求一列完整的三角函数公式
发布网友
发布时间：2022-04-30 08:01
共1个回答
热心网友
时间：2022-06-19 11:26
sin(-a)
=
-
sin(a)
cos(-a)
=
cos(a)
sin(π/2
-
a)
=
cos(a)
cos(π/2
-
a)
=
sin(a)
sin(π/2
+
a)
=
cos(a)
cos(π/2
+
a)
=
-
sin(a)
sin(π
-
a)
=
sin(a)
cos(π
-
a)
=
-
cos(a)
sin(π
+
a)
=
-
sin(a)
cos(π
+
a)
=
-
cos(a)
2.两角和与差的三角函数
sin(a
+
b)
=
sin(a)cos(b)
+
cos(α)sin(b)
cos(a
+
b)
=
cos(a)cos(b)
-
sin(a)sin(b)
sin(a
-
b)
=
sin(a)cos(b)
-
cos(a)sin(b)
cos(a
-
b)
=
cos(a)cos(b)
+
sin(a)sin(b)
tan(a
+
b)
=
[tan(a)
+
tan(b)]
/
[1
-
tan(a)tan(b)]
tan(a
-
b)
=
[tan(a)
-
tan(b)]
/
[1
+
tan(a)tan(b)]
3.和差化积公式
sin(a)
+
sin(b)
=
2sin[(a
+
b)/2]cos[(a
-
b)/2]
sin(a)
sin(b)
=
2cos[(a
+
b)/2]sin[(a
-
b)/2]
cos(a)
+
cos(b)
=
2cos[(a
+
b)/2]cos[(a
-
b)/2]
cos(a)
-
cos(b)
=
-
2sin[(a
+
b)/2]sin[(a
-
b)/2]
4.积化和差公式
sin(a)sin(b)
=
-
1/2[cos(a
+
b)
-
cos(a
-
b)]
cos(a)cos(b)
=
1/2[cos(a
+
b)
+
cos(a
-b)]
sin(a)cos(b)
=
1/2[sin(a
+
b)
+
sin(a
-
b)]
5.二倍角公式
sin(2a)
=
2sin(a)cos(b)
cos(2a)
=
cos2(a)
-
sin2(a)
=
2cos2(a)
-1=1
-
2sin2(a)
6.半角公式
sin2(a/2)
=
[1
-
cos(a)]
/
2
cos2(a/2)
=
[1
+
cos(a)]
/
2
tan(a/2)
=
[1
-
cos(a)]
/sin(a)
=
sina
/
[1
+
cos(a)]
7.万能公式
sin(a)
=
2tan(a/2)
/
[1+tan2(a/2)]
cos(a)
=
[1-tan2(a/2)]
/
[1+tan2(a/2)]
tan(a)
=
2tan(a/2)
/
[1-tan2(a/2)]
