反三角函数和三角函数的转换公式列一下~谢谢了~ 谁知道反三角函数的转换公式？

\u53cd\u4e09\u89d2\u51fd\u6570\u8f6c\u6362\u516c\u5f0f

\u53cd\u4e09\u89d2\u51fd\u6570\u516c\u5f0f:
arcsin(-x)=-arcsinx
arccos(-x)=\u220f\uff0darccosx
arctan(-x)=-arctanx
arccot(-x)=\u220f\uff0darccotx
arcsinx+arccosx=\u220f/2=arctanx+arccotx
sin(arcsinx)=x=cos(arccosx)=tan(arctanx)=cot(arccotx)
\u5f53x\u2208\u3014\u2014\u220f/2\uff0c\u220f/2\u3015\u65f6\uff0c\u6709arcsin(sinx)=x
\u5f53x\u2208\u30140,\u220f\u3015,arccos(cosx)=x
x\u2208(\u2014\u220f/2\uff0c\u220f/2),arctan(tanx)=x
x\u2208(0\uff0c\u220f),arccot(cotx)=x
x\u30090,arctanx=arctan1/x,arccotx\u7c7b\u4f3c
\u82e5(arctanx+arctany)\u2208(\u2014\u220f/2\uff0c\u220f/2),\u5219arctanx+arctany=arctan(x+y/1-xy)

\u540c\u89d2\u4e09\u89d2\u51fd\u6570\u7684\u57fa\u672c\u5173\u7cfb\u5f0f
\u5012\u6570\u5173\u7cfb: \u5546\u7684\u5173\u7cfb\uff1a \u5e73\u65b9\u5173\u7cfb\uff1a
tan\u03b1 \u00b7cot\u03b1\uff1d1
sin\u03b1 \u00b7csc\u03b1\uff1d1
cos\u03b1 \u00b7sec\u03b1\uff1d1 sin\u03b1/cos\u03b1\uff1dtan\u03b1\uff1dsec\u03b1/csc\u03b1
cos\u03b1/sin\u03b1\uff1dcot\u03b1\uff1dcsc\u03b1/sec\u03b1 sin2\u03b1\uff0bcos2\u03b1\uff1d1
1\uff0btan2\u03b1\uff1dsec2\u03b1
1\uff0bcot2\u03b1\uff1dcsc2\u03b1

\u8bf1\u5bfc\u516c\u5f0f
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

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\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\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

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

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

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\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

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
(\u5176\u4e2dk\u2208Z)

\u4e24\u89d2\u548c\u4e0e\u5dee\u7684\u4e09\u89d2\u51fd\u6570\u516c\u5f0f \u4e07\u80fd\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
2tan(\u03b1/2)
sin\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0btan2(\u03b1/2)

1\uff0dtan2(\u03b1/2)
cos\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0btan2(\u03b1/2)

2tan(\u03b1/2)
tan\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0dtan2(\u03b1/2)

\u534a\u89d2\u7684\u6b63\u5f26\u3001\u4f59\u5f26\u548c\u6b63\u5207\u516c\u5f0f \u4e09\u89d2\u51fd\u6570\u7684\u964d\u5e42\u516c\u5f0f

\u4e8c\u500d\u89d2\u7684\u6b63\u5f26\u3001\u4f59\u5f26\u548c\u6b63\u5207\u516c\u5f0f \u4e09\u500d\u89d2\u7684\u6b63\u5f26\u3001\u4f59\u5f26\u548c\u6b63\u5207\u516c\u5f0f
sin2\u03b1\uff1d2sin\u03b1cos\u03b1

cos2\u03b1\uff1dcos2\u03b1\uff0dsin2\u03b1\uff1d2cos2\u03b1\uff0d1\uff1d1\uff0d2sin2\u03b1

2tan\u03b1
tan2\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014
1\uff0dtan2\u03b1

sin3\u03b1\uff1d3sin\u03b1\uff0d4sin3\u03b1

cos3\u03b1\uff1d4cos3\u03b1\uff0d3cos\u03b1

3tan\u03b1\uff0dtan3\u03b1
tan3\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0d3tan2\u03b1

\u4e09\u89d2\u51fd\u6570\u7684\u548c\u5dee\u5316\u79ef\u516c\u5f0f \u4e09\u89d2\u51fd\u6570\u7684\u79ef\u5316\u548c\u5dee\u516c\u5f0f
\u03b1\uff0b\u03b2 \u03b1\uff0d\u03b2
sin\u03b1\uff0bsin\u03b2\uff1d2sin\u2014\uff0d\uff0d\u00b7cos\u2014\uff0d\u2014
2 2
\u03b1\uff0b\u03b2 \u03b1\uff0d\u03b2
sin\u03b1\uff0dsin\u03b2\uff1d2cos\u2014\uff0d\uff0d\u00b7sin\u2014\uff0d\u2014
2 2
\u03b1\uff0b\u03b2 \u03b1\uff0d\u03b2
cos\u03b1\uff0bcos\u03b2\uff1d2cos\u2014\uff0d\uff0d\u00b7cos\u2014\uff0d\u2014
2 2
\u03b1\uff0b\u03b2 \u03b1\uff0d\u03b2
cos\u03b1\uff0dcos\u03b2\uff1d\uff0d2sin\u2014\uff0d\uff0d\u00b7sin\u2014\uff0d\u2014
2 2 1
sin\u03b1 \u00b7cos\u03b2\uff1d-[sin\uff08\u03b1\uff0b\u03b2\uff09\uff0bsin\uff08\u03b1\uff0d\u03b2\uff09]
2
1
cos\u03b1 \u00b7sin\u03b2\uff1d-[sin\uff08\u03b1\uff0b\u03b2\uff09\uff0dsin\uff08\u03b1\uff0d\u03b2\uff09]
2
1
cos\u03b1 \u00b7cos\u03b2\uff1d-[cos\uff08\u03b1\uff0b\u03b2\uff09\uff0bcos\uff08\u03b1\uff0d\u03b2\uff09]
2
1
sin\u03b1 \u00b7sin\u03b2\uff1d\uff0d -[cos\uff08\u03b1\uff0b\u03b2\uff09\uff0dcos\uff08\u03b1\uff0d\u03b2\uff09]
2

\u53cd\u4e09\u89d2\u51fd\u6570\u516c\u5f0f:
arcsin(-x)=-arcsinx
arccos(-x)=\u220f\uff0darccosx
arctan(-x)=-arctanx
arccot(-x)=\u220f\uff0darccotx
arcsinx+arccosx=\u220f/2=arctanx+arccotx
sin(arcsinx)=x=cos(arccosx)=tan(arctanx)=cot(arccotx)
\u5f53x\u2208\u3014\u2014\u220f/2\uff0c\u220f/2\u3015\u65f6\uff0c\u6709arcsin(sinx)=x
\u5f53x\u2208\u30140,\u220f\u3015,arccos(cosx)=x
x\u2208(\u2014\u220f/2\uff0c\u220f/2),arctan(tanx)=x
x\u2208(0\uff0c\u220f),arccot(cotx)=x
x\u30090,arctanx=arctan1/x,arccotx\u7c7b\u4f3c
\u82e5(arctanx+arctany)\u2208(\u2014\u220f/2\uff0c\u220f/2),\u5219arctanx+arctany=arctan(x+y/1-xy)
\u540c\u89d2\u4e09\u89d2\u51fd\u6570\u7684\u57fa\u672c\u5173\u7cfb\u5f0f
\u5012\u6570\u5173\u7cfb:
\u5546\u7684\u5173\u7cfb\uff1a
\u5e73\u65b9\u5173\u7cfb\uff1a
tan\u03b1
\u00b7cot\u03b1\uff1d1
sin\u03b1
\u00b7csc\u03b1\uff1d1
cos\u03b1
\u00b7sec\u03b1\uff1d1
sin\u03b1/cos\u03b1\uff1dtan\u03b1\uff1dsec\u03b1/csc\u03b1
cos\u03b1/sin\u03b1\uff1dcot\u03b1\uff1dcsc\u03b1/sec\u03b1
sin2\u03b1\uff0bcos2\u03b1\uff1d1
1\uff0btan2\u03b1\uff1dsec2\u03b1
1\uff0bcot2\u03b1\uff1dcsc2\u03b1
\u8bf1\u5bfc\u516c\u5f0f
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
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\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\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
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
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
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\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
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
(\u5176\u4e2dk\u2208Z)
\u4e24\u89d2\u548c\u4e0e\u5dee\u7684\u4e09\u89d2\u51fd\u6570\u516c\u5f0f
\u4e07\u80fd\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
2tan(\u03b1/2)
sin\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0btan2(\u03b1/2)
1\uff0dtan2(\u03b1/2)
cos\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0btan2(\u03b1/2)
2tan(\u03b1/2)
tan\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0dtan2(\u03b1/2)
\u534a\u89d2\u7684\u6b63\u5f26\u3001\u4f59\u5f26\u548c\u6b63\u5207\u516c\u5f0f
\u4e09\u89d2\u51fd\u6570\u7684\u964d\u5e42\u516c\u5f0f
\u4e8c\u500d\u89d2\u7684\u6b63\u5f26\u3001\u4f59\u5f26\u548c\u6b63\u5207\u516c\u5f0f
\u4e09\u500d\u89d2\u7684\u6b63\u5f26\u3001\u4f59\u5f26\u548c\u6b63\u5207\u516c\u5f0f
sin2\u03b1\uff1d2sin\u03b1cos\u03b1
cos2\u03b1\uff1dcos2\u03b1\uff0dsin2\u03b1\uff1d2cos2\u03b1\uff0d1\uff1d1\uff0d2sin2\u03b1
2tan\u03b1
tan2\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014
1\uff0dtan2\u03b1
sin3\u03b1\uff1d3sin\u03b1\uff0d4sin3\u03b1
cos3\u03b1\uff1d4cos3\u03b1\uff0d3cos\u03b1
3tan\u03b1\uff0dtan3\u03b1
tan3\u03b1\uff1d\u2014\u2014\u2014\u2014\u2014\u2014
1\uff0d3tan2\u03b1
\u4e09\u89d2\u51fd\u6570\u7684\u548c\u5dee\u5316\u79ef\u516c\u5f0f
\u4e09\u89d2\u51fd\u6570\u7684\u79ef\u5316\u548c\u5dee\u516c\u5f0f
\u03b1\uff0b\u03b2
\u03b1\uff0d\u03b2
sin\u03b1\uff0bsin\u03b2\uff1d2sin\u2014\uff0d\uff0d\u00b7cos\u2014\uff0d\u2014
2
2
\u03b1\uff0b\u03b2
\u03b1\uff0d\u03b2
sin\u03b1\uff0dsin\u03b2\uff1d2cos\u2014\uff0d\uff0d\u00b7sin\u2014\uff0d\u2014
2
2
\u03b1\uff0b\u03b2
\u03b1\uff0d\u03b2
cos\u03b1\uff0bcos\u03b2\uff1d2cos\u2014\uff0d\uff0d\u00b7cos\u2014\uff0d\u2014
2
2
\u03b1\uff0b\u03b2
\u03b1\uff0d\u03b2
cos\u03b1\uff0dcos\u03b2\uff1d\uff0d2sin\u2014\uff0d\uff0d\u00b7sin\u2014\uff0d\u2014
2
2
1
sin\u03b1
\u00b7cos\u03b2\uff1d-[sin\uff08\u03b1\uff0b\u03b2\uff09\uff0bsin\uff08\u03b1\uff0d\u03b2\uff09]
2
1
cos\u03b1
\u00b7sin\u03b2\uff1d-[sin\uff08\u03b1\uff0b\u03b2\uff09\uff0dsin\uff08\u03b1\uff0d\u03b2\uff09]
2
1
cos\u03b1
\u00b7cos\u03b2\uff1d-[cos\uff08\u03b1\uff0b\u03b2\uff09\uff0bcos\uff08\u03b1\uff0d\u03b2\uff09]
2
1
sin\u03b1
\u00b7sin\u03b2\uff1d\uff0d
-[cos\uff08\u03b1\uff0b\u03b2\uff09\uff0dcos\uff08\u03b1\uff0d\u03b2\uff09]
2

解答过程所示：

反三角函数为反正弦arcsin x，反余弦arccos x，反正切arctan x，反余切arccot x，反正割arcsec x，反余割arccsc x这些函数的统称。

扩展资料：

1、反正弦函数：正弦函数y=sin x在[-π/2，π/2]上的反函数，叫做反正弦函数。记作arcsinx，表示一个正弦值为x的角，该角的范围在[-π/2，π/2]区间内。定义域[-1，1] ，值域[-π/2，π/2]。

2、反正弦函数（反三角函数之一）为正弦函数y=sinx（x∈[-½π,½π]）的反函数，记作y=arcsinx或siny=x（x∈[-1,1]）。

3、用x表示自变量，用y表示因变量（函数值）时，正弦函数

的反函数叫做反正弦函数。

参考资料来源：百度百科-反正弦函数

参考资料来源：百度百科-反三角函数

反三角函数公式:
arcsin(-x)=-arcsinx
arccos(-x)=∏－arccosx
arctan(-x)=-arctanx
arccot(-x)=∏－arccotx
arcsinx+arccosx=∏/2=arctanx+arccotx
sin(arcsinx)=x=cos(arccosx)=tan(arctanx)=cot(arccotx)
当x∈〔—∏/2，∏/2〕时，有arcsin(sinx)=x
当x∈〔0,∏〕,arccos(cosx)=x
x∈(—∏/2，∏/2),arctan(tanx)=x
x∈(0，∏),arccot(cotx)=x
x〉0,arctanx=arctan1/x,arccotx类似
若(arctanx+arctany)∈(—∏/2，∏/2),则arctanx+arctany=arctan(x+y/1-xy)

同角三角函数的基本关系式
倒数关系: 商的关系： 平方关系：
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

最佳答案里：

1＋tan2α＝sec2α
1＋cot2α＝csc2α
应该是
1 + tan^2 α=sec^2 α
1＋ cot^2 α＝csc^2 α

鍙嶄笁瑙掑嚱鏁板拰涓夎鍑芥暟鐨勮浆鎹㈠叕寮鍒椾竴涓媬璋㈣阿浜唦
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绛旓細渚嬪锛歝os60掳=1/2锛宎rccos1/2=60掳銆鍙嶄笁瑙掑嚱鏁鏄竴绉嶅熀鏈垵绛夊嚱鏁般傚畠鏄弽姝ｅ鸡arcsin x锛屽弽浣欏鸡arccos x锛屽弽姝ｅ垏arctan x锛屽弽浣欏垏arccot x锛屽弽姝ｅ壊arcsec x锛屽弽浣欏壊arccsc x杩欎簺鍑芥暟鐨缁熺О锛屽悇鑷〃绀哄叾鍙嶆寮︺佸弽浣欏鸡銆佸弽姝ｅ垏銆佸弽浣欏垏 锛屽弽姝ｅ壊锛屽弽浣欏壊涓簒鐨勮銆

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