求三角函数数值表(sin cos tan cot sec csc versin covers) 求初三的三角函数诱导公式（只有cos sin tan cot...
csc/sec\uff08\u4e09\u89d2\u51fd\u6570\uff09\u7684\u82f1\u6587\u5b8c\u6574\u62fc\u5199\u662f\u4ec0\u4e48\uff1f\u6b63\u5f26\uff1asine\uff08\u7b80\u5199sin\uff09[sain]
\u4f59\u5f26\uff1acosine\uff08\u7b80\u5199cos\uff09[kəusain]
\u6b63\u5207\uff1atangent\uff08\u7b80\u5199tan\uff09['tændʒənt]
\u4f59\u5207\uff1acotangent\uff08\u7b80\u5199cot\uff09['kəu'tændʒənt]
\u6b63\u5272\uff1asecant\uff08\u7b80\u5199sec\uff09['si:kənt]
\u4f59\u5272\uff1acosecant\uff08\u7b80\u5199csc\uff09['kau'si:kənt]
\u6b63\u77e2\uff1aversine\uff08\u7b80\u5199versin\uff09['və:sain]
\u4f59\u77e2\uff1aversed cosine\uff08\u7b80\u5199vercos\uff09['və:sə:d][kəusain]
\uff08\u5176\u4e2d\u6b63\u77e2\u548c\u4f59\u77e2\u662f\u4e24\u4e2a\u5df2\u7ecf\u5e9f\u9664\u7684\u51fd\u6570\uff09
\u00b7\u5e73\u65b9\u5173\u7cfb\uff1a
sin�0�5(\u03b1)+cos�0�5(\u03b1)=1 cos�0�5(a)=(1+cos2a)/2
tan�0�5(\u03b1)+1=sec�0�5(\u03b1) sin�0�5(a)=(1-cos2a)/2
cot�0�5(\u03b1)+1=csc�0�5(\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�0�5+B�0�5)^(1/2)sin(\u03b1+t)\uff0c\u5176\u4e2d
sint=B/(A�0�5+B�0�5)^(1/2)
cost=A/(A�0�5+B�0�5)^(1/2)
tant=B/A
Asin\u03b1+Bcos\u03b1=(A�0�5+B�0�5)^(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�0�5(\u03b1)-sin�0�5(\u03b1)=2cos�0�5(\u03b1)-1=1-2sin�0�5(\u03b1)
tan(2\u03b1)=2tan\u03b1/[1-tan�0�5(\u03b1)]
\u00b7\u4e09\u500d\u89d2\u516c\u5f0f\uff1a
sin(3\u03b1)=3sin\u03b1-4sin�0�6(\u03b1)
cos(3\u03b1)=4cos�0�6(\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�0�5(\u03b1)=(1-cos(2\u03b1))/2=versin(2\u03b1)/2
cos�0�5(\u03b1)=(1+cos(2\u03b1))/2=covers(2\u03b1)/2
tan�0�5(\u03b1)=(1-cos(2\u03b1))/(1+cos(2\u03b1))
\u00b7\u4e07\u80fd\u516c\u5f0f\uff1a
sin\u03b1=2tan(\u03b1/2)/[1+tan�0�5(\u03b1/2)]
cos\u03b1=[1-tan�0�5(\u03b1/2)]/[1+tan�0�5(\u03b1/2)]
tan\u03b1=2tan(\u03b1/2)/[1-tan�0�5(\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�0�5\u03b1
1-cos2\u03b1=2sin�0�5\u03b1
1+sin\u03b1=(sin\u03b1/2+cos\u03b1/2)�0�5
\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�0�5(\u03b1)+sin�0�5(\u03b1-2\u03c0/3)+sin�0�5(\u03b1+2\u03c0/3)=3/2
tanAtanBtan(A+B)+tanA+tanB-tan(A+B)=0
cosx+cos2x+...+cosnx= [sin(n+1)x+sinnx-sinx]/2sinx
\u8bc1\u660e\uff1a
\u5de6\u8fb9=2sinx(cosx+cos2x+...+cosnx)/2sinx
=[sin2x-0+sin3x-sinx+sin4x-sin2x+...+ sinnx-sin(n-2)x+sin(n+1)x-sin(n-1)x]/2sinx \uff08\u79ef\u5316\u548c\u5dee\uff09
=[sin(n+1)x+sinnx-sinx]/2sinx=\u53f3\u8fb9
\u7b49\u5f0f\u5f97\u8bc1
sinx+sin2x+...+sinnx= - [cos(n+1)x+cosnx-cosx-1]/2sinx
\u8bc1\u660e:
\u5de6\u8fb9=-2sinx[sinx+sin2x+...+sinnx]/(-2sinx)
=[cos2x-cos0+cos3x-cosx+...+cosnx-cos(n-2)x+cos(n+1)x-cos(n-1)x]/(-2sinx)
=- [cos(n+1)x+cosnx-cosx-1]/2sinx=\u53f3\u8fb9
\u7b49\u5f0f\u5f97\u8bc1
[\u7f16\u8f91\u672c\u6bb5]\u4e09\u89d2\u51fd\u6570\u7684\u8bf1\u5bfc\u516c\u5f0f
\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)
[\u7f16\u8f91\u672c\u6bb5]\u6b63\u4f59\u5f26\u5b9a\u7406
\u6b63\u5f26\u5b9a\u7406\u662f\u6307\u5728\u4e00\u4e2a\u4e09\u89d2\u5f62\u4e2d\uff0c\u5404\u8fb9\u548c\u5b83\u6240\u5bf9\u7684\u89d2\u7684\u6b63\u5f26\u7684\u6bd4\u76f8\u7b49\uff0c\u5373a/sinA=b/sinB=c/sinC=2R \uff0e
\u4f59\u5f26\u5b9a\u7406\u662f\u6307\u4e09\u89d2\u5f62\u4e2d\u4efb\u4f55\u4e00\u8fb9\u7684\u5e73\u65b9\u7b49\u4e8e\u5176\u5b83\u4e24\u8fb9\u7684\u5e73\u65b9\u548c\u51cf\u53bb\u8fd9\u4e24\u8fb9\u4e0e\u5b83\u4eec\u5939\u89d2\u7684\u4f59\u5f26\u7684\u79ef\u76842\u500d\uff0c\u5373a^2=b^2+c^2-2bc cosA
\u89d2A\u7684\u5bf9\u8fb9\u4e8e\u659c\u8fb9\u7684\u6bd4\u53eb\u505a\u89d2A\u7684\u6b63\u5f26\uff0c\u8bb0\u4f5csinA\uff0c\u5373sinA=\u89d2A\u7684\u5bf9\u8fb9/\u659c\u8fb9
\u659c\u8fb9\u4e0e\u90bb\u8fb9\u5939\u89d2a
sin=y/r
\u65e0\u8bbay>x\u6216y\u2264x
\u65e0\u8bbaa\u591a\u5927\u591a\u5c0f\u53ef\u4ee5\u4efb\u610f\u5927\u5c0f
\u6b63\u5f26\u7684\u6700\u5927\u503c\u4e3a1
cos：30°等于根号3/2，45°＝根号2/2，60°，等于1/2
tan;30°等于根号3/3,45°等于1,60°等于根号3。
扩展阅读：tansincos数值表图 ... 37sin cos tan数值表 ... sin计算表 ... sin cos tan特殊数值表 ... 勾股计算器 ... sin tan cos三角函数表 ... sin角度对照表 ... sin cos tan 关系对边图解 ... sin所有值一览表 ...
