能帮忙整理一下三角函数的公式吗？就是sin（a）的平方等于什么之类的sin,cos,tan之间的转换。急。谢谢。 sin cos tan之间的关系还有各种公式，麻烦列一下出来...
\u5e2e\u5fd9\u628a\u4e09\u89d2\u51fd\u6570\u516c\u5f0f\u7ed9\u5217\u4e3e\u4e00\u4e0b\uff0c\u8c22\u8c22\u3002\u53c2\u8003\u8d44\u6599http://baike.baidu.com/view/91555.htm
\u5305\u62ec\u4e09\u89d2\u51fd\u6570\u521d\u7b49\u9ad8\u7b49\u516c\u5f0f\u4ee5\u53ca\u77e5\u8bc6\u70b9\u548c\u5404\u79cd\u8ba1\u7b97\u4e60\u9898
\u5404\u79cd\u8003\u8bd5\u8981\u70b9\u548c\u7406\u89e3\u65b9\u5f0f
\u57fa\u672c\u521d\u7b49\u5185\u5bb9
\u5b83\u6709\u516d\u79cd\u57fa\u672c\u51fd\u6570(\u521d\u7b49\u57fa\u672c\u8868\u793a)\uff1a
\u51fd\u6570\u540d \u6b63\u5f26 \u4f59\u5f26 \u6b63\u5207 \u4f59\u5207 \u6b63\u5272 \u4f59\u5272
\u5728\u5e73\u9762\u76f4\u89d2\u5750\u6807\u7cfbxOy\u4e2d\uff0c\u4ece\u70b9O\u5f15\u51fa\u4e00\u6761\u5c04\u7ebfOP\uff0c\u8bbe\u65cb\u8f6c\u89d2\u4e3a\u03b8\uff0c\u8bbeOP=r\uff0cP\u70b9\u7684\u5750\u6807\u4e3a\uff08x\uff0cy\uff09\u6709
\u6b63\u5f26\u51fd\u6570 sin\u03b8=y/r
\u4f59\u5f26\u51fd\u6570 cos\u03b8=x/r
\u6b63\u5207\u51fd\u6570 tan\u03b8=y/x
\u4f59\u5207\u51fd\u6570 cot\u03b8=x/y
\u6b63\u5272\u51fd\u6570 sec\u03b8=r/x
\u4f59\u5272\u51fd\u6570 csc\u03b8=r/y
\uff08\u659c\u8fb9\u4e3ar\uff0c\u5bf9\u8fb9\u4e3ay\uff0c\u90bb\u8fb9\u4e3ax\u3002\uff09
\u4ee5\u53ca\u4e24\u4e2a\u4e0d\u5e38\u7528\uff0c\u5df2\u8d8b\u4e8e\u88ab\u6dd8\u6c70\u7684\u51fd\u6570\uff1a
\u6b63\u77e2\u51fd\u6570 versin\u03b8 =1-cos\u03b8
\u4f59\u77e2\u51fd\u6570 covers\u03b8 =1-sin\u03b8
\u540c\u89d2\u4e09\u89d2\u51fd\u6570\u95f4\u7684\u57fa\u672c\u5173\u7cfb\u5f0f\uff1a
[\u7f16\u8f91\u672c\u6bb5]
\u00b7\u5e73\u65b9\u5173\u7cfb\uff1a
sin^2(\u03b1)+cos^2(\u03b1)=1
tan^2(\u03b1)+1=sec^2(\u03b1)
cot^2(\u03b1)+1=csc^2(\u03b1)
\u00b7\u79ef\u7684\u5173\u7cfb\uff1a
sin\u03b1=tan\u03b1*cos\u03b1
cos\u03b1=cot\u03b1*sin\u03b1
tan\u03b1=sin\u03b1*sec\u03b1
cot\u03b1=cos\u03b1*csc\u03b1
sec\u03b1=tan\u03b1*csc\u03b1
csc\u03b1=sec\u03b1*cot\u03b1
\u00b7\u5012\u6570\u5173\u7cfb\uff1a
tan\u03b1\u00b7cot\u03b1=1
sin\u03b1\u00b7csc\u03b1=1
cos\u03b1\u00b7sec\u03b1=1
\u76f4\u89d2\u4e09\u89d2\u5f62ABC\u4e2d,
\u89d2A\u7684\u6b63\u5f26\u503c\u5c31\u7b49\u4e8e\u89d2A\u7684\u5bf9\u8fb9\u6bd4\u659c\u8fb9,
\u4f59\u5f26\u7b49\u4e8e\u89d2A\u7684\u90bb\u8fb9\u6bd4\u659c\u8fb9
\u6b63\u5207\u7b49\u4e8e\u5bf9\u8fb9\u6bd4\u90bb\u8fb9,
\u00b7\u4e09\u89d2\u51fd\u6570\u6052\u7b49\u53d8\u5f62\u516c\u5f0f
\u00b7\u4e24\u89d2\u548c\u4e0e\u5dee\u7684\u4e09\u89d2\u51fd\u6570\uff1a
cos(\u03b1+\u03b2)=cos\u03b1\u00b7cos\u03b2-sin\u03b1\u00b7sin\u03b2
cos(\u03b1-\u03b2)=cos\u03b1\u00b7cos\u03b2+sin\u03b1\u00b7sin\u03b2
sin(\u03b1\u00b1\u03b2)=sin\u03b1\u00b7cos\u03b2\u00b1cos\u03b1\u00b7sin\u03b2
tan(\u03b1+\u03b2)=(tan\u03b1+tan\u03b2)/(1-tan\u03b1\u00b7tan\u03b2)
tan(\u03b1-\u03b2)=(tan\u03b1-tan\u03b2)/(1+tan\u03b1\u00b7tan\u03b2)
\u00b7\u4e09\u89d2\u548c\u7684\u4e09\u89d2\u51fd\u6570\uff1a
sin(\u03b1+\u03b2+\u03b3)=sin\u03b1\u00b7cos\u03b2\u00b7cos\u03b3+cos\u03b1\u00b7sin\u03b2\u00b7cos\u03b3+cos\u03b1\u00b7cos\u03b2\u00b7sin\u03b3-sin\u03b1\u00b7sin\u03b2\u00b7sin\u03b3
cos(\u03b1+\u03b2+\u03b3)=cos\u03b1\u00b7cos\u03b2\u00b7cos\u03b3-cos\u03b1\u00b7sin\u03b2\u00b7sin\u03b3-sin\u03b1\u00b7cos\u03b2\u00b7sin\u03b3-sin\u03b1\u00b7sin\u03b2\u00b7cos\u03b3
tan(\u03b1+\u03b2+\u03b3)=(tan\u03b1+tan\u03b2+tan\u03b3-tan\u03b1\u00b7tan\u03b2\u00b7tan\u03b3)/(1-tan\u03b1\u00b7tan\u03b2-tan\u03b2\u00b7tan\u03b3-tan\u03b3\u00b7tan\u03b1)
\u00b7\u8f85\u52a9\u89d2\u516c\u5f0f\uff1a
Asin\u03b1+Bcos\u03b1=(A^2+B^2)^(1/2)sin(\u03b1+t)\uff0c\u5176\u4e2d
sint=B/(A^2+B^2)^(1/2)
cost=A/(A^2+B^2)^(1/2)
tant=B/A
Asin\u03b1+Bcos\u03b1=(A^2+B^2)^(1/2)cos(\u03b1-t)\uff0ctant=A/B
\u00b7\u500d\u89d2\u516c\u5f0f\uff1a
sin(2\u03b1)=2sin\u03b1\u00b7cos\u03b1=2/(tan\u03b1+cot\u03b1)
cos(2\u03b1)=cos^2(\u03b1)-sin^2(\u03b1)=2cos^2(\u03b1)-1=1-2sin^2(\u03b1)
tan(2\u03b1)=2tan\u03b1/[1-tan^2(\u03b1)]
\u00b7\u4e09\u500d\u89d2\u516c\u5f0f\uff1a
sin(3\u03b1)=3sin\u03b1-4sin^3(\u03b1)
cos(3\u03b1)=4cos^3(\u03b1)-3cos\u03b1
\u00b7\u534a\u89d2\u516c\u5f0f\uff1a
sin(\u03b1/2)=\u00b1\u221a((1-cos\u03b1)/2)
cos(\u03b1/2)=\u00b1\u221a((1+cos\u03b1)/2)
tan(\u03b1/2)=\u00b1\u221a((1-cos\u03b1)/(1+cos\u03b1))=sin\u03b1/(1+cos\u03b1)=(1-cos\u03b1)/sin\u03b1
\u00b7\u964d\u5e42\u516c\u5f0f
sin^2(\u03b1)=(1-cos(2\u03b1))/2=versin(2\u03b1)/2
cos^2(\u03b1)=(1+cos(2\u03b1))/2=covers(2\u03b1)/2
tan^2(\u03b1)=(1-cos(2\u03b1))/(1+cos(2\u03b1))
\u00b7\u4e07\u80fd\u516c\u5f0f\uff1a
sin\u03b1=2tan(\u03b1/2)/[1+tan^2(\u03b1/2)]
cos\u03b1=[1-tan^2(\u03b1/2)]/[1+tan^2(\u03b1/2)]
tan\u03b1=2tan(\u03b1/2)/[1-tan^2(\u03b1/2)]
\u00b7\u79ef\u5316\u548c\u5dee\u516c\u5f0f\uff1a
sin\u03b1\u00b7cos\u03b2=(1/2)[sin(\u03b1+\u03b2)+sin(\u03b1-\u03b2)]
cos\u03b1\u00b7sin\u03b2=(1/2)[sin(\u03b1+\u03b2)-sin(\u03b1-\u03b2)]
cos\u03b1\u00b7cos\u03b2=(1/2)[cos(\u03b1+\u03b2)+cos(\u03b1-\u03b2)]
sin\u03b1\u00b7sin\u03b2=-(1/2)[cos(\u03b1+\u03b2)-cos(\u03b1-\u03b2)]
\u00b7\u548c\u5dee\u5316\u79ef\u516c\u5f0f\uff1a
sin\u03b1+sin\u03b2=2sin[(\u03b1+\u03b2)/2]cos[(\u03b1-\u03b2)/2]
sin\u03b1-sin\u03b2=2cos[(\u03b1+\u03b2)/2]sin[(\u03b1-\u03b2)/2]
cos\u03b1+cos\u03b2=2cos[(\u03b1+\u03b2)/2]cos[(\u03b1-\u03b2)/2]
cos\u03b1-cos\u03b2=-2sin[(\u03b1+\u03b2)/2]sin[(\u03b1-\u03b2)/2]
\u00b7\u63a8\u5bfc\u516c\u5f0f
tan\u03b1+cot\u03b1=2/sin2\u03b1
tan\u03b1-cot\u03b1=-2cot2\u03b1
1+cos2\u03b1=2cos^2\u03b1
1-cos2\u03b1=2sin^2\u03b1
1+sin\u03b1=(sin\u03b1/2+cos\u03b1/2)^2
\u00b7\u5176\u4ed6\uff1a
sin\u03b1+sin(\u03b1+2\u03c0/n)+sin(\u03b1+2\u03c0*2/n)+sin(\u03b1+2\u03c0*3/n)+\u2026\u2026+sin[\u03b1+2\u03c0*(n-1)/n]=0
cos\u03b1+cos(\u03b1+2\u03c0/n)+cos(\u03b1+2\u03c0*2/n)+cos(\u03b1+2\u03c0*3/n)+\u2026\u2026+cos[\u03b1+2\u03c0*(n-1)/n]=0 \u4ee5\u53ca
sin^2(\u03b1)+sin^2(\u03b1-2\u03c0/3)+sin^2(\u03b1+2\u03c0/3)=3/2
tanAtanBtan(A+B)+tanA+tanB-tan(A+B)=0
\u4e09\u89d2\u51fd\u6570\u7684\u89d2\u5ea6\u6362\u7b97
[\u7f16\u8f91\u672c\u6bb5]
\u516c\u5f0f\u4e00\uff1a
\u8bbe\u03b1\u4e3a\u4efb\u610f\u89d2\uff0c\u7ec8\u8fb9\u76f8\u540c\u7684\u89d2\u7684\u540c\u4e00\u4e09\u89d2\u51fd\u6570\u7684\u503c\u76f8\u7b49\uff1a
sin\uff082k\u03c0\uff0b\u03b1\uff09\uff1dsin\u03b1
cos\uff082k\u03c0\uff0b\u03b1\uff09\uff1dcos\u03b1
tan\uff082k\u03c0\uff0b\u03b1\uff09\uff1dtan\u03b1
cot\uff082k\u03c0\uff0b\u03b1\uff09\uff1dcot\u03b1
\u516c\u5f0f\u4e8c\uff1a
\u8bbe\u03b1\u4e3a\u4efb\u610f\u89d2\uff0c\u03c0+\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff08\u03c0\uff0b\u03b1\uff09\uff1d\uff0dsin\u03b1
cos\uff08\u03c0\uff0b\u03b1\uff09\uff1d\uff0dcos\u03b1
tan\uff08\u03c0\uff0b\u03b1\uff09\uff1dtan\u03b1
cot\uff08\u03c0\uff0b\u03b1\uff09\uff1dcot\u03b1
\u516c\u5f0f\u4e09\uff1a
\u4efb\u610f\u89d2\u03b1\u4e0e -\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff08\uff0d\u03b1\uff09\uff1d\uff0dsin\u03b1
cos\uff08\uff0d\u03b1\uff09\uff1dcos\u03b1
tan\uff08\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1
cot\uff08\uff0d\u03b1\uff09\uff1d\uff0dcot\u03b1
\u516c\u5f0f\u56db\uff1a
\u5229\u7528\u516c\u5f0f\u4e8c\u548c\u516c\u5f0f\u4e09\u53ef\u4ee5\u5f97\u5230\u03c0-\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff08\u03c0\uff0d\u03b1\uff09\uff1dsin\u03b1
cos\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dcos\u03b1
tan\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1
cot\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dcot\u03b1
\u516c\u5f0f\u4e94\uff1a
\u5229\u7528\u516c\u5f0f\u4e00\u548c\u516c\u5f0f\u4e09\u53ef\u4ee5\u5f97\u52302\u03c0-\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff082\u03c0\uff0d\u03b1\uff09\uff1d\uff0dsin\u03b1
cos\uff082\u03c0\uff0d\u03b1\uff09\uff1dcos\u03b1
tan\uff082\u03c0\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1
cot\uff082\u03c0\uff0d\u03b1\uff09\uff1d\uff0dcot\u03b1
\u516c\u5f0f\u516d\uff1a
\u03c0/2\u00b1\u03b1\u53ca3\u03c0/2\u00b1\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff08\u03c0/2\uff0b\u03b1\uff09\uff1dcos\u03b1
cos\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dsin\u03b1
tan\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcot\u03b1
cot\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dtan\u03b1
sin\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dcos\u03b1
cos\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dsin\u03b1
tan\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dcot\u03b1
cot\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dtan\u03b1
sin\uff083\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcos\u03b1
cos\uff083\u03c0/2\uff0b\u03b1\uff09\uff1dsin\u03b1
tan\uff083\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcot\u03b1
cot\uff083\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dtan\u03b1
sin\uff083\u03c0/2\uff0d\u03b1\uff09\uff1d\uff0dcos\u03b1
cos\uff083\u03c0/2\uff0d\u03b1\uff09\uff1d\uff0dsin\u03b1
tan\uff083\u03c0/2\uff0d\u03b1\uff09\uff1dcot\u03b1
cot\uff083\u03c0/2\uff0d\u03b1\uff09\uff1dtan\u03b1
(\u4ee5\u4e0ak\u2208Z)
\u90e8\u5206\u9ad8\u7b49\u5185\u5bb9
[\u7f16\u8f91\u672c\u6bb5]
\u00b7\u9ad8\u7b49\u4ee3\u6570\u4e2d\u4e09\u89d2\u51fd\u6570\u7684\u6307\u6570\u8868\u793a(\u7531\u6cf0\u52d2\u7ea7\u6570\u6613\u5f97)\uff1a
sinx=[e^(ix)-e^(-ix)]/(2i)
cosx=[e^(ix)+e^(-ix)]/2
tanx=[e^(ix)-e^(-ix)]/[ie^(ix)+ie^(-ix)]
\u6cf0\u52d2\u5c55\u5f00\u6709\u65e0\u7a77\u7ea7\u6570\uff0ce^z=exp(z)\uff1d1\uff0bz/1\uff01\uff0bz^2/2\uff01\uff0bz^3/3\uff01\uff0bz^4/4\uff01\uff0b\u2026\uff0bz^n/n\uff01\uff0b\u2026
\u6b64\u65f6\u4e09\u89d2\u51fd\u6570\u5b9a\u4e49\u57df\u5df2\u63a8\u5e7f\u81f3\u6574\u4e2a\u590d\u6570\u96c6\u3002
\u00b7\u4e09\u89d2\u51fd\u6570\u4f5c\u4e3a\u5fae\u5206\u65b9\u7a0b\u7684\u89e3\uff1a
\u5bf9\u4e8e\u5fae\u5206\u65b9\u7a0b\u7ec4 y=-y'';y=y''''\uff0c\u6709\u901a\u89e3Q,\u53ef\u8bc1\u660e
Q=Asinx+Bcosx\uff0c\u56e0\u6b64\u4e5f\u53ef\u4ee5\u4ece\u6b64\u51fa\u53d1\u5b9a\u4e49\u4e09\u89d2\u51fd\u6570\u3002
\u8865\u5145\uff1a\u7531\u76f8\u5e94\u7684\u6307\u6570\u8868\u793a\u6211\u4eec\u53ef\u4ee5\u5b9a\u4e49\u4e00\u79cd\u7c7b\u4f3c\u7684\u51fd\u6570\u2014\u2014\u53cc\u66f2\u51fd\u6570\uff0c\u5176\u62e5\u6709\u5f88\u591a\u4e0e\u4e09\u89d2\u51fd\u6570\u7684\u7c7b\u4f3c\u7684\u6027\u8d28\uff0c\u4e8c\u8005\u76f8\u6620\u6210\u8da3\u3002
\u7279\u6b8a\u4e09\u89d2\u51fd\u6570\u503c
[\u7f16\u8f91\u672c\u6bb5]
a 0` 30` 45` 60` 90`
sina 0 1/2 \u221a2/2 \u221a3/2 1
cosa 1 \u221a3/2 \u221a2/2 1/2 0
tana 0 \u221a3/3 1 \u221a3 None
cota None \u221a3 1 \u221a3/3 0
\u4e09\u89d2\u51fd\u6570\u7684\u8ba1\u7b97
[\u7f16\u8f91\u672c\u6bb5]
\u5e42\u7ea7\u6570
c0+c1x+c2x2+...+cnxn+...=\u2211cnxn (n=0..\u221e)
c0+c1(x-a)+c2(x-a)2+...+cn(x-a)n+...=\u2211cn(x-a)n (n=0..\u221e)
\u5b83\u4eec\u7684\u5404\u9879\u90fd\u662f\u6b63\u6574\u6570\u5e42\u7684\u5e42\u51fd\u6570, \u5176\u4e2dc0,c1,c2,...cn...\u53caa\u90fd\u662f\u5e38\u6570, \u8fd9\u79cd\u7ea7\u6570\u79f0\u4e3a\u5e42\u7ea7\u6570.
\u6cf0\u52d2\u5c55\u5f00\u5f0f(\u5e42\u7ea7\u6570\u5c55\u5f00\u6cd5):
f(x)=f(a)+f'(a)/1!*(x-a)+f''(a)/2!*(x-a)2+...f(n)(a)/n!*(x-a)n+...
\u5b9e\u7528\u5e42\u7ea7\u6570\uff1a
ex = 1+x+x2/2!+x3/3!+...+xn/n!+...
ln(1+x)= x-x2/3+x3/3-...(-1)k-1*xk/k+... (|x|<1)
sin x = x-x3/3!+x5/5!-...(-1)k-1*x2k-1/(2k-1)!+... (-\u221e<x<\u221e)
cos x = 1-x2/2!+x4/4!-...(-1)k*x2k/(2k)!+... (-\u221e<x<\u221e)
arcsin x = x + 1/2*x3/3 + 1*3/(2*4)*x5/5 + ... (|x|<1)
arccos x = \u03c0 - ( x + 1/2*x3/3 + 1*3/(2*4)*x5/5 + ... ) (|x|<1)
arctan x = x - x^3/3 + x^5/5 - ... (x\u22641)
sinh x = x+x3/3!+x5/5!+...(-1)k-1*x2k-1/(2k-1)!+... (-\u221e<x<\u221e)
cosh x = 1+x2/2!+x4/4!+...(-1)k*x2k/(2k)!+... (-\u221e<x<\u221e)
arcsinh x = x - 1/2*x3/3 + 1*3/(2*4)*x5/5 - ... (|x|<1)
arctanh x = x + x^3/3 + x^5/5 + ... (|x|<1)
\u5728\u89e3\u521d\u7b49\u4e09\u89d2\u51fd\u6570\u65f6\uff0c\u53ea\u9700\u8bb0\u4f4f\u516c\u5f0f\u4fbf\u53ef\u8f7b\u677e\u4f5c\u7b54\uff0c\u5728\u7ade\u8d5b\u4e2d\uff0c\u5f80\u5f80\u4f1a\u7528\u5230\u4e0e\u56fe\u50cf\u7ed3\u5408\u7684\u65b9\u6cd5\u6c42\u4e09\u89d2\u51fd\u6570\u503c\u3001\u4e09\u89d2\u51fd\u6570\u4e0d\u7b49\u5f0f\u3001\u9762\u79ef\u7b49\u7b49\u3002
---
\u5085\u7acb\u53f6\u7ea7\u6570(\u4e09\u89d2\u7ea7\u6570)
f(x)=a0/2+\u2211(n=0..\u221e) (ancosnx+bnsinnx)
a0=1/\u03c0\u222b(\u03c0..-\u03c0) (f(x))dx
an=1/\u03c0\u222b(\u03c0..-\u03c0) (f(x)cosnx)dx
bn=1/\u03c0\u222b(\u03c0..-\u03c0) (f(x)sinnx)dx
\u516c\u5f0f\u4e00\uff1a
\u8bbe\u03b1\u4e3a\u4efb\u610f\u89d2\uff0c\u7ec8\u8fb9\u76f8\u540c\u7684\u89d2\u7684\u540c\u4e00\u4e09\u89d2\u51fd\u6570\u7684\u503c\u76f8\u7b49\uff1a
sin\uff082k\u03c0\uff0b\u03b1\uff09\uff1dsin\u03b1
cos\uff082k\u03c0\uff0b\u03b1\uff09\uff1dcos\u03b1
tan\uff082k\u03c0\uff0b\u03b1\uff09\uff1dtan\u03b1
cot\uff082k\u03c0\uff0b\u03b1\uff09\uff1dcot\u03b1
\u516c\u5f0f\u4e8c\uff1a
\u8bbe\u03b1\u4e3a\u4efb\u610f\u89d2\uff0c\u03c0+\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff08\u03c0\uff0b\u03b1\uff09\uff1d\uff0dsin\u03b1
cos\uff08\u03c0\uff0b\u03b1\uff09\uff1d\uff0dcos\u03b1
tan\uff08\u03c0\uff0b\u03b1\uff09\uff1dtan\u03b1
cot\uff08\u03c0\uff0b\u03b1\uff09\uff1dcot\u03b1
\u516c\u5f0f\u4e09\uff1a
\u4efb\u610f\u89d2\u03b1\u4e0e -\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff08\uff0d\u03b1\uff09\uff1d\uff0dsin\u03b1
cos\uff08\uff0d\u03b1\uff09\uff1dcos\u03b1
tan\uff08\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1
cot\uff08\uff0d\u03b1\uff09\uff1d\uff0dcot\u03b1
\u516c\u5f0f\u56db\uff1a
\u5229\u7528\u516c\u5f0f\u4e8c\u548c\u516c\u5f0f\u4e09\u53ef\u4ee5\u5f97\u5230\u03c0-\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff08\u03c0\uff0d\u03b1\uff09\uff1dsin\u03b1
cos\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dcos\u03b1
tan\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1
cot\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dcot\u03b1
\u516c\u5f0f\u4e94\uff1a
\u5229\u7528\u516c\u5f0f\u4e00\u548c\u516c\u5f0f\u4e09\u53ef\u4ee5\u5f97\u52302\u03c0-\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff082\u03c0\uff0d\u03b1\uff09\uff1d\uff0dsin\u03b1
cos\uff082\u03c0\uff0d\u03b1\uff09\uff1dcos\u03b1
tan\uff082\u03c0\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1
cot\uff082\u03c0\uff0d\u03b1\uff09\uff1d\uff0dcot\u03b1
\u516c\u5f0f\u516d\uff1a
\u03c0/2\u00b1\u03b1\u53ca3\u03c0/2\u00b1\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a
sin\uff08\u03c0/2\uff0b\u03b1\uff09\uff1dcos\u03b1
cos\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dsin\u03b1
tan\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcot\u03b1
cot\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dtan\u03b1
sin\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dcos\u03b1
cos\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dsin\u03b1
tan\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dcot\u03b1
cot\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dtan\u03b1
sin\uff083\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcos\u03b1
cos\uff083\u03c0/2\uff0b\u03b1\uff09\uff1dsin\u03b1
tan\uff083\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcot\u03b1
cot\uff083\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dtan\u03b1
sin\uff083\u03c0/2\uff0d\u03b1\uff09\uff1d\uff0dcos\u03b1
cos\uff083\u03c0/2\uff0d\u03b1\uff09\uff1d\uff0dsin\u03b1
tan\uff083\u03c0/2\uff0d\u03b1\uff09\uff1dcot\u03b1
cot\uff083\u03c0/2\uff0d\u03b1\uff09\uff1dtan\u03b1
(\u4ee5\u4e0ak\u2208Z)
sin0=0
sin\u03c0/6=0.5
sin\u03c0/4=\u4e8c\u5206\u4e4b\u6839\u53f72
sin\u03c0/3=\u4e8c\u5206\u4e4b\u6839\u53f73
sin\u03c0/2=1
cos0=1
cos\u03c0/6=\u4e8c\u5206\u4e4b\u6839\u53f73
cos\u03c0/4=\u4e8c\u5206\u4e4b\u6839\u53f72
cos\u03c0/3=0.5
cos\u03c0/2=0
tan0=0
tan\u03c0/6=\u4e09\u5206\u4e4b\u6839\u53f73
tan\u03c0/4=1
tan\u03c0/3=\u6839\u53f73
tan\u03c0/2\u65e0\u5b9e\u4e49
cot0 \u65e0\u5b9e\u4e49
cot\u03c0/6=\u6839\u53f73
cot\u03c0/4=1
cot\u03c0/3=\u4e09\u5206\u4e4b\u6839\u53f73
cotv/2=0
O(\u2229_\u2229)O~
\u518d\u7ed9\u4f60\u53d1\u4e00\u4e9b\u8f85\u52a9\u516c\u5f0f
\u4e00\uff09\u4e24\u89d2\u548c\u5dee\u516c\u5f0f \uff08\u5199\u7684\u90fd\u8981\u8bb0\uff09
sin(A+B)=sinAcosB+cosAsinB
sin(A-B)=sinAcosB-sinBcosA
cos(A+B)=cosAcosB-sinAsinB
cos(A-B)=cosAcosB+sinAsinB
tan(A+B)=(tanA+tanB)/(1-tanAtanB)
tan(A-B)=(tanA-tanB)/(1+tanAtanB)
\u4e8c\uff09\u7528\u4ee5\u4e0a\u516c\u5f0f\u53ef\u63a8\u51fa\u4e0b\u5217\u4e8c\u500d\u89d2\u516c\u5f0f
tan2A=2tanA/[1-(tanA)^2]
cos2a=(cosa)^2-(sina)^2=2(cosa)^2 -1=1-2(sina)^2
\uff08\u4e0a\u9762\u8fd9\u4e2a\u4f59\u5f26\u7684\u5f88\u91cd\u8981\uff09
sin2A=2sinA*cosA
\u4e09\uff09\u534a\u89d2\u7684\u53ea\u9700\u8bb0\u4f4f\u8fd9\u4e2a\uff1a
tan(A/2)=(1-cosA)/sinA=sinA/(1+cosA)
\u56db\uff09\u7528\u4e8c\u500d\u89d2\u4e2d\u7684\u4f59\u5f26\u53ef\u63a8\u51fa\u964d\u5e42\u516c\u5f0f
(sinA)^2=(1-cos2A)/2
(cosA)^2=(1+cos2A)/2
\u4e94\uff09\u7528\u4ee5\u4e0a\u964d\u5e42\u516c\u5f0f\u53ef\u63a8\u51fa\u4ee5\u4e0b\u5e38\u7528\u7684\u5316\u7b80\u516c\u5f0f
1-cosA=sin^(A/2)*2
1-sinA=cos^(A/2)*2
\u4e00\uff09\u4e24\u89d2\u548c\u5dee\u516c\u5f0f \uff08\u5199\u7684\u90fd\u8981\u8bb0\uff09
sin(A+B)=sinAcosB+cosAsinB
sin(A-B)=sinAcosB-sinBcosA
cos(A+B)=cosAcosB-sinAsinB
cos(A-B)=cosAcosB+sinAsinB
tan(A+B)=(tanA+tanB)/(1-tanAtanB)
tan(A-B)=(tanA-tanB)/(1+tanAtanB)
\u4e8c\uff09\u7528\u4ee5\u4e0a\u516c\u5f0f\u53ef\u63a8\u51fa\u4e0b\u5217\u4e8c\u500d\u89d2\u516c\u5f0f
tan2A=2tanA/[1-(tanA)^2]
cos2a=(cosa)^2-(sina)^2=2(cosa)^2 -1=1-2(sina)^2
\uff08\u4e0a\u9762\u8fd9\u4e2a\u4f59\u5f26\u7684\u5f88\u91cd\u8981\uff09
sin2A=2sinA*cosA
\u4e09\uff09\u534a\u89d2\u7684\u53ea\u9700\u8bb0\u4f4f\u8fd9\u4e2a\uff1a
tan(A/2)=(1-cosA)/sinA=sinA/(1+cosA)
\u56db\uff09\u7528\u4e8c\u500d\u89d2\u4e2d\u7684\u4f59\u5f26\u53ef\u63a8\u51fa\u964d\u5e42\u516c\u5f0f
(sinA)^2=(1-cos2A)/2
(cosA)^2=(1+cos2A)/2
\u4e94\uff09\u7528\u4ee5\u4e0a\u964d\u5e42\u516c\u5f0f\u53ef\u63a8\u51fa\u4ee5\u4e0b\u5e38\u7528\u7684\u5316\u7b80\u516c\u5f0f
1-cosA=sin^(A/2)*2
1-sinA=cos^(A/2)*2
\u540c\u89d2\u4e09\u89d2\u51fd\u6570\u57fa\u672c\u5173\u7cfb
\u2488\u540c\u89d2\u4e09\u89d2\u51fd\u6570\u7684\u57fa\u672c\u5173\u7cfb\u5f0f
\u5012\u6570\u5173\u7cfb:
tan\u03b1 \u00b7cot\u03b1\uff1d1
sin\u03b1 \u00b7csc\u03b1\uff1d1
cos\u03b1 \u00b7sec\u03b1\uff1d1
\u5546\u7684\u5173\u7cfb\uff1a
sin\u03b1/cos\u03b1\uff1dtan\u03b1\uff1dsec\u03b1/csc\u03b1
cos\u03b1/sin\u03b1\uff1dcot\u03b1\uff1dcsc\u03b1/sec\u03b1
\u5e73\u65b9\u5173\u7cfb\uff1a
sin^2(\u03b1)\uff0bcos^2(\u03b1)\uff1d1
1\uff0btan^2(\u03b1)\uff1dsec^2(\u03b1)
1\uff0bcot^2(\u03b1)\uff1dcsc^2(\u03b1)
\u5e0c\u671b\u80fd\u5e2e\u5230\u4f60\uff0c\u8bf7\u91c7\u7eb3\u6b63\u786e\u7b54\u6848\uff0c\u70b9\u51fb\u3010\u91c7\u7eb3\u7b54\u6848\u3011\uff0c\u8c22\u8c22 ^_^
1、(sina)^2+(cosa)^2=1
tana=(sina)/(cosa)
2、二倍角公式：
cos(2a）=2(cosa)^2-1=1-2(sina)^2
sin2a=2sina cosa
3、万能公式：
tan2a=2tana /1-（tana)^2
cos2a=1-(tana)^2/1+(tana)^2
sin2a=2tana/1+(tana)^2
竞赛中会有：
1、半角公式：
(sina)^2=( 1-cos2a)/2
(cosa)^2=(1+cos2a)/2
tana=(1-cos2a)/sin2a=sin2a/(1+cos2a)
2、三倍角公式：
sin(3α) = 3sinα-4sin^3α = 4sinα·sin(60°+α)sin(60°-α)
cos(3α) = 4cos^3α-3cosα = 4cosα·cos(60°+α)cos(60°-α)
tan(3α) = (3tanα-tan^3α)/(1-3tan^2α) = tanαtan(60°+α)tan(-α)60°
3、和差化积：
sin a+sinb=2sin[(a+b)/2]·cos[(a-b)/2]
sin a-sin b=2cos[(a+b)/2]·sin[(a-b)/2]
cosa+cosb=2cos[(a+b)/2]·cos[(a-b)/2]
cosa-cosb=-2sin[(a+b)/2]·sin[(a-b)/2]
4、积化和差：
sinasinb=-[cos(a+b)-cos(a-b)]/2
cosacosb=[cos(a+b)+cos(a-b)]/2
sinacosb=[sin(a+b)+sin(a-b)]/2
cosasinb=[sin(a+b)-sin(a-b)]/2
绛旓細涓夎鍑芥暟璇卞鍏紡 璇卞鍏紡涓锛氱粓杈圭浉鍚岀殑瑙掔殑鍚屼竴涓夎鍑芥暟鐨鍊肩浉绛 璁疚变负浠绘剰閿愯锛屽姬搴﹀埗涓鐨勮鐨勮〃绀猴細sin(2k蟺+伪)=sin伪(k鈭圸)cos(2k蟺+伪)=cos伪(k鈭圸)tan(2k蟺+伪)=tan伪(k鈭圸)cot(2k蟺+伪)=cot伪(k鈭圸)璇卞鍏紡浜岋細蟺+伪鐨勪笁瑙掑嚱鏁板间笌伪鐨勪笁瑙掑嚱鏁板间箣闂寸殑鍏崇郴 璁疚变负...
绛旓細涓夎鍑芥暟鐨璇卞鍏紡(鍏叕寮) 鍏紡涓 sin(-伪) = -sin伪 cos(-伪) = cos伪 tan (-伪)=-tan伪 鍏紡浜宻in(蟺/2-伪) = cos伪 cos(蟺/2-伪) = sin伪 鍏紡涓 sin(蟺/2+伪) = cos伪 cos(蟺/2+伪) = -sin伪 鍏紡鍥泂in(蟺-伪) = sin伪 cos(蟺-伪) = -cos伪 鍏紡浜攕in(蟺+伪) = ...
绛旓細涓夎鍑芥暟鏄暟瀛︿腑閲嶈鐨勪竴涓垎鏀紝瀹冨寘鎷寮︺佷綑寮︺佹鍒囥佷綑鍒囥佹鍓插拰浣欏壊绛夊叚涓熀鏈嚱鏁般傝繖浜涘嚱鏁板湪瑙ｅ喅鍑犱綍闂銆佺墿鐞嗛棶棰樹互鍙婂伐绋嬮棶棰樼瓑鏂归潰鏈夌潃骞挎硾鐨勫簲鐢ㄣ備负浜嗘柟渚垮涔犲拰鐞嗚В锛屾垜浠彲浠ュ皢涓夎鍑芥暟鐨勫叕寮杩涜鍒嗙被鍜鏁寸悊銆傞鍏堬紝鎴戜滑鍙互鎸夌収鍑芥暟鐨勭被鍨嬭繘琛屽垎绫汇傛牴鎹畾涔夛紝涓夎鍑芥暟鍙互鍒嗕负姝ｅ鸡鍑芥暟銆...
绛旓細锛堝叚杈瑰舰璁板繂娉曪細鍥惧舰缁撴瀯鈥滀笂寮︿腑鍒囦笅鍓诧紝宸︽鍙充綑涓棿1鈥濓紱璁板繂鏂规硶鈥滃瑙掔嚎涓婁袱涓鍑芥暟鐨绉负1锛涢槾褰变笁瑙掑舰涓婁袱椤剁偣鐨涓夎鍑芥暟鍊肩殑骞虫柟鍜岀瓑浜庝笅椤剁偣鐨勪笁瑙掑嚱鏁板肩殑骞虫柟锛涗换鎰忎竴椤剁偣鐨勪笁瑙掑嚱鏁板肩瓑浜庣浉閭讳袱涓《鐐圭殑涓夎鍑芥暟鍊肩殑涔樼Н銆傗濓級璇卞鍏紡锛堝彛璇:濂囧彉鍋朵笉鍙橈紝绗﹀彿鐪嬭薄闄愩傦級sin锛堬紞伪锛夛紳锛...
绛旓細涓夎鍑芥暟 璁句笁瑙掑舰abc涓轰竴鐩磋涓夎褰(a:楂樺害b:璺ㄥ害c:鏂滃害),鍏跺叚涓笁瑙掑嚱鏁板涓:Sin胃=a/c=楂樺害/鏂滃害 Cos胃=b/c=璺ㄥ害/鏂滃害 Tan胃=a/b=楂樺害/璺ㄥ害 Cot胃=b/a=楂樺害/璺ㄥ害 Sec胃=c/b=鏂滃害/璺ㄥ害 Csc胃=c/a=鏂滃害/楂樺害 姹傛枩鐜鐨勫叕寮灏辨槸:Tan胃=楂樺害/璺ㄥ害 浠ヤ笂鍏紡鍙埄鐢ㄧ瀛﹀嚱鏁拌绠楀櫒鏉ユ眰...
绛旓細鍚岃涓夎鍑芥暟鐨8涓鍏紡鍒嗗埆鏄笁涓掓暟鍏崇郴,涓や釜鍟嗘暟鍏崇郴,涓変釜骞虫柟鍏崇郴銆備笅闈㈠氨鍜屾垜涓璧蜂簡瑙涓涓鍚э紝渚涘ぇ瀹跺弬鑰冦傚悓瑙掍笁瑙掑嚱鏁版湁鍝8涓叕寮 锛1锛夊掓暟鍏崇郴 1.sina*csca=1(a鈮爇蟺,k鈭圸锛2.cosa*scsa=1(a鈮爇蟺+蟺/2,k鈭圸)3.tana*cota=1(a鈮爇蟺/2,k鈭圸)(2)鍟嗘暟鍏崇郴 4.tana=sina/...
绛旓細鏁板涓夎鍑芥暟鐭ヨ瘑鐐鏁寸悊锛1銆佹暟瀛涓夎鍑芥暟鐨瀹氫箟銆2銆佹暟瀛︿笁瑙掑嚱鏁伴噸鐐鍏紡銆3銆佺壒娈婅鐨勬暟瀛︿笁瑙掑嚱鏁板笺4銆佹暟瀛︿笁瑙掑嚱鏁板叧绯诲叕寮忋傞氬父鐨勪笁瑙掑嚱鏁版槸鍦ㄥ钩闈㈢洿瑙掑潗鏍囩郴涓畾涔夌殑锛屽叾瀹氫箟鍩熶负鏁翠釜瀹炴暟鍩熴傚彟涓绉嶅畾涔夋槸鍦ㄧ洿瑙掍笁瑙掑舰涓紝浣嗗苟涓嶅畬鍏ㄣ1銆佹暟瀛︿笁瑙掑嚱鏁扮殑瀹氫箟 鍦ㄧ洿瑙掍笁瑙掑舰涓紝瀵光垹伪鑰岃█锛屾湁...
绛旓細= (tan伪 + tan尾) / [1 - tan伪路tan尾]tan(伪-尾)</ = (tan伪 - tan尾) / [1 + tan伪路tan尾]鎺屾彙浜嗚繖浜鍏紡锛屼綘鐨涓夎鍑芥暟瀛︿範涔嬭矾灏嗘洿涓虹晠閫氥傚鏋滀綘瑙夊緱杩欎簺鍐呭鏈甯姪锛屼笉濡ㄧ偣鍑昏禐銆佹敹钘忥紝杩欎笉浠呮槸瀵规垜鏁寸悊宸ヤ綔鐨勮偗瀹氾紝涔熸槸鎴戞寔缁垎浜洿澶氭暟瀛︾簿鍗庣殑鍔ㄥ姏婧愭硥銆傛劅璋綘鐨勬敮鎸侊紒
绛旓細涓よ鍜屼笌宸殑涓夎鍑芥暟鍏紡 涓囪兘鍏紡 sin锛埼憋紜尾锛夛紳sin伪cos尾锛媍os伪sin尾 sin锛埼憋紞尾锛夛紳sin伪cos尾锛峜os伪sin尾 cos锛埼憋紜尾锛夛紳cos伪cos尾锛峴in伪sin尾 cos锛埼憋紞尾锛夛紳cos伪cos尾锛媠in伪sin尾 tan伪锛媡an尾 tan锛埼憋紜尾锛夛紳鈥斺斺1锛峵an伪 路tan尾 tan伪锛峵an尾 tan锛埼憋紞尾锛...
绛旓細2蟺锛嵨憋級锛濓紞sin伪 cos锛2蟺锛嵨憋級锛漜os伪 tan锛2蟺锛嵨憋級锛濓紞tan伪 cot锛2蟺锛嵨憋級锛濓紞cot伪 sin锛2k蟺锛嬑憋級锛漵in伪 cos锛2k蟺锛嬑憋級锛漜os伪 tan锛2k蟺锛嬑憋級锛漷an伪 cot锛2k蟺锛嬑憋級锛漜ot伪 (鍏朵腑k鈭圸) 涓よ鍜屼笌宸殑涓夎鍑芥暟鍏紡涓囪兘鍏紡 sin锛埼憋紜尾锛夛紳sin伪cos尾锛媍os伪...
