跪求 计算cos(-25π/12)的值,谢谢
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/20 07:13:32
跪求 计算cos(-25π/12)的值,谢谢
跪求 计算cos(-25π/12)的值,谢谢
跪求 计算cos(-25π/12)的值,谢谢
cos(-25π/12)=cos(1/12π)=cos(15°)=cos(45°-30°)=cos30°cos45°+sin30°sin45°=4分之(根6加根2)
0.99348267009461174019660389923773
cos(-25π/12)=cos(-2π-1/12π)=cos(-1/12π)=cos(1/12π)约等于0.99348267009461174019660389923773
cos(-25π/12)=cos(-375°)=cos(-15°)=cos(30°-45°)=cos30°cos45°+sin30°sin45°
cos(-25π/12)=cos(25π/12)=cos(π/12)(因为周期是2pi)
=根号((1+cos(PI/6))/2)
=根号((1+0.5根号3)/2)
(√6+√2)/4
(√6+√2)/4
cos(-25π/12)=cos(2π+1/12π)
=cos(2π)*cos(1/12π)-sin(2π)*sin(1/12π)
=cos(1/12π)
=cos(π/3-π/4)
=cos(π/3)*cos(π/4)+sin(π/3)*sin(π/4)
最后的答案时4分之(根2+根6), 不好意思不会打根号。
给点分吧
cos(-25π/12)=cos(25π/12)=cos(2π+π/12)
=cos(π/12)
cos(π/12)^2=[1+cos(π/6)]/2
cos(-25π/12) 等于(1+根号3/2)再开根号
约等于 0.99348267009461174019660389923773