求09届高一所有数学公式（必修一和必修四）！ 求高一数学必修一 必修四全部公式

\u6025\u6c42\u3002\u9ad8\u4e00\u6570\u5b66\u5fc5\u4fee\u4e00\u548c\u5fc5\u4fee\u56db\u7684\u516c\u5f0f\u603b\u7ed3\uff01\u4e07\u5206\u611f\u8c22

\u4f60\u597d\uff0c\u7684\u786e\u4e0d\u5168\uff0c\u6211\u67e5\u627e\u540e\u8865\u5168\u4e86
\u5fc5\u4fee\u4e00\uff1a
\u4e58\u6cd5\u4e0e\u56e0\u5f0f\u5206\u89e3
a^2-b^2=(a+b)(a-b)
a^3+b^3=(a+b)(a^2-ab+b^2) 
a^3-b^3=(a-b(a^2+ab+b^2)
\u4e09\u89d2\u4e0d\u7b49\u5f0f |a+b|\u2264|a|+|b| |a-b|\u2264|a|+|b| |a|\u2264b-b\u2264a\u2264b
|a-b|\u2265|a|-|b| -|a|\u2264a\u2264|a|
\u4e00\u5143\u4e8c\u6b21\u65b9\u7a0b\u7684\u89e3 -b+\u221a(b^2-4ac)/2a -b-\u221a(b^2-4ac)/2a
\u6839\u4e0e\u7cfb\u6570\u7684\u5173\u7cfb X1+X2=-b/a X1*X2=c/a \u6ce8\uff1a\u97e6\u8fbe\u5b9a\u7406
\u5224\u522b\u5f0f
b^2-4ac=0 \u6ce8\uff1a\u65b9\u7a0b\u6709\u4e24\u4e2a\u76f8\u7b49\u7684\u5b9e\u6839
b^2-4ac>0 \u6ce8\uff1a\u65b9\u7a0b\u6709\u4e24\u4e2a\u4e0d\u7b49\u7684\u5b9e\u6839 �
b^2-4ac<0 \u6ce8\uff1a\u65b9\u7a0b\u6ca1\u6709\u5b9e\u6839\uff0c\u6709\u5171\u8f6d\u590d\u6570\u6839
\u4e09\u89d2\u51fd\u6570\u516c\u5f0f
\u4e24\u89d2\u548c\u516c\u5f0f
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)
cot(A+B)=(cotAcotB-1)/(cotB+cotA) �
cot(A-B)=(cotAcotB+1)/(cotB-cotA)
\u500d\u89d2\u516c\u5f0f
tan2A=2tanA/[1-(tanA)^2]
cos2a=(cosa)^2-(sina)^2=2(cosa)^2 -1=1-2(sina)^2
\u534a\u89d2\u516c\u5f0f
sin(A/2)=\u221a((1-cosA)/2) sin(A/2)=-\u221a((1-cosA)/2)
cos(A/2)=\u221a((1+cosA)/2) cos(A/2)=-\u221a((1+cosA)/2)
tan(A/2)=\u221a((1-cosA)/((1+cosA)) tan(A/2)=-\u221a((1-cosA)/((1+cosA))
cot(A/2)=\u221a((1+cosA)/((1-cosA)) cot(A/2)=-\u221a((1+cosA)/((1-cosA)) �
\u548c\u5dee\u5316\u79ef
2sinAcosB=sin(A+B)+sin(A-B)
2cosAsinB=sin(A+B)-sin(A-B) )
2cosAcosB=cos(A+B)-sin(A-B)
-2sinAsinB=cos(A+B)-cos(A-B)
sinA+sinB=2sin((A+B)/2)cos((A-B)/2
cosA+cosB=2cos((A+B)/2)sin((A-B)/2)
tanA+tanB=sin(A+B)/cosAcosB

\u67d0\u4e9b\u6570\u5217\u524dn\u9879\u548c
1+2+3+4+5+6+7+8+9+\u2026+n=n(n+1)/2
1+3+5+7+9+11+13+15+\u2026+(2n-1)=n2 \u001e
2+4+6+8+10+12+14+\u2026+(2n)=n(n+1) 5
1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+\u2026+n^2=n(n+1)(2n+1)/6
1^3+2^3+3^3+4^3+5^3+6^3+\u2026n^3=n2(n+1)2/4
1*2+2*3+3*4+4*5+5*6+6*7+\u2026+n(n+1)=n(n+1)(n+2)/3
\u6b63\u5f26\u5b9a\u7406 a/sinA=b/sinB=c/sinC=2R \u6ce8\uff1a \u5176\u4e2d R \u8868\u793a\u4e09\u89d2\u5f62\u7684\u5916\u63a5\u5706\u534a\u5f84
\u4f59\u5f26\u5b9a\u7406 b^2=a^2+c^2-2accosB \u6ce8\uff1a\u89d2B\u662f\u8fb9a\u548c\u8fb9c\u7684\u5939\u89d2
\u5706\u7684\u6807\u51c6\u65b9\u7a0b (x-a)^2+(y-b)^2=^r2 \u6ce8\uff1a\uff08a,b\uff09\u662f\u5706\u5fc3\u5750\u6807 \u001f
\u5706\u7684\u4e00\u822c\u65b9\u7a0b x^2+y^2+Dx+Ey+F=0 \u6ce8\uff1aD^2+E^2-4F>0
\u629b\u7269\u7ebf\u6807\u51c6\u65b9\u7a0b y^2=2px y^2=-2px x^2=2py x^2=-2py
\u76f4\u68f1\u67f1\u4fa7\u9762\u79ef S=c*h \u659c\u68f1\u67f1\u4fa7\u9762\u79ef S=c'*h
\u6b63\u68f1\u9525\u4fa7\u9762\u79ef S=1/2c*h' \u6b63\u68f1\u53f0\u4fa7\u9762\u79ef S=1/2(c+c')h'
\u5706\u53f0\u4fa7\u9762\u79ef S=1/2(c+c')l=pi(R+r)l \u7403\u7684\u8868\u9762\u79ef S=4pi*r2
\u5706\u67f1\u4fa7\u9762\u79ef S=c*h=2pi*h \u5706\u9525\u4fa7\u9762\u79ef S=1/2*c*l=pi*r*l
\u5f27\u957f\u516c\u5f0f l=a*r a\u662f\u5706\u5fc3\u89d2\u7684\u5f27\u5ea6\u6570r >0 \u6247\u5f62\u9762\u79ef\u516c\u5f0f s=1/2*l*r
\u9525\u4f53\u4f53\u79ef\u516c\u5f0f V=1/3*S*H \u5706\u9525\u4f53\u4f53\u79ef\u516c\u5f0f V=1/3*pi*r2h �
\u659c\u68f1\u67f1\u4f53\u79ef V=S'L \u6ce8\uff1a\u5176\u4e2d,S'\u662f\u76f4\u622a\u9762\u9762\u79ef\uff0c L\u662f\u4fa7\u68f1\u957f
\u67f1\u4f53\u4f53\u79ef\u516c\u5f0f V=s*h \u5706\u67f1\u4f53 V=pi*r2h
\u5fc5\u4fee\u56db\uff1a\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)


\u8bf1\u5bfc\u516c\u5f0f\u8bb0\u5fc6\u53e3\u8bc0

\u203b\u89c4\u5f8b\u603b\u7ed3\u203b
\u4e0a\u9762\u8fd9\u4e9b\u8bf1\u5bfc\u516c\u5f0f\u53ef\u4ee5\u6982\u62ec\u4e3a\uff1a
\u5bf9\u4e8ek\u00b7\u03c0/2\u00b1\u03b1(k\u2208Z)\u7684\u4e2a\u4e09\u89d2\u51fd\u6570\u503c\uff0c
\u2460\u5f53k\u662f\u5076\u6570\u65f6\uff0c\u5f97\u5230\u03b1\u7684\u540c\u540d\u51fd\u6570\u503c\uff0c\u5373\u51fd\u6570\u540d\u4e0d\u6539\u53d8\uff1b
\u2461\u5f53k\u662f\u5947\u6570\u65f6\uff0c\u5f97\u5230\u03b1\u76f8\u5e94\u7684\u4f59\u51fd\u6570\u503c\uff0c\u5373sin\u2192cos;cos\u2192sin;tan\u2192cot,cot\u2192tan.
\uff08\u5947\u53d8\u5076\u4e0d\u53d8\uff09
\u7136\u540e\u5728\u524d\u9762\u52a0\u4e0a\u628a\u03b1\u770b\u6210\u9510\u89d2\u65f6\u539f\u51fd\u6570\u503c\u7684\u7b26\u53f7\u3002
\uff08\u7b26\u53f7\u770b\u8c61\u9650\uff09

\u4f8b\u5982\uff1a
sin(2\u03c0\uff0d\u03b1)\uff1dsin(4\u00b7\u03c0/2\uff0d\u03b1)\uff0ck\uff1d4\u4e3a\u5076\u6570\uff0c\u6240\u4ee5\u53d6sin\u03b1\u3002
\u5f53\u03b1\u662f\u9510\u89d2\u65f6\uff0c2\u03c0\uff0d\u03b1\u2208(270\u00b0\uff0c360\u00b0)\uff0csin(2\u03c0\uff0d\u03b1)\uff1c0\uff0c\u7b26\u53f7\u4e3a\u201c\uff0d\u201d\u3002
\u6240\u4ee5sin(2\u03c0\uff0d\u03b1)\uff1d\uff0dsin\u03b1

\u4e0a\u8ff0\u7684\u8bb0\u5fc6\u53e3\u8bc0\u662f\uff1a
\u5947\u53d8\u5076\u4e0d\u53d8\uff0c\u7b26\u53f7\u770b\u8c61\u9650\u3002
\u516c\u5f0f\u53f3\u8fb9\u7684\u7b26\u53f7\u4e3a\u628a\u03b1\u89c6\u4e3a\u9510\u89d2\u65f6\uff0c\u89d2k\u00b7360\u00b0+\u03b1\uff08k\u2208Z\uff09\uff0c-\u03b1\u3001180\u00b0\u00b1\u03b1\uff0c360\u00b0-\u03b1
\u6240\u5728\u8c61\u9650\u7684\u539f\u4e09\u89d2\u51fd\u6570\u503c\u7684\u7b26\u53f7\u53ef\u8bb0\u5fc6
\u6c34\u5e73\u8bf1\u5bfc\u540d\u4e0d\u53d8\uff1b\u7b26\u53f7\u770b\u8c61\u9650\u3002
\u5404\u79cd\u4e09\u89d2\u51fd\u6570\u5728\u56db\u4e2a\u8c61\u9650\u7684\u7b26\u53f7\u5982\u4f55\u5224\u65ad\uff0c\u4e5f\u53ef\u4ee5\u8bb0\u4f4f\u53e3\u8bc0\u201c\u4e00\u5168\u6b63\uff1b\u4e8c\u6b63\u5f26\uff1b\u4e09\u4e3a\u5207\uff1b\u56db\u4f59\u5f26\u201d\uff0e
\u8fd9\u5341\u4e8c\u5b57\u53e3\u8bc0\u7684\u610f\u601d\u5c31\u662f\u8bf4\uff1a
\u7b2c\u4e00\u8c61\u9650\u5185\u4efb\u4f55\u4e00\u4e2a\u89d2\u7684\u56db\u79cd\u4e09\u89d2\u51fd\u6570\u503c\u90fd\u662f\u201c\uff0b\u201d\uff1b
\u7b2c\u4e8c\u8c61\u9650\u5185\u53ea\u6709\u6b63\u5f26\u662f\u201c\uff0b\u201d\uff0c\u5176\u4f59\u5168\u90e8\u662f\u201c\uff0d\u201d\uff1b
\u7b2c\u4e09\u8c61\u9650\u5185\u5207\u51fd\u6570\u662f\u201c\uff0b\u201d\uff0c\u5f26\u51fd\u6570\u662f\u201c\uff0d\u201d\uff1b
\u7b2c\u56db\u8c61\u9650\u5185\u53ea\u6709\u4f59\u5f26\u662f\u201c\uff0b\u201d\uff0c\u5176\u4f59\u5168\u90e8\u662f\u201c\uff0d\u201d\uff0e



\u5176\u4ed6\u4e09\u89d2\u51fd\u6570\u77e5\u8bc6\uff1a



\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)


\u540c\u89d2\u4e09\u89d2\u51fd\u6570\u5173\u7cfb\u516d\u89d2\u5f62\u8bb0\u5fc6\u6cd5

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\uff082\uff09\u5546\u6570\u5173\u7cfb\uff1a\u516d\u8fb9\u5f62\u4efb\u610f\u4e00\u9876\u70b9\u4e0a\u7684\u51fd\u6570\u503c\u7b49\u4e8e\u4e0e\u5b83\u76f8\u90bb\u7684\u4e24\u4e2a\u9876\u70b9\u4e0a\u51fd\u6570\u503c\u7684\u4e58\u79ef\u3002
\uff08\u4e3b\u8981\u662f\u4e24\u6761\u865a\u7ebf\u4e24\u7aef\u7684\u4e09\u89d2\u51fd\u6570\u503c\u7684\u4e58\u79ef\uff09\u3002\u7531\u6b64\uff0c\u53ef\u5f97\u5546\u6570\u5173\u7cfb\u5f0f\u3002
\uff083\uff09\u5e73\u65b9\u5173\u7cfb\uff1a\u5728\u5e26\u6709\u9634\u5f71\u7ebf\u7684\u4e09\u89d2\u5f62\u4e2d\uff0c\u4e0a\u9762\u4e24\u4e2a\u9876\u70b9\u4e0a\u7684\u4e09\u89d2\u51fd\u6570\u503c\u7684\u5e73\u65b9\u548c\u7b49\u4e8e\u4e0b\u9762\u9876\u70b9\u4e0a\u7684\u4e09\u89d2\u51fd\u6570\u503c\u7684\u5e73\u65b9\u3002



\u4e24\u89d2\u548c\u5dee\u516c\u5f0f

\u2489\u4e24\u89d2\u548c\u4e0e\u5dee\u7684\u4e09\u89d2\u51fd\u6570\u516c\u5f0f
sin\uff08\u03b1\uff0b\u03b2\uff09\uff1dsin\u03b1cos\u03b2\uff0bcos\u03b1sin\u03b2
sin\uff08\u03b1\uff0d\u03b2\uff09\uff1dsin\u03b1cos\u03b2\uff0dcos\u03b1sin\u03b2
cos\uff08\u03b1\uff0b\u03b2\uff09\uff1dcos\u03b1cos\u03b2\uff0dsin\u03b1sin\u03b2
cos\uff08\u03b1\uff0d\u03b2\uff09\uff1dcos\u03b1cos\u03b2\uff0bsin\u03b1sin\u03b2

tan\u03b1\uff0btan\u03b2
tan\uff08\u03b1\uff0b\u03b2\uff09\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0dtan\u03b1 \u00b7tan\u03b2

tan\u03b1\uff0dtan\u03b2
tan\uff08\u03b1\uff0d\u03b2\uff09\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0btan\u03b1 \u00b7tan\u03b2


\u500d\u89d2\u516c\u5f0f

\u248a\u4e8c\u500d\u89d2\u7684\u6b63\u5f26\u3001\u4f59\u5f26\u548c\u6b63\u5207\u516c\u5f0f\uff08\u5347\u5e42\u7f29\u89d2\u516c\u5f0f\uff09
sin2\u03b1\uff1d2sin\u03b1cos\u03b1
cos2\u03b1\uff1dcos^2(\u03b1)\uff0dsin^2(\u03b1)\uff1d2cos^2(\u03b1)\uff0d1\uff1d1\uff0d2sin^2(\u03b1)

2tan\u03b1
tan2\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014
1\uff0dtan^2(\u03b1)


\u534a\u89d2\u516c\u5f0f

\u248b\u534a\u89d2\u7684\u6b63\u5f26\u3001\u4f59\u5f26\u548c\u6b63\u5207\u516c\u5f0f\uff08\u964d\u5e42\u6269\u89d2\u516c\u5f0f\uff09

1\uff0dcos\u03b1
sin^2(\u03b1/2)\uff1d\u2014\u2014\u2014\u2014\u2014
2

1\uff0bcos\u03b1
cos^2(\u03b1/2)\uff1d\u2014\u2014\u2014\u2014\u2014
2

1\uff0dcos\u03b1
tan^2(\u03b1/2)\uff1d\u2014\u2014\u2014\u2014\u2014
1\uff0bcos\u03b1


\u4e07\u80fd\u516c\u5f0f

\u248c\u4e07\u80fd\u516c\u5f0f
2tan(\u03b1/2)
sin\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0btan^2(\u03b1/2)

1\uff0dtan^2(\u03b1/2)
cos\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0btan^2(\u03b1/2)

2tan(\u03b1/2)
tan\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0dtan^2(\u03b1/2)

\u4e07\u80fd\u516c\u5f0f\u63a8\u5bfc

\u9644\u63a8\u5bfc\uff1a
sin2\u03b1=2sin\u03b1cos\u03b1=2sin\u03b1cos\u03b1/(cos^2(\u03b1)+sin^2(\u03b1))......*\uff0c
\uff08\u56e0\u4e3acos^2(\u03b1)+sin^2(\u03b1)=1\uff09
\u518d\u628a*\u5206\u5f0f\u4e0a\u4e0b\u540c\u9664cos^2(\u03b1)\uff0c\u53ef\u5f97sin2\u03b1\uff1dtan2\u03b1/(1\uff0btan^2(\u03b1))
\u7136\u540e\u7528\u03b1/2\u4ee3\u66ff\u03b1\u5373\u53ef\u3002
\u540c\u7406\u53ef\u63a8\u5bfc\u4f59\u5f26\u7684\u4e07\u80fd\u516c\u5f0f\u3002\u6b63\u5207\u7684\u4e07\u80fd\u516c\u5f0f\u53ef\u901a\u8fc7\u6b63\u5f26\u6bd4\u4f59\u5f26\u5f97\u5230\u3002


\u4e09\u500d\u89d2\u516c\u5f0f

\u248d\u4e09\u500d\u89d2\u7684\u6b63\u5f26\u3001\u4f59\u5f26\u548c\u6b63\u5207\u516c\u5f0f
sin3\u03b1\uff1d3sin\u03b1\uff0d4sin^3(\u03b1)
cos3\u03b1\uff1d4cos^3(\u03b1)\uff0d3cos\u03b1

3tan\u03b1\uff0dtan^3(\u03b1)
tan3\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0d3tan^2(\u03b1)


\u4e09\u500d\u89d2\u516c\u5f0f\u63a8\u5bfc

\u9644\u63a8\u5bfc\uff1a
tan3\u03b1\uff1dsin3\u03b1/cos3\u03b1
\uff1d(sin2\u03b1cos\u03b1\uff0bcos2\u03b1sin\u03b1)/(cos2\u03b1cos\u03b1-sin2\u03b1sin\u03b1)
\uff1d(2sin\u03b1cos^2(\u03b1)\uff0bcos^2(\u03b1)sin\u03b1\uff0dsin^3(\u03b1))/(cos^3(\u03b1)\uff0dcos\u03b1sin^2(\u03b1)\uff0d2sin^2(\u03b1)cos\u03b1)
\u4e0a\u4e0b\u540c\u9664\u4ee5cos^3(\u03b1)\uff0c\u5f97\uff1a
tan3\u03b1\uff1d(3tan\u03b1\uff0dtan^3(\u03b1))/(1-3tan^2(\u03b1))

sin3\u03b1\uff1dsin(2\u03b1\uff0b\u03b1)\uff1dsin2\u03b1cos\u03b1\uff0bcos2\u03b1sin\u03b1
\uff1d2sin\u03b1cos^2(\u03b1)\uff0b(1\uff0d2sin^2(\u03b1))sin\u03b1
\uff1d2sin\u03b1\uff0d2sin^3(\u03b1)\uff0bsin\u03b1\uff0d2sin^2(\u03b1)
\uff1d3sin\u03b1\uff0d4sin^3(\u03b1)

cos3\u03b1\uff1dcos(2\u03b1\uff0b\u03b1)\uff1dcos2\u03b1cos\u03b1\uff0dsin2\u03b1sin\u03b1
\uff1d(2cos^2(\u03b1)\uff0d1)cos\u03b1\uff0d2cos\u03b1sin^2(\u03b1)
\uff1d2cos^3(\u03b1)\uff0dcos\u03b1\uff0b(2cos\u03b1\uff0d2cos^3(\u03b1))
\uff1d4cos^3(\u03b1)\uff0d3cos\u03b1
\u5373
sin3\u03b1\uff1d3sin\u03b1\uff0d4sin^3(\u03b1)
cos3\u03b1\uff1d4cos^3(\u03b1)\uff0d3cos\u03b1

\u4e09\u500d\u89d2\u516c\u5f0f\u8054\u60f3\u8bb0\u5fc6

\u8bb0\u5fc6\u65b9\u6cd5\uff1a\u8c10\u97f3\u3001\u8054\u60f3
\u6b63\u5f26\u4e09\u500d\u89d2\uff1a3\u5143 \u51cf 4\u51433\u89d2\uff08\u6b20\u503a\u4e86(\u88ab\u51cf\u6210\u8d1f\u6570)\uff0c\u6240\u4ee5\u8981\u201c\u6323\u94b1\u201d(\u97f3\u4f3c\u201c\u6b63\u5f26\u201d)\uff09
\u4f59\u5f26\u4e09\u500d\u89d2\uff1a4\u51433\u89d2 \u51cf 3\u5143\uff08\u51cf\u5b8c\u4e4b\u540e\u8fd8\u6709\u201c\u4f59\u201d\uff09
\u2606\u2606\u6ce8\u610f\u51fd\u6570\u540d\uff0c\u5373\u6b63\u5f26\u7684\u4e09\u500d\u89d2\u90fd\u7528\u6b63\u5f26\u8868\u793a\uff0c\u4f59\u5f26\u7684\u4e09\u500d\u89d2\u90fd\u7528\u4f59\u5f26\u8868\u793a\u3002


\u548c\u5dee\u5316\u79ef\u516c\u5f0f

\u248e\u4e09\u89d2\u51fd\u6570\u7684\u548c\u5dee\u5316\u79ef\u516c\u5f0f

\u03b1\uff0b\u03b2 \u03b1\uff0d\u03b2
sin\u03b1\uff0bsin\u03b2\uff1d2sin\u2014----\u00b7cos\u2014---
2 2

\u03b1\uff0b\u03b2 \u03b1\uff0d\u03b2
sin\u03b1\uff0dsin\u03b2\uff1d2cos\u2014----\u00b7sin\u2014----
2 2

\u03b1\uff0b\u03b2 \u03b1\uff0d\u03b2
cos\u03b1\uff0bcos\u03b2\uff1d2cos\u2014-----\u00b7cos\u2014-----
2 2

\u03b1\uff0b\u03b2 \u03b1\uff0d\u03b2
cos\u03b1\uff0dcos\u03b2\uff1d\uff0d2sin\u2014-----\u00b7sin\u2014-----
2 2

\u79ef\u5316\u548c\u5dee\u516c\u5f0f

\u248f\u4e09\u89d2\u51fd\u6570\u7684\u79ef\u5316\u548c\u5dee\u516c\u5f0f
sin\u03b1 \u00b7cos\u03b2\uff1d0.5[sin\uff08\u03b1\uff0b\u03b2\uff09\uff0bsin\uff08\u03b1\uff0d\u03b2\uff09]
cos\u03b1 \u00b7sin\u03b2\uff1d0.5[sin\uff08\u03b1\uff0b\u03b2\uff09\uff0dsin\uff08\u03b1\uff0d\u03b2\uff09]
cos\u03b1 \u00b7cos\u03b2\uff1d0.5[cos\uff08\u03b1\uff0b\u03b2\uff09\uff0bcos\uff08\u03b1\uff0d\u03b2\uff09]
sin\u03b1 \u00b7sin\u03b2\uff1d\uff0d 0.5[cos\uff08\u03b1\uff0b\u03b2\uff09\uff0dcos\uff08\u03b1\uff0d\u03b2\uff09]


\u548c\u5dee\u5316\u79ef\u516c\u5f0f\u63a8\u5bfc

\u9644\u63a8\u5bfc\uff1a
\u9996\u5148,\u6211\u4eec\u77e5\u9053sin(a+b)=sina*cosb+cosa*sinb,sin(a-b)=sina*cosb-cosa*sinb
\u6211\u4eec\u628a\u4e24\u5f0f\u76f8\u52a0\u5c31\u5f97\u5230sin(a+b)+sin(a-b)=2sina*cosb
\u6240\u4ee5,sina*cosb=(sin(a+b)+sin(a-b))/2
\u540c\u7406,\u82e5\u628a\u4e24\u5f0f\u76f8\u51cf,\u5c31\u5f97\u5230cosa*sinb=(sin(a+b)-sin(a-b))/2
\u540c\u6837\u7684,\u6211\u4eec\u8fd8\u77e5\u9053cos(a+b)=cosa*cosb-sina*sinb,cos(a-b)=cosa*cosb+sina*sinb
\u6240\u4ee5,\u628a\u4e24\u5f0f\u76f8\u52a0,\u6211\u4eec\u5c31\u53ef\u4ee5\u5f97\u5230cos(a+b)+cos(a-b)=2cosa*cosb
\u6240\u4ee5\u6211\u4eec\u5c31\u5f97\u5230,cosa*cosb=(cos(a+b)+cos(a-b))/2
\u540c\u7406,\u4e24\u5f0f\u76f8\u51cf\u6211\u4eec\u5c31\u5f97\u5230sina*sinb=-(cos(a+b)-cos(a-b))/2
\u8fd9\u6837,\u6211\u4eec\u5c31\u5f97\u5230\u4e86\u79ef\u5316\u548c\u5dee\u7684\u56db\u4e2a\u516c\u5f0f:
sina*cosb=(sin(a+b)+sin(a-b))/2
cosa*sinb=(sin(a+b)-sin(a-b))/2
cosa*cosb=(cos(a+b)+cos(a-b))/2
sina*sinb=-(cos(a+b)-cos(a-b))/2
\u597d,\u6709\u4e86\u79ef\u5316\u548c\u5dee\u7684\u56db\u4e2a\u516c\u5f0f\u4ee5\u540e,\u6211\u4eec\u53ea\u9700\u4e00\u4e2a\u53d8\u5f62,\u5c31\u53ef\u4ee5\u5f97\u5230\u548c\u5dee\u5316\u79ef\u7684\u56db\u4e2a\u516c\u5f0f.
\u6211\u4eec\u628a\u4e0a\u8ff0\u56db\u4e2a\u516c\u5f0f\u4e2d\u7684a+b\u8bbe\u4e3ax,a-b\u8bbe\u4e3ay,\u90a3\u4e48a=(x+y)/2,b=(x-y)/2
\u628aa,b\u5206\u522b\u7528x,y\u8868\u793a\u5c31\u53ef\u4ee5\u5f97\u5230\u548c\u5dee\u5316\u79ef\u7684\u56db\u4e2a\u516c\u5f0f:
sinx+siny=2sin((x+y)/2)*cos((x-y)/2)
sinx-siny=2cos((x+y)/2)*sin((x-y)/2)
cosx+cosy=2cos((x+y)/2)*cos((x-y)/2)
cosx-cosy=-2sin((x+y)/2)*sin((x-y)/2)


\u5411\u91cf\u7684\u8fd0\u7b97
\u52a0\u6cd5\u8fd0\u7b97
AB\uff0bBC\uff1dAC\uff0c\u8fd9\u79cd\u8ba1\u7b97\u6cd5\u5219\u53eb\u505a\u5411\u91cf\u52a0\u6cd5\u7684\u4e09\u89d2\u5f62\u6cd5\u5219\u3002
\u5df2\u77e5\u4e24\u4e2a\u4ece\u540c\u4e00\u70b9O\u51fa\u53d1\u7684\u4e24\u4e2a\u5411\u91cfOA\u3001OB\uff0c\u4ee5OA\u3001OB\u4e3a\u90bb\u8fb9\u4f5c\u5e73\u884c\u56db\u8fb9\u5f62OACB\uff0c\u5219\u4ee5O\u4e3a\u8d77\u70b9\u7684\u5bf9\u89d2\u7ebfOC\u5c31\u662f\u5411\u91cfOA\u3001OB\u7684\u548c\uff0c\u8fd9\u79cd\u8ba1\u7b97\u6cd5\u5219\u53eb\u505a\u5411\u91cf\u52a0\u6cd5\u7684\u5e73\u884c\u56db\u8fb9\u5f62\u6cd5\u5219\u3002
\u5bf9\u4e8e\u96f6\u5411\u91cf\u548c\u4efb\u610f\u5411\u91cfa\uff0c\u6709\uff1a0\uff0ba\uff1da\uff0b0\uff1da\u3002
|a\uff0bb|\u2264|a|\uff0b|b|\u3002
\u5411\u91cf\u7684\u52a0\u6cd5\u6ee1\u8db3\u6240\u6709\u7684\u52a0\u6cd5\u8fd0\u7b97\u5b9a\u5f8b\u3002

\u51cf\u6cd5\u8fd0\u7b97
\u4e0ea\u957f\u5ea6\u76f8\u7b49\uff0c\u65b9\u5411\u76f8\u53cd\u7684\u5411\u91cf\uff0c\u53eb\u505aa\u7684\u76f8\u53cd\u5411\u91cf\uff0c\uff0d(\uff0da)\uff1da\uff0c\u96f6\u5411\u91cf\u7684\u76f8\u53cd\u5411\u91cf\u4ecd\u7136\u662f\u96f6\u5411\u91cf\u3002
\uff081\uff09a\uff0b(\uff0da)\uff1d(\uff0da)\uff0ba\uff1d0\uff082\uff09a\uff0db\uff1da\uff0b(\uff0db)\u3002

\u6570\u4e58\u8fd0\u7b97
\u5b9e\u6570\u03bb\u4e0e\u5411\u91cfa\u7684\u79ef\u662f\u4e00\u4e2a\u5411\u91cf\uff0c\u8fd9\u79cd\u8fd0\u7b97\u53eb\u505a\u5411\u91cf\u7684\u6570\u4e58\uff0c\u8bb0\u4f5c\u03bba\uff0c|\u03bba|\uff1d|\u03bb||a|\uff0c\u5f53\u03bb > 0\u65f6\uff0c\u03bba\u7684\u65b9\u5411\u548ca\u7684\u65b9\u5411\u76f8\u540c\uff0c\u5f53\u03bb < 0\u65f6\uff0c\u03bba\u7684\u65b9\u5411\u548ca\u7684\u65b9\u5411\u76f8\u53cd\uff0c\u5f53\u03bb = 0\u65f6\uff0c\u03bba = 0\u3002
\u8bbe\u03bb\u3001\u03bc\u662f\u5b9e\u6570\uff0c\u90a3\u4e48\uff1a\uff081\uff09(\u03bb\u03bc)a = \u03bb(\u03bca)\uff082\uff09(\u03bb + \u03bc)a = \u03bba + \u03bca\uff083\uff09\u03bb(a \u00b1 b) = \u03bba \u00b1 \u03bbb\uff084\uff09(\uff0d\u03bb)a =\uff0d(\u03bba) = \u03bb(\uff0da)\u3002

\u5411\u91cf\u7684\u52a0\u6cd5\u8fd0\u7b97\u3001\u51cf\u6cd5\u8fd0\u7b97\u3001\u6570\u4e58\u8fd0\u7b97\u7edf\u79f0\u7ebf\u6027\u8fd0\u7b97\u3002

\u5411\u91cf\u7684\u6570\u91cf\u79ef
\u5df2\u77e5\u4e24\u4e2a\u975e\u96f6\u5411\u91cfa\u3001b\uff0c\u90a3\u4e48|a||b|cos \u03b8\u53eb\u505aa\u4e0eb\u7684\u6570\u91cf\u79ef\u6216\u5185\u79ef\uff0c\u8bb0\u4f5ca•b\uff0c\u03b8\u662fa\u4e0eb\u7684\u5939\u89d2\uff0c|a|cos \u03b8\uff08|b|cos \u03b8\uff09\u53eb\u505a\u5411\u91cfa\u5728b\u65b9\u5411\u4e0a\uff08b\u5728a\u65b9\u5411\u4e0a\uff09\u7684\u6295\u5f71\u3002\u96f6\u5411\u91cf\u4e0e\u4efb\u610f\u5411\u91cf\u7684\u6570\u91cf\u79ef\u4e3a0\u3002
a•b\u7684\u51e0\u4f55\u610f\u4e49\uff1a\u6570\u91cf\u79efa•b\u7b49\u4e8ea\u7684\u957f\u5ea6|a|\u4e0eb\u5728a\u7684\u65b9\u5411\u4e0a\u7684\u6295\u5f71|b|cos \u03b8\u7684\u4e58\u79ef\u3002
\u4e24\u4e2a\u5411\u91cf\u7684\u6570\u91cf\u79ef\u7b49\u4e8e\u5b83\u4eec\u5bf9\u5e94\u5750\u6807\u7684\u4e58\u79ef\u7684\u548c\u3002

\u4e09\u89d2\u51fd\u6570\u516c\u5f0f
\u4e24\u89d2\u548c\u516c\u5f0f 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) ctg(A+B)=(ctgActgB-1)/(ctgB+ctgA) ctg(A-B)=(ctgActgB+1)/(ctgB-ctgA)
\u500d\u89d2\u516c\u5f0f tan2A=2tanA/(1-tan2A) ctg2A=(ctg2A-1)/2ctga cos2a=cos2a-sin2a=2cos2a-1=1-2sin2a
\u534a\u89d2\u516c\u5f0f sin(A/2)=\u221a((1-cosA)/2) sin(A/2)=-\u221a((1-cosA)/2) cos(A/2)=\u221a((1+cosA)/2) cos(A/2)=-\u221a((1+cosA)/2) tan(A/2)=\u221a((1-cosA)/((1+cosA)) tan(A/2)=-\u221a((1-cosA)/((1+cosA)) ctg(A/2)=\u221a((1+cosA)/((1-cosA)) ctg(A/2)=-\u221a((1+cosA)/((1-cosA))
\u79ef\u5316\u548c\u5dee 2sinAcosB=sin(A+B)+sin(A-B)
2cosAsinB=sin(A+B)-sin(A-B)
2cosAcosB=cos(A+B)-sin(A-B)
-2sinAsinB=cos(A+B)-cos(A-B)
\u548c\u5dee\u5316\u79ef sinA+sinB=2sin((A+B)/2)cos((A-B)/2
cosA+cosB=2cos((A+B)/2)sin((A-B)/2)
tanA+tanB=sin(A+B)/cosAcosB
tanA-tanB=sin(A-B)/cosAcosB
ctgA+ctgB=sin(A+B)/sinAsinB
-ctgA+ctgB=sin(A+B)/sinAsin
\u96c6\u5408\u4e0e\u51fd\u6570\u6982\u5ff5
\u4e00,\u96c6\u5408\u6709\u5173\u6982\u5ff5
1,\u96c6\u5408\u7684\u542b\u4e49:\u67d0\u4e9b\u6307\u5b9a\u7684\u5bf9\u8c61\u96c6\u5728\u4e00\u8d77\u5c31\u6210\u4e3a\u4e00\u4e2a\u96c6\u5408,\u5176\u4e2d\u6bcf\u4e00\u4e2a\u5bf9\u8c61\u53eb\u5143\u7d20.
2,\u96c6\u5408\u7684\u4e2d\u5143\u7d20\u7684\u4e09\u4e2a\u7279\u6027:
1.\u5143\u7d20\u7684\u786e\u5b9a\u6027; 2.\u5143\u7d20\u7684\u4e92\u5f02\u6027; 3.\u5143\u7d20\u7684\u65e0\u5e8f\u6027
\u8bf4\u660e:(1)\u5bf9\u4e8e\u4e00\u4e2a\u7ed9\u5b9a\u7684\u96c6\u5408,\u96c6\u5408\u4e2d\u7684\u5143\u7d20\u662f\u786e\u5b9a\u7684,\u4efb\u4f55\u4e00\u4e2a\u5bf9\u8c61\u6216\u8005\u662f\u6216\u8005\u4e0d\u662f\u8fd9\u4e2a\u7ed9\u5b9a\u7684\u96c6\u5408\u7684\u5143\u7d20.
(2)\u4efb\u4f55\u4e00\u4e2a\u7ed9\u5b9a\u7684\u96c6\u5408\u4e2d,\u4efb\u4f55\u4e24\u4e2a\u5143\u7d20\u90fd\u662f\u4e0d\u540c\u7684\u5bf9\u8c61,\u76f8\u540c\u7684\u5bf9\u8c61\u5f52\u5165\u4e00\u4e2a\u96c6\u5408\u65f6,\u4ec5\u7b97\u4e00\u4e2a\u5143\u7d20.
(3)\u96c6\u5408\u4e2d\u7684\u5143\u7d20\u662f\u5e73\u7b49\u7684,\u6ca1\u6709\u5148\u540e\u987a\u5e8f,\u56e0\u6b64\u5224\u5b9a\u4e24\u4e2a\u96c6\u5408\u662f\u5426\u4e00\u6837,\u4ec5\u9700\u6bd4\u8f83\u5b83\u4eec\u7684\u5143\u7d20\u662f\u5426\u4e00\u6837,\u4e0d\u9700\u8003\u67e5\u6392\u5217\u987a\u5e8f\u662f\u5426\u4e00\u6837.
(4)\u96c6\u5408\u5143\u7d20\u7684\u4e09\u4e2a\u7279\u6027\u4f7f\u96c6\u5408\u672c\u8eab\u5177\u6709\u4e86\u786e\u5b9a\u6027\u548c\u6574\u4f53\u6027.






\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
3,\u96c6\u5408\u7684\u8868\u793a:{ \u2026 } \u5982{\u6211\u6821\u7684\u7bee\u7403\u961f\u5458},{\u592a\u5e73\u6d0b,\u5927\u897f\u6d0b,\u5370\u5ea6\u6d0b,\u5317\u51b0\u6d0b}
1. \u7528\u62c9\u4e01\u5b57\u6bcd\u8868\u793a\u96c6\u5408:a={\u6211\u6821\u7684\u7bee\u7403\u961f\u5458},b={1,2,3,4,5}
2.\u96c6\u5408\u7684\u8868\u793a\u65b9\u6cd5:\u5217\u4e3e\u6cd5\u4e0e\u63cf\u8ff0\u6cd5.
\u6ce8\u610f\u554a:\u5e38\u7528\u6570\u96c6\u53ca\u5176\u8bb0\u6cd5:
\u975e\u8d1f\u6574\u6570\u96c6(\u5373\u81ea\u7136\u6570\u96c6) \u8bb0\u4f5c:n
\u6b63\u6574\u6570\u96c6 n*\u6216 n+ \u6574\u6570\u96c6z \u6709\u7406\u6570\u96c6q \u5b9e\u6570\u96c6r
\u5173\u4e8e"\u5c5e\u4e8e"\u7684\u6982\u5ff5
\u96c6\u5408\u7684\u5143\u7d20\u901a\u5e38\u7528\u5c0f\u5199\u7684\u62c9\u4e01\u5b57\u6bcd\u8868\u793a,\u5982:a\u662f\u96c6\u5408a\u7684\u5143\u7d20,\u5c31\u8bf4a\u5c5e\u4e8e\u96c6\u5408a \u8bb0\u4f5c a\u2208a ,\u76f8\u53cd,a\u4e0d\u5c5e\u4e8e\u96c6\u5408a \u8bb0\u4f5c a(a
\u5217\u4e3e\u6cd5:\u628a\u96c6\u5408\u4e2d\u7684\u5143\u7d20\u4e00\u4e00\u5217\u4e3e\u51fa\u6765,\u7136\u540e\u7528\u4e00\u4e2a\u5927\u62ec\u53f7\u62ec\u4e0a.
\u63cf\u8ff0\u6cd5:\u5c06\u96c6\u5408\u4e2d\u7684\u5143\u7d20\u7684\u516c\u5171\u5c5e\u6027\u63cf\u8ff0\u51fa\u6765,\u5199\u5728\u5927\u62ec\u53f7\u5185\u8868\u793a\u96c6\u5408\u7684\u65b9\u6cd5.\u7528\u786e\u5b9a\u7684\u6761\u4ef6\u8868\u793a\u67d0\u4e9b\u5bf9\u8c61\u662f\u5426\u5c5e\u4e8e\u8fd9\u4e2a\u96c6\u5408\u7684\u65b9\u6cd5.
\u2460\u8bed\u8a00\u63cf\u8ff0\u6cd5:\u4f8b:{\u4e0d\u662f\u76f4\u89d2\u4e09\u89d2\u5f62\u7684\u4e09\u89d2\u5f62}
\u2461\u6570\u5b66\u5f0f\u5b50\u63cf\u8ff0\u6cd5:\u4f8b:\u4e0d\u7b49\u5f0fx-3]2\u7684\u89e3\u96c6\u662f{x(r| x-3]2}\u6216{x| x-3]2}
4,\u96c6\u5408\u7684\u5206\u7c7b:
1.\u6709\u9650\u96c6 \u542b\u6709\u6709\u9650\u4e2a\u5143\u7d20\u7684\u96c6\u5408
2.\u65e0\u9650\u96c6 \u542b\u6709\u65e0\u9650\u4e2a\u5143\u7d20\u7684\u96c6\u5408
3.\u7a7a\u96c6 \u4e0d\u542b\u4efb\u4f55\u5143\u7d20\u7684\u96c6\u5408 \u4f8b:{x|x2=-5}
\u4e8c,\u96c6\u5408\u95f4\u7684\u57fa\u672c\u5173\u7cfb
1."\u5305\u542b"\u5173\u7cfb\u2014\u5b50\u96c6
\u6ce8\u610f:\u6709\u4e24\u79cd\u53ef\u80fd(1)a\u662fb\u7684\u4e00\u90e8\u5206,;(2)a\u4e0eb\u662f\u540c\u4e00\u96c6\u5408.
\u53cd\u4e4b: \u96c6\u5408a\u4e0d\u5305\u542b\u4e8e\u96c6\u5408b,\u6216\u96c6\u5408b\u4e0d\u5305\u542b\u96c6\u5408a,\u8bb0\u4f5cab\u6216ba
2."\u76f8\u7b49"\u5173\u7cfb(5\u22655,\u4e145\u22645,\u52195=5)
\u5b9e\u4f8b:\u8bbe a={x|x2-1=0} b={-1,1} "\u5143\u7d20\u76f8\u540c"
\u7ed3\u8bba:\u5bf9\u4e8e\u4e24\u4e2a\u96c6\u5408a\u4e0eb,\u5982\u679c\u96c6\u5408a\u7684\u4efb\u4f55\u4e00\u4e2a\u5143\u7d20\u90fd\u662f\u96c6\u5408b\u7684\u5143\u7d20,\u540c\u65f6,\u96c6\u5408b\u7684\u4efb\u4f55\u4e00\u4e2a\u5143\u7d20\u90fd\u662f\u96c6\u5408a\u7684\u5143\u7d20,\u6211\u4eec\u5c31\u8bf4\u96c6\u5408a\u7b49\u4e8e\u96c6\u5408b,\u5373:a=b
\u2460 \u4efb\u4f55\u4e00\u4e2a\u96c6\u5408\u662f\u5b83\u672c\u8eab\u7684\u5b50\u96c6.a(a
\u2461\u771f\u5b50\u96c6:\u5982\u679ca(b,\u4e14a( b\u90a3\u5c31\u8bf4\u96c6\u5408a\u662f\u96c6\u5408b\u7684\u771f\u5b50\u96c6,\u8bb0\u4f5cab(\u6216ba)
\u2462\u5982\u679c a(b, b(c ,\u90a3\u4e48 a(c
\u2463 \u5982\u679ca(b \u540c\u65f6 b(a \u90a3\u4e48a=b
3. \u4e0d\u542b\u4efb\u4f55\u5143\u7d20\u7684\u96c6\u5408\u53eb\u505a\u7a7a\u96c6,\u8bb0\u4e3a\u03c6
\u89c4\u5b9a: \u7a7a\u96c6\u662f\u4efb\u4f55\u96c6\u5408\u7684\u5b50\u96c6, \u7a7a\u96c6\u662f\u4efb\u4f55\u975e\u7a7a\u96c6\u5408\u7684\u771f\u5b50\u96c6.
\u4e09,\u96c6\u5408\u7684\u8fd0\u7b97
1.\u4ea4\u96c6\u7684\u5b9a\u4e49:\u4e00\u822c\u5730,\u7531\u6240\u6709\u5c5e\u4e8ea\u4e14\u5c5e\u4e8eb\u7684\u5143\u7d20\u6240\u7ec4\u6210\u7684\u96c6\u5408,\u53eb\u505aa,b\u7684\u4ea4\u96c6.
\u8bb0\u4f5ca\u2229b(\u8bfb\u4f5c"a\u4ea4b"),\u5373a\u2229b={x|x\u2208a,\u4e14x\u2208b}.
2,\u5e76\u96c6\u7684\u5b9a\u4e49:\u4e00\u822c\u5730,\u7531\u6240\u6709\u5c5e\u4e8e\u96c6\u5408a\u6216\u5c5e\u4e8e\u96c6\u5408b\u7684\u5143\u7d20\u6240\u7ec4\u6210\u7684\u96c6\u5408,\u53eb\u505aa,b\u7684\u5e76\u96c6.\u8bb0\u4f5c:a\u222ab(\u8bfb\u4f5c"a\u5e76b"),\u5373a\u222ab={x|x\u2208a,\u6216x\u2208b}.
3,\u4ea4\u96c6\u4e0e\u5e76\u96c6\u7684\u6027\u8d28:a\u2229a = a, a\u2229\u03c6= \u03c6, a\u2229b = b\u2229a,a\u222aa = a,a\u222a\u03c6= a ,a\u222ab = b\u222aa.
4,\u5168\u96c6\u4e0e\u8865\u96c6
(1)\u8865\u96c6:\u8bbes\u662f\u4e00\u4e2a\u96c6\u5408,a\u662fs\u7684\u4e00\u4e2a\u5b50\u96c6(\u5373),\u7531s\u4e2d\u6240\u6709\u4e0d\u5c5e\u4e8ea\u7684\u5143\u7d20\u7ec4\u6210\u7684\u96c6\u5408,\u53eb\u505as\u4e2d\u5b50\u96c6a\u7684\u8865\u96c6(\u6216\u4f59\u96c6)
\u8bb0\u4f5c: csa \u5373 csa ={x ( x(s\u4e14 x(a}
(2)\u5168\u96c6:\u5982\u679c\u96c6\u5408s\u542b\u6709\u6211\u4eec\u6240\u8981\u7814\u7a76\u7684\u5404\u4e2a\u96c6\u5408\u7684\u5168\u90e8\u5143\u7d20,\u8fd9\u4e2a\u96c6\u5408\u5c31\u53ef\u4ee5\u770b\u4f5c\u4e00\u4e2a\u5168\u96c6.\u901a\u5e38\u7528u\u6765\u8868\u793a.
(3)\u6027\u8d28:\u2474cu(c ua)=a \u2475(c ua)\u2229a=\u03c6 \u2476(cua)\u222aa=u

三角函数公式

两角和公式
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)
ctg(A+B)=(ctgActgB-1)/(ctgB+ctgA) ctg(A-B)=(ctgActgB+1)/(ctgB-ctgA)

倍角公式
tan2A=2tanA/(1-tan2A) ctg2A=(ctg2A-1)/2ctga
cos2a=cos2a-sin2a=2cos2a-1=1-2sin2a

半角公式
sin(A/2)=√((1-cosA)/2) sin(A/2)=-√((1-cosA)/2)
cos(A/2)=√((1+cosA)/2) cos(A/2)=-√((1+cosA)/2)
tan(A/2)=√((1-cosA)/((1+cosA)) tan(A/2)=-√((1-cosA)/((1+cosA))
ctg(A/2)=√((1+cosA)/((1-cosA)) ctg(A/2)=-√((1+cosA)/((1-cosA))

和差化积
2sinAcosB=sin(A+B)+sin(A-B) 2cosAsinB=sin(A+B)-sin(A-B)
2cosAcosB=cos(A+B)-sin(A-B) -2sinAsinB=cos(A+B)-cos(A-B)
sinA+sinB=2sin((A+B)/2)cos((A-B)/2 cosA+cosB=2cos((A+B)/2)sin((A-B)/2)
tanA+tanB=sin(A+B)/cosAcosB tanA-tanB=sin(A-B)/cosAcosB
ctgA+ctgBsin(A+B)/sinAsinB -ctgA+ctgBsin(A+B)/sinAsinB

某些数列前n项和
1+2+3+4+5+6+7+8+9+…+n=n(n+1)/2 1+3+5+7+9+11+13+15+…+(2n-1)=n2
2+4+6+8+10+12+14+…+(2n)=n(n+1) 12+22+32+42+52+62+72+82+…+n2=n(n+1)(2n+1)/6
13+23+33+43+53+63+…n3=n2(n+1)2/4 1*2+2*3+3*4+4*5+5*6+6*7+…+n(n+1)=n(n+1)(n+2)/3

正弦定理 a/sinA=b/sinB=c/sinC=2R 注： 其中 R 表示三角形的外接圆半径

余弦定理 b2=a2+c2-2accosB 注：角B是边a和边c的夹角

弧长公式 l=a*r a是圆心角的弧度数r >0 扇形面积公式 s=1/2*l*r

乘法与因式分 a2-b2=(a+b)(a-b) a3+b3=(a+b)(a2-ab+b2) a3-b3=(a-b(a2+ab+b2)

三角不等式 |a+b|≤|a|+|b| |a-b|≤|a|+|b| |a|≤b<=>-b≤a≤b

|a-b|≥|a|-|b| -|a|≤a≤|a|

一元二次方程的解 -b+√(b2-4ac)/2a -b-√(b2-4ac)/2a

根与系数的关系 X1+X2=-b/a X1*X2=c/a 注：韦达定理

判别式
b2-4ac=0 注：方程有两个相等的实根
b2-4ac>0 注：方程有两个不等的实根
b2-4ac<0 注：方程没有实根，有共轭复数根
降幂公式
（sin^2）x=1-cos2x/2
（cos^2）x=i=cos2x/2

万能公式
令tan(a/2)=t
sina=2t/(1+t^2)
cosa=(1-t^2)/(1+t^2)
tana=2t/(1-t^2)

对数的性质及推导
用^表示乘方，用log(a)(b)表示以a为底，b的对数
*表示乘号，/表示除号

定义式：
若a^n=b(a>0且a≠1)
则n=log(a)(b)

基本性质：
1.a^(log(a)(b))=b
2.log(a)(MN)=log(a)(M)+log(a)(N);
3.log(a)(M/N)=log(a)(M)-log(a)(N);
4.log(a)(M^n)=nlog(a)(M)

推导
1.这个就不用推了吧，直接由定义式可得(把定义式中的[n=log(a)(b)]带入a^n=b)

2.
MN=M*N
由基本性质1(换掉M和N)
a^[log(a)(MN)] = a^[log(a)(M)] * a^[log(a)(N)]
由指数的性质
a^[log(a)(MN)] = a^{[log(a)(M)] + [log(a)(N)]}
又因为指数函数是单调函数，所以
log(a)(MN) = log(a)(M) + log(a)(N)

3.与2类似处理
MN=M/N
由基本性质1(换掉M和N)
a^[log(a)(M/N)] = a^[log(a)(M)] / a^[log(a)(N)]
由指数的性质
a^[log(a)(M/N)] = a^{[log(a)(M)] - [log(a)(N)]}
又因为指数函数是单调函数，所以
log(a)(M/N) = log(a)(M) - log(a)(N)

4.与2类似处理
M^n=M^n
由基本性质1(换掉M)
a^[log(a)(M^n)] = {a^[log(a)(M)]}^n
由指数的性质
a^[log(a)(M^n)] = a^{[log(a)(M)]*n}
又因为指数函数是单调函数，所以
log(a)(M^n)=nlog(a)(M)

其他性质：

性质一：换底公式
log(a)(N)=log(b)(N) / log(b)(a)

推导如下
N = a^[log(a)(N)]
a = b^[log(b)(a)]

综合两式可得
N = {b^[log(b)(a)]}^[log(a)(N)] = b^{[log(a)(N)]*[log(b)(a)]}

又因为N=b^[log(b)(N)]
所以
b^[log(b)(N)] = b^{[log(a)(N)]*[log(b)(a)]}
所以
log(b)(N) = [log(a)(N)]*[log(b)(a)] {这步不明白或有疑问看上面的}
所以log(a)(N)=log(b)(N) / log(b)(a)

性质二：（不知道什么名字）
log(a^n)(b^m)=m/n*[log(a)(b)]

推导如下
由换底公式[lnx是log(e)(x),e称作自然对数的底]
log(a^n)(b^m)=ln(a^n) / ln(b^n)
由基本性质4可得
log(a^n)(b^m) = [n*ln(a)] / [m*ln(b)] = (m/n)*{[ln(a)] / [ln(b)]}
再由换底公式
log(a^n)(b^m)=m/n*[log(a)(b)]
--------------------------------------------（性质及推导 完 ）

公式三:
log(a)(b)=1/log(b)(a)

证明如下:
由换底公式 log(a)(b)=log(b)(b)/log(b)(a) ----取以b为底的对数,log(b)(b)=1
=1/log(b)(a)
还可变形得:
log(a)(b)*log(b)(a)=1

三角函数的和差化积公式
sinα＋sinβ＝2sin（α＋β）/2·cos（α－β）/2
sinα－sinβ＝2cos（α＋β）/2·sin（α－β）/2
cosα＋cosβ＝2cos（α＋β）/2·cos（α－β）/2
cosα－cosβ＝－2sin（α＋β）/2·sin（α－β）/2

三角函数的积化和差公式
sinα ·cosβ＝1/2 [sin（α＋β）＋sin（α－β）]
cosα ·sinβ＝1/2 [sin（α＋β）－sin（α－β）]
cosα ·cosβ＝1/2 [cos（α＋β）＋cos（α－β）]
sinα ·sinβ＝-1/2 [cos（α＋β）－cos（α－β）]

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