请问谁知道高中所有三角函数公式？ 三角函数公式 高中所有的
\u6c42\u9ad8\u4e2d\u4e09\u89d2\u51fd\u6570\u4e2d\u6240\u6709\u7684\u516c\u5f0fsin^2(\u03b1)+cos^2(\u03b1)=1 cos^2a=(1+cos2a)/2
tan^2(\u03b1)+1=sec^2(\u03b1) sin^2a=(1-cos2a)/2
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
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\u89d2\u5ea6\u6362\u7b97
\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
\u7f16\u8f91\u672c\u6bb5\u90e8\u5206\u9ad8\u7b49\u5185\u5bb9
\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
\u7f16\u8f91\u672c\u6bb5\u7279\u6b8a\u4e09\u89d2\u51fd\u6570\u503c
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
\u7f16\u8f91\u672c\u6bb5\u4e09\u89d2\u51fd\u6570\u7684\u8ba1\u7b97
\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
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\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
\u4e09\u89d2\u51fd\u6570\u7684\u6570\u503c\u7b26\u53f7
\u6b63\u5f26 \u4e00\uff0c\u4e8c\u4e3a\u6b63\uff0c \u4e09\uff0c\u56db\u4e3a\u8d1f
\u4f59\u5f26 \u4e00\uff0c\u56db\u4e3a\u6b63 \u4e8c\uff0c\u4e09\u4e3a\u8d1f
\u6b63\u5207 \u4e00\uff0c\u4e09\u4e3a\u6b63 \u4e8c\uff0c\u56db\u4e3a\u8d1f
\u7f16\u8f91\u672c\u6bb5\u4e09\u89d2\u51fd\u6570\u5b9a\u4e49\u57df\u548c\u503c\u57df
sin(x),cos(x)\u7684\u5b9a\u4e49\u57df\u4e3aR,\u503c\u57df\u4e3a\u3014-1,1\u3015
tan(x)\u7684\u5b9a\u4e49\u57df\u4e3ax\u4e0d\u7b49\u4e8e\u03c0/2+k\u03c0,\u503c\u57df\u4e3aR
cot(x)\u7684\u5b9a\u4e49\u57df\u4e3ax\u4e0d\u7b49\u4e8ek\u03c0,\u503c\u57df\u4e3aR
1\u3001sin(A+B) = sinAcosB+cosAsinB\uff1b
2\u3001sin(A-B) = sinAcosB-cosAsinB\uff1b
3\u3001cos(A+B) = cosAcosB-sinAsinB\uff1b
4\u3001cos(A-B) = cosAcosB+sinAsinB\uff1b
5\u3001tan(A+B) = (tanA+tanB)/(1-tanAtanB)\uff1b
6\u3001tan(A-B) = (tanA-tanB)/(1+tanAtanB)\uff1b
7\u3001cot(A+B) = (cotAcotB-1)/(cotB+cotA)\uff1b
8\u3001cot(A-B) = (cotAcotB+1)/(cotB-cotA)\u3002
\u4e09\u89d2\u51fd\u6570\u5e94\u7528\uff1a
\u4e09\u89d2\u51fd\u6570\u4e00\u822c\u7528\u4e8e\u8ba1\u7b97\u4e09\u89d2\u5f62\u4e2d\u672a\u77e5\u957f\u5ea6\u7684\u8fb9\u548c\u672a\u77e5\u7684\u89d2\u5ea6\uff0c\u5728\u5bfc\u822a\u3001\u5de5\u7a0b\u5b66\u4ee5\u53ca\u7269\u7406\u5b66\u65b9\u9762\u90fd\u6709\u5e7f\u6cdb\u7684\u7528\u9014\u3002\u53e6\u5916\uff0c\u4ee5\u4e09\u89d2\u51fd\u6570\u4e3a\u6a21\u7248\uff0c\u53ef\u4ee5\u5b9a\u4e49\u4e00\u7c7b\u76f8\u4f3c\u7684\u51fd\u6570\uff0c\u53eb\u505a\u53cc\u66f2\u51fd\u6570\u3002\u5e38\u89c1\u7684\u53cc\u66f2\u51fd\u6570\u4e5f\u88ab\u79f0\u4e3a\u53cc\u66f2\u6b63\u5f26\u51fd\u6570\u3001\u53cc\u66f2\u4f59\u5f26\u51fd\u6570\u7b49\u7b49\u3002
\u4e09\u89d2\u51fd\u6570\uff08\u4e5f\u53eb\u505a\u5706\u51fd\u6570\uff09\u662f\u89d2\u7684\u51fd\u6570\uff1b\u5b83\u4eec\u5728\u7814\u7a76\u4e09\u89d2\u5f62\u548c\u5efa\u6a21\u5468\u671f\u73b0\u8c61\u548c\u8bb8\u591a\u5176\u4ed6\u5e94\u7528\u4e2d\u662f\u5f88\u91cd\u8981\u7684\u3002\u4e09\u89d2\u51fd\u6570\u901a\u5e38\u5b9a\u4e49\u4e3a\u5305\u542b\u8fd9\u4e2a\u89d2\u7684\u76f4\u89d2\u4e09\u89d2\u5f62\u7684\u4e24\u4e2a\u8fb9\u7684\u6bd4\u7387\uff0c\u4e5f\u53ef\u4ee5\u7b49\u4ef7\u7684\u5b9a\u4e49\u4e3a\u5355\u4f4d\u5706\u4e0a\u7684\u5404\u79cd\u7ebf\u6bb5\u7684\u957f\u5ea6\u3002\u66f4\u73b0\u4ee3\u7684\u5b9a\u4e49\u628a\u5b83\u4eec\u8868\u8fbe\u4e3a\u65e0\u7a77\u7ea7\u6570\u6216\u7279\u5b9a\u5fae\u5206\u65b9\u7a0b\u7684\u89e3\uff0c\u5141\u8bb8\u5b83\u4eec\u6269\u5c55\u5230\u4efb\u610f\u6b63\u6570\u548c\u8d1f\u6570\u503c\uff0c\u751a\u81f3\u662f\u590d\u6570\u503c\u3002
cos(α+β)=cosαcosβ-sinαsinβ
cos(α-β)=cosαcosβ+sinαsinβ
sin(α+β)=sinαcosβ+cosαsinβ
sin(α-β)=sinαcosβ-cosαsinβ
tαn(α+β)=(tαnα+tαnβ)/(1-tαnαtαnβ)
tαn(α-β)=(tαnα+tαnβ)/(1+tαnαtαnβ)
1.万能公式
令tan(a/2)=t
sina=2t/(1+t^2)
cosa=(1-t^2)/(1+t^2)
tana=2t/(1-t^2)
2.辅助角公式
asint+bcost=(a^2+b^2)^(1/2)sin(t+r)
cosr=a/[(a^2+b^2)^(1/2)]
sinr=b/[(a^2+b^2)^(1/2)]
tanr=b/a
3.三倍角公式
sin(3a)=3sina-4(sina)^3
cos(3a)=4(cosa)^3-3cosa
tan(3a)=[3tana-(tana)^3]/[1-3(tana^2)]
4.积化和差
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
5.积化和差
sina+sinb=2sin[(a+b)/2]cos[(a-b)/2]
sina-sinb=2sin[(a-b)/2]cos[(a+b)/2]
cosa+cosb=2cos[(a+b)/2]cos[(a-b)/2]
cosa-cosb=-2sin[(a+b)/2]sin[(a-b)/2]
买本今年的备考词典全包括在内.
祝你成功.
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)
cot(A+B)=(cotAcotB-1)/(cotB+cotA) �
cot(A-B)=(cotAcotB+1)/(cotB-cotA) 倍角公式
tan2A=2tanA/[1-(tanA)^2]
cos2a=(cosa)^2-(sina)^2=2(cosa)^2 -1=1-2(sina)^2
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))
cot(A/2)=√((1+cosA)/((1-cosA)) cot(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
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 b^2=a^2+c^2-2accosB 注：角B是边a和边c的夹角
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绛旓細sin(A+B) = sinAcosB+cosAsinB sin(A-B) = sinAcosB-cosAsinB 楂樹腑闇鎺屾彙鐨涓夎鍑芥暟鍏紡涓昏鏈夛紙鍏跺疄鍦ㄦ暟瀛﹁鏈繀淇洓閲岄兘鏈夌殑锛夛細 1涓よ鍜屽叕 sin(A+B) = sinAcosB+cosAsinB sin(A-B) = sinAcosB-cosAsinB cos(A+B) = cosAcosB-sinAsinB cos(A-B) = cosAcosB+sinAsinB tan(A...
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绛旓細楂樹腑闃舵鐨涓夎鍑芥暟鍏紡鏄暟瀛﹀涔犱腑闈炲父閲嶈鐨勯儴鍒嗭紝鍏朵腑涓昏鍖呮嫭鍚岃涓夎鍑芥暟鐨勫熀鏈叧绯汇佷袱瑙掑拰涓庡樊鐨勪笁瑙掑嚱鏁板叕寮忋佽緟鍔╄鍏紡銆佸嶈鍏紡銆佷笁鍊嶈鍏紡銆佸崐瑙掑叕寮忎互鍙婂拰宸寲绉叕寮忕瓑銆傜浉鍏崇殑鍐呭濡備笅锛1銆佸悓瑙掍笁瑙掑嚱鏁扮殑鍩烘湰鍏崇郴寮忥細sin^2锛坸锛+cos^2锛坸锛=1銆傝繖涓叕寮忚〃绀哄湪浠讳綍涓涓搴涓嬶紝姝ｅ鸡...
绛旓細鐩磋涓夎褰BC涓,瑙扐鐨勬寮﹀煎氨绛変簬瑙扐鐨勫杈规瘮鏂滆竟,浣欏鸡绛変簬瑙扐鐨勯偦杈规瘮鏂滆竟 姝ｅ垏绛変簬瀵硅竟姣旈偦杈,路涓夎鍑芥暟鎭掔瓑鍙樺舰鍏紡 路涓よ鍜屼笌宸殑涓夎鍑芥暟锛歝os(伪+尾)=cos伪路cos尾-sin伪路sin尾 cos(伪-尾)=cos伪路cos尾+sin伪路sin尾 sin(伪卤尾)=sin伪路cos尾卤cos伪路sin尾 tan(伪+尾)=...
绛旓細cos(60掳+a)] =4cosacos(60掳-a)cos(60掳+a) 涓婅堪涓ゅ紡鐩告瘮鍙緱 tan3a=tanatan(60掳-a)tan(60掳+a) 鐜板垪鍑鍏紡濡備笅: sin2伪=2sin伪cos伪 tan2伪=2tan伪/(1-tan伪 ) cos2伪=cos伪-sin伪=2cos伪-1=1-2sin伪 鍙埆杞昏杩欎簺瀛楃,瀹冧滑鍦ㄦ暟瀛﹀涔犱腑浼氳捣鍒伴噸瑕佷綔鐢,鍖呮嫭鍦ㄤ竴浜涘浘鍍忛棶棰樺拰鍑芥暟...
绛旓細涓夎鍑芥暟鐨勮瀵鍏紡(鍏叕寮) 鍏紡涓: sin(-伪) = -sin伪 cos(-伪) = cos伪 tan (-伪)=-tan伪 鍏紡浜: sin(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: sin(蟺-伪) = sin伪 cos(蟺-伪) = -cos伪 鍏紡浜: sin...
