怎么手算三角函数 手算三角函数
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\u7531\u4e8e\u4e09\u89d2\u51fd\u6570\u7684\u5468\u671f\u6027\uff0c\u5b83\u5e76\u4e0d\u5177\u6709\u5355\u503c\u51fd\u6570\u610f\u4e49\u4e0a\u7684\u53cd\u51fd\u6570\u3002
\u4e09\u89d2\u51fd\u6570\u5728\u590d\u6570\u4e2d\u6709\u8f83\u4e3a\u91cd\u8981\u7684\u5e94\u7528\u3002\u5728\u7269\u7406\u5b66\u4e2d\uff0c\u4e09\u89d2\u51fd\u6570\u4e5f\u662f\u5e38\u7528\u7684\u5de5\u5177\u3002
\u57fa\u672c\u521d\u7b49\u5185\u5bb9
\u5b83\u6709\u516d\u79cd\u57fa\u672c\u51fd\u6570(\u521d\u7b49\u57fa\u672c\u8868\u793a)\uff1a
\u51fd\u6570\u540d \u6b63\u5f26 \u4f59\u5f26 \u6b63\u5207 \u4f59\u5207 \u6b63\u5272 \u4f59\u5272
\u6b63\u5f26\u51fd\u6570 sin\u03b8=y/r
\u4f59\u5f26\u51fd\u6570 cos\u03b8=x/r
\u6b63\u5207\u51fd\u6570 tan\u03b8=y/x
\u4f59\u5207\u51fd\u6570 cot\u03b8=x/y
\u6b63\u5272\u51fd\u6570 sec\u03b8=r/x
\u4f59\u5272\u51fd\u6570 csc\u03b8=r/y
\u4ee5\u53ca\u4e24\u4e2a\u4e0d\u5e38\u7528\uff0c\u5df2\u8d8b\u4e8e\u88ab\u6dd8\u6c70\u7684\u51fd\u6570\uff1a
\u6b63\u77e2\u51fd\u6570 versin\u03b8 =1-cos\u03b8
\u4f59\u77e2\u51fd\u6570 vercos\u03b8 =1-sin\u03b8
\u540c\u89d2\u4e09\u89d2\u51fd\u6570\u95f4\u7684\u57fa\u672c\u5173\u7cfb\u5f0f\uff1a
\u00b7\u5e73\u65b9\u5173\u7cfb\uff1a
sin^2(\u03b1)+cos^2(\u03b1)=1
tan^2(\u03b1)+1=sec^2(\u03b1)
cot^2(\u03b1)+1=csc^2(\u03b1)
\u00b7\u79ef\u7684\u5173\u7cfb\uff1a
sin\u03b1=tan\u03b1*cos\u03b1 cos\u03b1=cot\u03b1*sin\u03b1
tan\u03b1=sin\u03b1*sec\u03b1 cot\u03b1=cos\u03b1*csc\u03b1
sec\u03b1=tan\u03b1*csc\u03b1 csc\u03b1=sec\u03b1*cot\u03b1
\u00b7\u5012\u6570\u5173\u7cfb\uff1a
tan\u03b1\u00b7cot\u03b1=1
sin\u03b1\u00b7csc\u03b1=1
cos\u03b1\u00b7sec\u03b1=1
\u4e09\u89d2\u51fd\u6570\u6052\u7b49\u53d8\u5f62\u516c\u5f0f\uff1a
\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\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)
\u00b7\u500d\u89d2\u516c\u5f0f\uff1a
sin(2\u03b1)=2sin\u03b1\u00b7cos\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
sin3\u03b1=3sin\u03b1-4sin^3(\u03b1)
cos3\u03b1=4cos^3(\u03b1)-3cos\u03b1
\u00b7\u534a\u89d2\u516c\u5f0f\uff1a
sin^2(\u03b1/2)=(1-cos\u03b1)/2
cos^2(\u03b1/2)=(1+cos\u03b1)/2
tan^2(\u03b1/2)=(1-cos\u03b1)/(1+cos\u03b1)
tan(\u03b1/2)=sin\u03b1/(1+cos\u03b1)=(1-cos\u03b1)/sin\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\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
\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)]/2
cosx=[e^(ix)+e^(-ix)]/2
tanx=[e^(ix)-e^(-ix)]/[^(ix)+e^(-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
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\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
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\u4e09\u89d2\u51fd\u6570
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\u57fa\u672c\u521d\u7b49\u5185\u5bb9
\u5b83\u6709\u516d\u79cd\u57fa\u672c\u51fd\u6570(\u521d\u7b49\u57fa\u672c\u8868\u793a)\uff1a
\u51fd\u6570\u540d \u6b63\u5f26 \u4f59\u5f26 \u6b63\u5207 \u4f59\u5207 \u6b63\u5272 \u4f59\u5272
\u6b63\u5f26\u51fd\u6570 sin\u03b8=y/r
\u4f59\u5f26\u51fd\u6570 cos\u03b8=x/r
\u6b63\u5207\u51fd\u6570 tan\u03b8=y/x
\u4f59\u5207\u51fd\u6570 cot\u03b8=x/y
\u6b63\u5272\u51fd\u6570 sec\u03b8=r/x
\u4f59\u5272\u51fd\u6570 csc\u03b8=r/y
\u4ee5\u53ca\u4e24\u4e2a\u4e0d\u5e38\u7528\uff0c\u5df2\u8d8b\u4e8e\u88ab\u6dd8\u6c70\u7684\u51fd\u6570\uff1a
\u6b63\u77e2\u51fd\u6570 versin\u03b8 =1-cos\u03b8
\u4f59\u77e2\u51fd\u6570 vercos\u03b8 =1-sin\u03b8
\u540c\u89d2\u4e09\u89d2\u51fd\u6570\u95f4\u7684\u57fa\u672c\u5173\u7cfb\u5f0f\uff1a
\u00b7\u5e73\u65b9\u5173\u7cfb\uff1a
sin^2(\u03b1)+cos^2(\u03b1)=1
tan^2(\u03b1)+1=sec^2(\u03b1)
cot^2(\u03b1)+1=csc^2(\u03b1)
\u00b7\u79ef\u7684\u5173\u7cfb\uff1a
sin\u03b1=tan\u03b1*cos\u03b1 cos\u03b1=cot\u03b1*sin\u03b1
tan\u03b1=sin\u03b1*sec\u03b1 cot\u03b1=cos\u03b1*csc\u03b1
sec\u03b1=tan\u03b1*csc\u03b1 csc\u03b1=sec\u03b1*cot\u03b1
\u00b7\u5012\u6570\u5173\u7cfb\uff1a
tan\u03b1\u00b7cot\u03b1=1
sin\u03b1\u00b7csc\u03b1=1
cos\u03b1\u00b7sec\u03b1=1
\u4e09\u89d2\u51fd\u6570\u6052\u7b49\u53d8\u5f62\u516c\u5f0f\uff1a
\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\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)
\u00b7\u500d\u89d2\u516c\u5f0f\uff1a
sin(2\u03b1)=2sin\u03b1\u00b7cos\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
sin3\u03b1=3sin\u03b1-4sin^3(\u03b1)
cos3\u03b1=4cos^3(\u03b1)-3cos\u03b1
\u00b7\u534a\u89d2\u516c\u5f0f\uff1a
sin^2(\u03b1/2)=(1-cos\u03b1)/2
cos^2(\u03b1/2)=(1+cos\u03b1)/2
tan^2(\u03b1/2)=(1-cos\u03b1)/(1+cos\u03b1)
tan(\u03b1/2)=sin\u03b1/(1+cos\u03b1)=(1-cos\u03b1)/sin\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\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
\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)]/2
cosx=[e^(ix)+e^(-ix)]/2
tanx=[e^(ix)-e^(-ix)]/[^(ix)+e^(-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
规定F=SIN(X),F1=SIN~(X)(即SIN(X)的一阶导数)
F2为SIN(X)的二阶导数,
................
Fn为SIN(X)的N阶导数.
则由泰勒定理可得F=SIN(X)=1+[F1/(1!)]*X+[F2/(2!)]*(X平方)+...+[Fn/(n!)]*(X的n次方) n的植随你选取的精确度不同而不同,n越大,结果越精确
sin40°(tan10°-根号3)
=sin40*sin10/cos10-根号3*sin40
=sin40*cos80/cos10-根号3*sin40
=sin40*cos80/2sin40*cos40-根号3*sin40
=(1/2)*cos80/cos40-根号3*sin40
=[(1/2)*cos80-根号3sin40*sin40]/cos40
=[(1/2)cos80-根号3/2*sin80]/cos40
=(sin30*cos80-cos30*sin80)/cos40
=sin(30-80)/cos40
=-sin50/cos40
=-1
县把角度化为弧度
然后代入下面的公式
他有无穷多项，你可以算到需要的精度为止
sinx=x-x^3/3!+x^5/5!-x^7/7!+……+(-1)^(k-1)*x^(2k-1)/(2k-1)!+……
cosx=1-x^2/2!+x^4/4!-x^6/6!+……+(-1)^k*x^(2k)/(2k)!+……
sin40'=sin(30'+10')=sin30'cos10'+cos30'sin10'----------(1)
cos10'=sin80'=2sin40'*cos40'---------------(2)
sin10'^2+cos10'^2=1 ----------------(3)
sin40'^2+cos40'^2=1 ----------------(4)
(1)(2)(3)(4)四个方程四个未知数就可以求解.
COS50'=SIN40'
解：cos50°=cos(90°-40°)=cos90°cos40°+sin90°sin40°，而cos90°=0、sin90°=1,得原式＝sin40°.
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