两角和与差的正余弦正切2高一数学教案

讲解新课：两角和与差的正弦1
推导sin((+()=cos[
(((+()]=cos[(
(()((]=cos(
(()cos(+sin(
(()sin(=sin(cos(+cos(sin(即：
（S(+(）以((代(得：
（S(((）2
公式的分析，结构解剖，求值和恒等变形
教学重点：由两角和的余弦公式推导出两角和的正弦公式教学难点：进行简单的三角函数式的化简，余弦，嘱记三，求下列各式的值：1(sin75(2(sin13(cos17(+cos13(sin17(解：1(原式=sin(30(+45()=sin30(cos45(+cos30(sin45(=
2(原式=sin(13(+17()=sin30(=
例2求证：cos(+
sin(=2sin(
+()证一（构造辅助角）：左边=2(
cos(+
sin()=2(sin
cos(+cos
sin()=2sin(
+()=右边证二：右边=2(sin
cos(+cos
sin()=2(
cos(+
sin()=cos(+
sin(=左边例3已知sin((+()=
，复习引入：1．两角和与差的余弦公式:

2．求cos75(的值解：cos75(=cos(45(+30()=cos45(cos30((sin45(sin30(=
3．计算：cos65(cos115((cos25(sin115(解：原式=cos65(cos115((sin65(sin115(=cos(65(+115()=cos180(=(14
计算：(cos70(cos20(+sin110(sin20(原式=(cos70(cos20(+sin70(sin20(=(cos(70(+20()=05．已知锐角(，讲解范例：例1不查表，课题：4
6两角和与差的正弦，并进而推得两角和的正弦公式，并运用进行简单的三角函数式的化简，(满足cos(=
cos((+()=
求cos(
解：∵cos(=
∴sin(=
又∵cos((+()=
<0∴(+(为钝角∴sin((+()=
∴cos(=cos[((+()((]=cos((+()cos(+sin((+()sin(=
（角变换技巧）二，正切（2）教学目的：能由两角和的余弦公式推导出两角和的正弦公式，求值和恒等变形授课类型：新授课课时安排：1课时教具：多媒体，实物投影仪教学过程：一，sin(((()=
求
的值解：∵sin((+()=
∴sin(cos(+cos(sin(=
，