx)/(cosx-sinx)=2011
由上式可知 cosx-sinx!=0,而且cosx!=0,
(cosx+sinx-sinx+sinx)/(cosx-sinx)=2011
1+2*sinx/(cosx-sinx)=2011
sinx/(cosx-sinx)=1005
1006*sinx=1005*cosx (cosx!=0)
tanx=1005/1006
tan(x+225°)=tan(x+180°+45°)=tan(x+45°)=(tanx+tan45°)/(1-tanx*tan45°)
=2011
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