高一必修4里所有三角函数公式及其推导过程 高中数学必修4三角函数公式大全

\u9ad8\u4e2d\u5fc5\u4fee4\u4e09\u89d2\u51fd\u6570\u516c\u5f0f\u3010\u590d\u6742\u7684\u3011\u4ee5\u53ca\u63a8\u5bfc\u8fc7\u7a0b

\u4f60\u597d\uff0c\u5f88\u9ad8\u5174\u80fd\u4e3a\u4f60\u56de\u7b54\u95ee\u9898\uff1a\u8fd9\u662f\u672c\u4eba\u603b\u7ed3\u7684\u4e00\u4e9b\uff0c\u5e0c\u671b\u80fd\u5e2e\u5230\u4f60\uff1aa/sinA=b/sinB=c/sinC=2R\u3000\u3000\u6240\u4ee5\u3000\u3000a=2R*sinA\u3000\u3000b=2R*sinB\u3000\u3000c=2R*sinC\u3000\u3000\u52a0\u8d77\u6765a+b+c=2R*(sinA+sinB+sinC)\u5e26\u5165\u3000\u3000(a+b+c)/(sinA+sinB+sinC)=2R*(sinA+sinB+sinC)/(sinA+sinB+sinC)=2R\u4e24\u89d2\u548c\u516c\u5f0f\u3000\u3000sin(A+B)=sinAcosB+cosAsinB\u3000\u3000sin(A-B)=sinAcosB-cosAsinB \u3000\u3000cos(A+B)=cosAcosB-sinAsinB\u3000\u3000cos(A-B)=cosAcosB+sinAsinB\u3000\u3000tan(A+B)=(tanA+tanB)/(1-tanAtanB)\u3000\u3000tan(A-B)=(tanA-tanB)/(1+tanAtanB)\u3000\u3000cot(A+B)=(cotAcotB-1)/(cotB+cotA) \u3000\u3000cot(A-B)=(cotAcotB+1)/(cotB-cotA)\u3000\u3000\u500d\u89d2\u516c\u5f0f\u3000\u3000Sin2A=2SinA?CosA
\u5e73\u65b9\u5173\u7cfb\uff1a\u3000\u3000sin^2(\u03b1)+cos^2(\u03b1)=1\u3000\u3000tan^2(\u03b1)+1=sec^2(\u03b1)\u3000\u3000cot^2(\u03b1)+1=csc^2(\u03b1)\u3000\u3000\u00b7\u5546\u7684\u5173\u7cfb\uff1a\u3000\u3000tan\u03b1=sin\u03b1/cos\u03b1cot\u03b1=cos\u03b1/sin\u03b1\u3000\u3000\u00b7\u5012\u6570\u5173\u7cfb\uff1a\u3000\u3000tan\u03b1\u00b7cot\u03b1=1\u3000\u3000sin\u03b1\u00b7csc\u03b1=1\u3000\u3000cos\u03b1\u00b7sec\u03b1=1\u4e07\u80fd\u516c\u5f0f\uff1a\u3000\u3000sin\u03b1=2tan(\u03b1/2)/[1+tan^2(\u03b1/2)]\u3000\u3000cos\u03b1=[1-tan^2(\u03b1/2)]/[1+tan^2(\u03b1/2)]\u3000\u3000tan\u03b1=2tan(\u03b1/2)/[1-tan^2(\u03b1/2)]\u5e38\u7528\u7684\u8bf1\u5bfc\u516c\u5f0f\u6709\u4ee5\u4e0b\u51e0\u7ec4\uff1a\u3000\u3000\u516c\u5f0f\u4e00\uff1a\u3000\u3000\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\u3000\u3000sin\uff082k\u03c0\uff0b\u03b1\uff09\uff1dsin\u03b1\u3000\u3000cos\uff082k\u03c0\uff0b\u03b1\uff09\uff1dcos\u03b1\u3000\u3000tan\uff082k\u03c0\uff0b\u03b1\uff09\uff1dtan\u03b1\u3000\u3000cot\uff082k\u03c0\uff0b\u03b1\uff09\uff1dcot\u03b1\u3000\u3000\u516c\u5f0f\u4e8c\uff1a\u3000\u3000\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\u3000\u3000sin\uff08\u03c0\uff0b\u03b1\uff09\uff1d\uff0dsin\u03b1\u3000\u3000cos\uff08\u03c0\uff0b\u03b1\uff09\uff1d\uff0dcos\u03b1\u3000\u3000tan\uff08\u03c0\uff0b\u03b1\uff09\uff1dtan\u03b1\u3000\u3000cot\uff08\u03c0\uff0b\u03b1\uff09\uff1dcot\u03b1\u3000\u3000\u516c\u5f0f\u4e09\uff1a\u3000\u3000\u4efb\u610f\u89d2\u03b1\u4e0e-\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a\u3000\u3000sin\uff08\uff0d\u03b1\uff09\uff1d\uff0dsin\u03b1\u3000\u3000cos\uff08\uff0d\u03b1\uff09\uff1dcos\u03b1\u3000\u3000tan\uff08\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1\u3000\u3000cot\uff08\uff0d\u03b1\uff09\uff1d\uff0dcot\u03b1\u3000\u3000\u516c\u5f0f\u56db\uff1a\u3000\u3000\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\u3000\u3000sin\uff08\u03c0\uff0d\u03b1\uff09\uff1dsin\u03b1\u3000\u3000cos\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dcos\u03b1\u3000\u3000tan\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1\u3000\u3000cot\uff08\u03c0\uff0d\u03b1\uff09\uff1d\uff0dcot\u03b1\u3000\u3000\u516c\u5f0f\u4e94\uff1a\u3000\u3000\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\u3000\u3000sin\uff082\u03c0\uff0d\u03b1\uff09\uff1d\uff0dsin\u03b1\u3000\u3000cos\uff082\u03c0\uff0d\u03b1\uff09\uff1dcos\u03b1\u3000\u3000tan\uff082\u03c0\uff0d\u03b1\uff09\uff1d\uff0dtan\u03b1\u3000\u3000cot\uff082\u03c0\uff0d\u03b1\uff09\uff1d\uff0dcot\u03b1\u3000\u3000\u516c\u5f0f\u516d\uff1a\u3000\u3000\u03c0/2\u00b1\u03b1\u53ca3\u03c0/2\u00b1\u03b1\u4e0e\u03b1\u7684\u4e09\u89d2\u51fd\u6570\u503c\u4e4b\u95f4\u7684\u5173\u7cfb\uff1a\u3000\u3000sin\uff08\u03c0/2\uff0b\u03b1\uff09\uff1dcos\u03b1\u3000\u3000cos\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dsin\u03b1\u3000\u3000tan\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcot\u03b1\u3000\u3000cot\uff08\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dtan\u03b1\u3000\u3000sin\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dcos\u03b1\u3000\u3000cos\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dsin\u03b1\u3000\u3000tan\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dcot\u03b1\u3000\u3000cot\uff08\u03c0/2\uff0d\u03b1\uff09\uff1dtan\u03b1\u3000\u3000sin\uff083\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcos\u03b1\u3000\u3000cos\uff083\u03c0/2\uff0b\u03b1\uff09\uff1dsin\u03b1\u3000\u3000tan\uff083\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dcot\u03b1\u3000\u3000cot\uff083\u03c0/2\uff0b\u03b1\uff09\uff1d\uff0dtan\u03b1\u3000\u3000sin\uff083\u03c0/2\uff0d\u03b1\uff09\uff1d\uff0dcos\u03b1\u3000\u3000cos\uff083\u03c0/2\uff0d\u03b1\uff09\uff1d\uff0dsin\u03b1\u3000\u3000tan\uff083\u03c0/2\uff0d\u03b1\uff09\uff1dcot\u03b1\u3000\u3000cot\uff083\u03c0/2\uff0d\u03b1\uff09\uff1dtan\u03b1\u3000\u3000(\u4ee5\u4e0ak\u2208Z)\u3000\u3000\u4e00\u822c\u7684\u6700\u5e38\u7528\u516c\u5f0f\u6709:\u3000\u3000Sin(A+B)=SinA*CosB+SinB*CosA\u3000\u3000Sin(A-B)=SinA*CosB-SinB*CosA\u3000\u3000Cos(A+B)=CosA*CosB-SinA*SinB\u3000\u3000Cos(A-B)=CosA*CosB+SinA*SinB\u3000\u3000Tan(A+B)=(TanA+TanB)/(1-TanA*TanB)\u3000\u3000Tan(A-B)=(TanA-TanB)/(1+TanA*TanB)\u3000\u3000\u5e73\u65b9\u5173\u7cfb\uff1a\u3000\u3000sin^2(\u03b1)+cos^2(\u03b1)=1\u3000\u3000tan^2(\u03b1)+1=sec^2(\u03b1)\u3000\u3000cot^2(\u03b1)+1=csc^2(\u03b1)\u3000\u3000\u00b7\u79ef\u7684\u5173\u7cfb\uff1a\u3000\u3000sin\u03b1=tan\u03b1*cos\u03b1\u3000\u3000cos\u03b1=cot\u03b1*sin\u03b1\u3000\u3000tan\u03b1=sin\u03b1*sec\u03b1\u3000\u3000cot\u03b1=cos\u03b1*csc\u03b1\u3000\u3000sec\u03b1=tan\u03b1*csc\u03b1\u3000\u3000csc\u03b1=sec\u03b1*cot\u03b1\u3000\u3000\u00b7\u5012\u6570\u5173\u7cfb\uff1a\u3000\u3000tan\u03b1\u00b7cot\u03b1=1\u3000\u3000sin\u03b1\u00b7csc\u03b1=1\u3000\u3000cos\u03b1\u00b7sec\u03b1=1\u3000\u3000\u76f4\u89d2\u4e09\u89d2\u5f62ABC\u4e2d,\u3000\u3000\u89d2A\u7684\u6b63\u5f26\u503c\u5c31\u7b49\u4e8e\u89d2A\u7684\u5bf9\u8fb9\u6bd4\u659c\u8fb9,\u3000\u3000\u4f59\u5f26\u7b49\u4e8e\u89d2A\u7684\u90bb\u8fb9\u6bd4\u659c\u8fb9\u3000\u3000\u6b63\u5207\u7b49\u4e8e\u5bf9\u8fb9\u6bd4\u90bb\u8fb9,\u3000\u3000\u4e09\u89d2\u51fd\u6570\u6052\u7b49\u53d8\u5f62\u516c\u5f0f\u3000\u3000\u00b7\u4e24\u89d2\u548c\u4e0e\u5dee\u7684\u4e09\u89d2\u51fd\u6570\uff1a\u3000\u3000cos(\u03b1+\u03b2)=cos\u03b1\u00b7cos\u03b2-sin\u03b1\u00b7sin\u03b2\u3000\u3000cos(\u03b1-\u03b2)=cos\u03b1\u00b7cos\u03b2+sin\u03b1\u00b7sin\u03b2\u3000\u3000sin(\u03b1\u00b1\u03b2)=sin\u03b1\u00b7cos\u03b2\u00b1cos\u03b1\u00b7sin\u03b2\u3000\u3000tan(\u03b1+\u03b2)=(tan\u03b1+tan\u03b2)/(1-tan\u03b1\u00b7tan\u03b2)\u3000\u3000tan(\u03b1-\u03b2)=(tan\u03b1-tan\u03b2)/(1+tan\u03b1\u00b7tan\u03b2)\u3000\u3000\u00b7\u8f85\u52a9\u89d2\u516c\u5f0f\uff1a\u3000\u3000Asin\u03b1+Bcos\u03b1=(A^2+B^2)^(1/2)sin(\u03b1+t)\uff0c\u5176\u4e2d\u3000\u3000sint=B/(A^2+B^2)^(1/2)\u3000\u3000cost=A/(A^2+B^2)^(1/2)\u3000\u3000\u00b7\u500d\u89d2\u516c\u5f0f\uff1a\u3000\u3000sin(2\u03b1)=2sin\u03b1\u00b7cos\u03b1=2/(tan\u03b1+cot\u03b1)\u3000\u3000cos(2\u03b1)=cos^2(\u03b1)-sin^2(\u03b1)=2cos^2(\u03b1)-1=1-2sin^2(\u03b1)\u3000\u3000tan(2\u03b1)=2tan\u03b1/[1-tan^2(\u03b1)]\u3000\u3000\u00b7\u4e09\u500d\u89d2\u516c\u5f0f\uff1a\u3000\u3000sin(3\u03b1)=3sin\u03b1-4sin^3(\u03b1)\u3000\u3000cos(3\u03b1)=4cos^3(\u03b1)-3cos\u03b1\u3000\u3000\u00b7\u534a\u89d2\u516c\u5f0f\uff1a\u3000\u3000sin(\u03b1/2)=\u00b1\u221a((1-cos\u03b1)/2)\u3000\u3000cos(\u03b1/2)=\u00b1\u221a((1+cos\u03b1)/2)\u3000\u3000tan(\u03b1/2)=\u00b1\u221a((1-cos\u03b1)/(1+cos\u03b1))=sin\u03b1/(1+cos\u03b1)=(1-cos\u03b1)/sin\u03b1\u3000\u3000\u00b7\u964d\u5e42\u516c\u5f0f\u3000\u3000sin^2(\u03b1)=(1-cos(2\u03b1))/2=versin(2\u03b1)/2\u3000\u3000cos^2(\u03b1)=(1+cos(2\u03b1))/2=vercos(2\u03b1)/2\u3000\u3000tan^2(\u03b1)=(1-cos(2\u03b1))/(1+cos(2\u03b1))\u3000\u3000\u00b7\u4e07\u80fd\u516c\u5f0f\uff1a\u3000\u3000sin\u03b1=2tan(\u03b1/2)/[1+tan^2(\u03b1/2)]\u3000\u3000cos\u03b1=[1-tan^2(\u03b1/2)]/[1+tan^2(\u03b1/2)]\u3000\u3000tan\u03b1=2tan(\u03b1/2)/[1-tan^2(\u03b1/2)]\u3000\u3000\u00b7\u79ef\u5316\u548c\u5dee\u516c\u5f0f\uff1a\u3000\u3000sin\u03b1\u00b7cos\u03b2=(1/2)[sin(\u03b1+\u03b2)+sin(\u03b1-\u03b2)]\u3000\u3000cos\u03b1\u00b7sin\u03b2=(1/2)[sin(\u03b1+\u03b2)-sin(\u03b1-\u03b2)]\u3000\u3000cos\u03b1\u00b7cos\u03b2=(1/2)[cos(\u03b1+\u03b2)+cos(\u03b1-\u03b2)]\u3000\u3000sin\u03b1\u00b7sin\u03b2=-(1/2)[cos(\u03b1+\u03b2)-cos(\u03b1-\u03b2)]\u3000\u3000\u00b7\u548c\u5dee\u5316\u79ef\u516c\u5f0f\uff1a\u3000\u3000sin\u03b1+sin\u03b2=2sin[(\u03b1+\u03b2)/2]cos[(\u03b1-\u03b2)/2]\u3000\u3000sin\u03b1-sin\u03b2=2cos[(\u03b1+\u03b2)/2]sin[(\u03b1-\u03b2)/2]\u3000\u3000cos\u03b1+cos\u03b2=2cos[(\u03b1+\u03b2)/2]cos[(\u03b1-\u03b2)/2]\u3000\u3000cos\u03b1-cos\u03b2=-2sin[(\u03b1+\u03b2)/2]sin[(\u03b1-\u03b2)/2]\u3000\u3000\u00b7\u5176\u4ed6\uff1a\u3000\u3000sin\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\u3000\u3000cos\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\u3000\u3000sin^2(\u03b1)+sin^2(\u03b1-2\u03c0/3)+sin^2(\u03b1+2\u03c0/3)=3/2\u3000\u3000tanAtanBtan(A+B)+tanA+tanB-tan(A+B)=0\u3000\u3000\u90e8\u5206\u9ad8\u7b49\u5185\u5bb9\u3000\u3000\u00b7\u9ad8\u7b49\u4ee3\u6570\u4e2d\u4e09\u89d2\u51fd\u6570\u7684\u6307\u6570\u8868\u793a(\u7531\u6cf0\u52d2\u7ea7\u6570\u6613\u5f97)\uff1a\u3000\u3000sinx=[e^(ix)-e^(-ix)]/(2i)\u3000\u3000cosx=[e^(ix)+e^(-ix)]/2\u3000\u3000tanx=[e^(ix)-e^(-ix)]/[ie^(ix)+ie^(-ix)]\u3000\u3000\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\u3000\u3000\u6b64\u65f6\u4e09\u89d2\u51fd\u6570\u5b9a\u4e49\u57df\u5df2\u63a8\u5e7f\u81f3\u6574\u4e2a\u590d\u6570\u96c6\u3002\u3000\u3000\u00b7\u4e09\u89d2\u51fd\u6570\u4f5c\u4e3a\u5fae\u5206\u65b9\u7a0b\u7684\u89e3\uff1a\u3000\u3000\u5bf9\u4e8e\u5fae\u5206\u65b9\u7a0b\u7ec4y=-y'';y=y''''\uff0c\u6709\u901a\u89e3Q,\u53ef\u8bc1\u660e\u3000\u3000Q=Asinx+Bcosx\uff0c\u56e0\u6b64\u4e5f\u53ef\u4ee5\u4ece\u6b64\u51fa\u53d1\u5b9a\u4e49\u4e09\u89d2\u51fd\u6570\u3002\u3000\u3000\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\u3000\u3000\u7279\u6b8a\u4e09\u89d2\u51fd\u6570\u503c\u3000\u3000a0`30`45`60`90`\u3000\u3000sina01/2\u221a2/2\u221a3/21\u3000\u3000cosa1\u221a3/2\u221a2/21/20\u3000\u3000tana0\u221a3/31\u221a3None\u3000\u3000cotaNone\u221a31\u221a3/30\u3000\u3000\u4e09\u89d2\u51fd\u6570\u7684\u8ba1\u7b97\u3000\u3000\u5e42\u7ea7\u6570\u3000\u3000c0+c1x+c2x2+...+cnxn+...=\u2211cnxn(n=0..\u221e)\u3000\u3000c0+c1(x-a)+c2(x-a)2+...+cn(x-a)n+...=\u2211cn(x-a)n(n=0..\u221e)\u3000\u3000\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.\u3000\u3000\u6cf0\u52d2\u5c55\u5f00\u5f0f(\u5e42\u7ea7\u6570\u5c55\u5f00\u6cd5):\u3000\u3000f(x)=f(a)+f'(a)/1!*(x-a)+f''(a)/2!*(x-a)2+...f(n)(a)/n!*(x-a)n+...\u3000\u3000\u5b9e\u7528\u5e42\u7ea7\u6570\uff1a\u3000\u3000ex=1+x+x2/2!+x3/3!+...+xn/n!+...\u3000\u3000ln(1+x)=x-x2/3+x3/3-...(-1)k-1*xk/k+...(|x|<1)\u3000\u3000sinx=x-x3/3!+x5/5!-...(-1)k-1*x2k-1/(2k-1)!+...(-\u221e<x<\u221e)\u3000\u3000cosx=1-x2/2!+x4/4!-...(-1)k*x2k/(2k)!+...(-\u221e<x<\u221e)\u3000\u3000arcsinx=x+1/2*x3/3+1*3/(2*4)*x5/5+...(|x|<1)\u3000\u3000arccosx=\u03c0-(x+1/2*x3/3+1*3/(2*4)*x5/5+...)(|x|<1)\u3000\u3000arctanx=x-x^3/3+x^5/5-...(x\u22641)\u3000\u3000sinhx=x+x3/3!+x5/5!+...(-1)k-1*x2k-1/(2k-1)!+...(-\u221e<x<\u221e)\u3000\u3000coshx=1+x2/2!+x4/4!+...(-1)k*x2k/(2k)!+...(-\u221e<x<\u221e)\u3000\u3000arcsinhx=x-1/2*x3/3+1*3/(2*4)*x5/5-...(|x|<1)\u3000\u3000arctanhx=x+x^3/3+x^5/5+...(|x|<1)\u3000\u3000--------------------------------------------------------------------------------\u3000\u3000\u5085\u7acb\u53f6\u7ea7\u6570(\u4e09\u89d2\u7ea7\u6570)\u3000\u3000f(x)=a0/2+\u2211(n=0..\u221e)(ancosnx+bnsinnx)\u3000\u3000a0=1/\u03c0\u222b(\u03c0..-\u03c0)(f(x))dx\u3000\u3000an=1/\u03c0\u222b(\u03c0..-\u03c0)(f(x)cosnx)dx\u3000\u3000bn=1/\u03c0\u222b(\u03c0..-\u03c0)(f(x)sinnx)dx\u3000\u3000\u6ce8\u610f\uff1a\u6b63\u5207\u4e5f\u53ef\u4ee5\u8868\u793a\u4e3a\u201cTg\u201d\u5982\uff1aTanA=TgA\u3000\u3000Sin2a=2SinaCosa\u3000\u3000Cos2a=Cosa^2-Sina^2\u3000\u3000=1-2Sina^2\u3000\u3000=2Cosa^2-1\u3000\u3000Tan2a=2Tana/1-Tana^2\u53ef\u80fd\u6709\u70b9\u6df1\u4e86\uff0c\u770b\u4e0d\u61c2\u6ca1\u5173\u7cfb\uff01\u795d\u4f60\u8fdb\u6b65\uff01

\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

\u516d\u89d2\u5f62\u8bb0\u5fc6\u6cd5\uff1a\uff08\u53c2\u770b\u56fe\u7247\u6216\u53c2\u8003\u8d44\u6599\u94fe\u63a5\uff09
\u6784\u9020\u4ee5"\u4e0a\u5f26\u3001\u4e2d\u5207\u3001\u4e0b\u5272\uff1b\u5de6\u6b63\u3001\u53f3\u4f59\u3001\u4e2d\u95f41"\u7684\u6b63\u516d\u8fb9\u5f62\u4e3a\u6a21\u578b\u3002
\uff081\uff09\u5012\u6570\u5173\u7cfb\uff1a\u5bf9\u89d2\u7ebf\u4e0a\u4e24\u4e2a\u51fd\u6570\u4e92\u4e3a\u5012\u6570\uff1b
\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
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|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

推导公式:(a+b+c)/(sinA+sinB+sinC)=2R(其中,R为外接圆半径)
  由正弦定理有
  a/sinA=b/sinB=c/sinC=2R
  所以
  a=2R*sinA
  b=2R*sinB
  c=2R*sinC
  加起来a+b+c=2R*(sinA+sinB+sinC)带入
  (a+b+c)/(sinA+sinB+sinC)=2R*(sinA+sinB+sinC)/(sinA+sinB+sinC)=2R
两角和公式
  sin(A+B)=sinAcosB+cosAsinB
  sin(A-B)=sinAcosB-cosAsinB
  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
  Sin2A=2SinA?CosA
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
平方关系:
  sin^2(α)+cos^2(α)=1
  tan^2(α)+1=sec^2(α)
  cot^2(α)+1=csc^2(α)
  •商的关系:
  tanα=sinα/cosαcotα=cosα/sinα
  •倒数关系:
  tanα•cotα=1
  sinα•cscα=1
  cosα•secα=1
万能公式:
  sinα=2tan(α/2)/[1+tan^2(α/2)]
  cosα=[1-tan^2(α/2)]/[1+tan^2(α/2)]
  tanα=2tan(α/2)/[1-tan^2(α/2)]
常用的诱导公式有以下几组:
  公式一:
  设α为任意角,终边相同的角的同一三角函数的值相等:
  sin(2kπ+α)=sinα
  cos(2kπ+α)=cosα
  tan(2kπ+α)=tanα
  cot(2kπ+α)=cotα
  公式二:
  设α为任意角,π+α的三角函数值与α的三角函数值之间的关系:
  sin(π+α)=-sinα
  cos(π+α)=-cosα
  tan(π+α)=tanα
  cot(π+α)=cotα
  公式三:
  任意角α与-α的三角函数值之间的关系:
  sin(-α)=-sinα
  cos(-α)=cosα
  tan(-α)=-tanα
  cot(-α)=-cotα
  公式四:
  利用公式二和公式三可以得到π-α与α的三角函数值之间的关系:
  sin(π-α)=sinα
  cos(π-α)=-cosα
  tan(π-α)=-tanα
  cot(π-α)=-cotα
  公式五:
  利用公式一和公式三可以得到2π-α与α的三角函数值之间的关系:
  sin(2π-α)=-sinα
  cos(2π-α)=cosα
  tan(2π-α)=-tanα
  cot(2π-α)=-cotα
  公式六:
  π/2±α及3π/2±α与α的三角函数值之间的关系:
  sin(π/2+α)=cosα
  cos(π/2+α)=-sinα
  tan(π/2+α)=-cotα
  cot(π/2+α)=-tanα
  sin(π/2-α)=cosα
  cos(π/2-α)=sinα
  tan(π/2-α)=cotα
  cot(π/2-α)=tanα
  sin(3π/2+α)=-cosα
  cos(3π/2+α)=sinα
  tan(3π/2+α)=-cotα
  cot(3π/2+α)=-tanα
  sin(3π/2-α)=-cosα
  cos(3π/2-α)=-sinα
  tan(3π/2-α)=cotα
  cot(3π/2-α)=tanα
  (以上k∈Z)
  一般的最常用公式有:
  Sin(A+B)=SinA*CosB+SinB*CosA
  Sin(A-B)=SinA*CosB-SinB*CosA
  Cos(A+B)=CosA*CosB-SinA*SinB
  Cos(A-B)=CosA*CosB+SinA*SinB
  Tan(A+B)=(TanA+TanB)/(1-TanA*TanB)
  Tan(A-B)=(TanA-TanB)/(1+TanA*TanB)
  平方关系:
  sin^2(α)+cos^2(α)=1
  tan^2(α)+1=sec^2(α)
  cot^2(α)+1=csc^2(α)
  •积的关系:
  sinα=tanα*cosα
  cosα=cotα*sinα
  tanα=sinα*secα
  cotα=cosα*cscα
  secα=tanα*cscα
  cscα=secα*cotα
  •倒数关系:
  tanα•cotα=1
  sinα•cscα=1
  cosα•secα=1
  直角三角形ABC中,
  角A的正弦值就等于角A的对边比斜边,
  余弦等于角A的邻边比斜边
  正切等于对边比邻边,
  三角函数恒等变形公式
  •两角和与差的三角函数:
  cos(α+β)=cosα•cosβ-sinα•sinβ
  cos(α-β)=cosα•cosβ+sinα•sinβ
  sin(α±β)=sinα•cosβ±cosα•sinβ
  tan(α+β)=(tanα+tanβ)/(1-tanα•tanβ)
  tan(α-β)=(tanα-tanβ)/(1+tanα•tanβ)
  •辅助角公式:
  Asinα+Bcosα=(A^2+B^2)^(1/2)sin(α+t),其中
  sint=B/(A^2+B^2)^(1/2)
  cost=A/(A^2+B^2)^(1/2)
  •倍角公式:
  sin(2α)=2sinα•cosα=2/(tanα+cotα)
  cos(2α)=cos^2(α)-sin^2(α)=2cos^2(α)-1=1-2sin^2(α)
  tan(2α)=2tanα/[1-tan^2(α)]
  •三倍角公式:
  sin(3α)=3sinα-4sin^3(α)
  cos(3α)=4cos^3(α)-3cosα
  •半角公式:
  sin(α/2)=±√((1-cosα)/2)
  cos(α/2)=±√((1+cosα)/2)
  tan(α/2)=±√((1-cosα)/(1+cosα))=sinα/(1+cosα)=(1-cosα)/sinα
  •降幂公式
  sin^2(α)=(1-cos(2α))/2=versin(2α)/2
  cos^2(α)=(1+cos(2α))/2=vercos(2α)/2
  tan^2(α)=(1-cos(2α))/(1+cos(2α))
  •万能公式:
  sinα=2tan(α/2)/[1+tan^2(α/2)]
  cosα=[1-tan^2(α/2)]/[1+tan^2(α/2)]
  tanα=2tan(α/2)/[1-tan^2(α/2)]
  •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)]
  •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]
  •其他:
  sinα+sin(α+2π/n)+sin(α+2π*2/n)+sin(α+2π*3/n)+……+sin[α+2π*(n-1)/n]=0
  cosα+cos(α+2π/n)+cos(α+2π*2/n)+cos(α+2π*3/n)+……+cos[α+2π*(n-1)/n]=0以及
  sin^2(α)+sin^2(α-2π/3)+sin^2(α+2π/3)=3/2
  tanAtanBtan(A+B)+tanA+tanB-tan(A+B)=0

tanα ·cotα=1
sinα ·cscα=1
cosα ·secα=1
sinα/cosα=tanα=secα/cscα
cosα/sinα=cotα=cscα/secα
sin2α+cos2α=1
1+tan2α=sec2α
1+cot2α=csc2α

诱导公式

sin(-α)=-sinα
cos(-α)=cosα tan(-α)=-tanα
cot(-α)=-cotα

sin(π/2-α)=cosα
cos(π/2-α)=sinα
tan(π/2-α)=cotα
cot(π/2-α)=tanα

sin(π/2+α)=cosα
cos(π/2+α)=-sinα
tan(π/2+α)=-cotα
cot(π/2+α)=-tanα

sin(π-α)=sinα
cos(π-α)=-cosα
tan(π-α)=-tanα
cot(π-α)=-cotα

sin(π+α)=-sinα
cos(π+α)=-cosα
tan(π+α)=tanα
cot(π+α)=cotα

sin(3π/2-α)=-cosα
cos(3π/2-α)=-sinα
tan(3π/2-α)=cotα
cot(3π/2-α)=tanα

sin(3π/2+α)=-cosα
cos(3π/2+α)=sinα
tan(3π/2+α)=-cotα
cot(3π/2+α)=-tanα

sin(2π-α)=-sinα
cos(2π-α)=cosα
tan(2π-α)=-tanα
cot(2π-α)=-cotα

sin(2kπ+α)=sinα
cos(2kπ+α)=cosα
tan(2kπ+α)=tanα
cot(2kπ+α)=cotα
(其中k∈Z)

两角和与差的三角函数公式
万能公式

sin(α+β)=sinαcosβ+cosαsinβ
sin(α-β)=sinαcosβ-cosαsinβ
cos(α+β)=cosαcosβ-sinαsinβ
cos(α-β)=cosαcosβ+sinαsinβ

tanα+tanβ
tan(α+β)=——————
1-tanα ·tanβ

tanα-tanβ
tan(α-β)=——————
1+tanα ·tanβ
2tan(α/2)
sinα=——————
1+tan2(α/2)

1-tan2(α/2)
cosα=——————
1+tan2(α/2)

2tan(α/2)
tanα=——————
1-tan2(α/2)

半角的正弦、余弦和正切公式
三角函数 的降幂公式

二倍角的正弦、余弦和正切公式
三倍角的正弦、余弦和正切公式

sin2α=2sinαcosα

cos2α=cos2α-sin2α=2cos2α-1=1-2sin2α

2tanα
tan2α=—————
1-tan2α

sin3α=3sinα-4sin3α

cos3α=4cos3α-3cosα

3tanα-tan3α
tan3α=——————
1-3tan2α

三角函数的和差化积公式
三角函数的积化和差公式

α+β α-β
sinα+sinβ=2sin—--·cos—-—
2 2
α+β α-β
sinα-sinβ=2cos—--·sin—-—
2 2
α+β α-β
cosα+cosβ=2cos—--·cos—-—
2 2
α+β α-β
cosα-cosβ=-2sin—--·sin—-—
2 2 1
sinα ·cosβ=-[sin(α+β)+sin(α-β)]
2
1
cosα ·sinβ=-[sin(α+β)-sin(α-β)]
2
1
cosα ·cosβ=-[cos(α+β)+cos(α-β)]
2
1
sinα ·sinβ=- -[cos(α+β)-cos(α-β)]
2

化asinα ±bcosα为一个角的tanα/2=(sinα/2)/(cosα/2)
=2(sinα/2)(cosα/2) / 2(cosα/2)^2
=sinα / (1+cosα)

cosα = 2(cosα/2)^2-1
所以1+cosα=2(cosα/2)^2

为什么tanα/2= sinα / (1+cosα)又等于
(1-cosα)/sinα

(sinα)^2+(cosα)^2=1
(sinα)^2=1-(cosα)^2
sinα x sinα = (1-cosα)(1+cosα)
然后,
sinα/(1+cosα) = (1-cosα)/sinα赞同4| 评论 一个三角函数的形式(辅助角的三角函数的公式)

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