两道三角恒等变换题目(需要解释)一.(1/sin10°)-(根号3/sin80°)=二.若tanα=1/3,则cos2α=
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/28 05:15:19
![两道三角恒等变换题目(需要解释)一.(1/sin10°)-(根号3/sin80°)=二.若tanα=1/3,则cos2α=](/uploads/image/z/7209094-22-4.jpg?t=%E4%B8%A4%E9%81%93%E4%B8%89%E8%A7%92%E6%81%92%E7%AD%89%E5%8F%98%E6%8D%A2%E9%A2%98%E7%9B%AE%EF%BC%88%E9%9C%80%E8%A6%81%E8%A7%A3%E9%87%8A%EF%BC%89%E4%B8%80.%EF%BC%881%2Fsin10%C2%B0%EF%BC%89-%EF%BC%88%E6%A0%B9%E5%8F%B73%2Fsin80%C2%B0%29%3D%E4%BA%8C.%E8%8B%A5tan%CE%B1%3D1%2F3%2C%E5%88%99cos2%CE%B1%3D)
两道三角恒等变换题目(需要解释)一.(1/sin10°)-(根号3/sin80°)=二.若tanα=1/3,则cos2α=
两道三角恒等变换题目(需要解释)
一.(1/sin10°)-(根号3/sin80°)=
二.若tanα=1/3,则cos2α=
两道三角恒等变换题目(需要解释)一.(1/sin10°)-(根号3/sin80°)=二.若tanα=1/3,则cos2α=
1/sin10°-√3/sin80°
=1/sin10°-√3/cos10°
=(cos10°-√3sin10°)/sin10°cos10°
=2(1/2cos10°-√3/2*sin10°)/sin10°cos10°
=2cos(10°+60°)/sin10°cos10°
=4cos70°/sin20°
=4sin20° /sin20°
=4
tanα=1/3,
cos2α
=[1-(tanα)^2]/[1+(tanα)^2]
=[1-(1/3)^2]/[1+(1/3)^2]
=(8/9)/(10/9)
=8/10
=4/5