tan(5π+a)=-2且cosa>0则sin(a-π)=sin(-π/3)+2sin4/3π+3sin2/3π=
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 18:30:55
![tan(5π+a)=-2且cosa>0则sin(a-π)=sin(-π/3)+2sin4/3π+3sin2/3π=](/uploads/image/z/8574041-65-1.jpg?t=tan%285%CF%80%2Ba%29%3D-2%E4%B8%94cosa%3E0%E5%88%99sin%28a-%CF%80%29%3Dsin%28-%CF%80%2F3%29%2B2sin4%2F3%CF%80%2B3sin2%2F3%CF%80%3D)
tan(5π+a)=-2且cosa>0则sin(a-π)=sin(-π/3)+2sin4/3π+3sin2/3π=
tan(5π+a)=-2且cosa>0则sin(a-π)=
sin(-π/3)+2sin4/3π+3sin2/3π=
tan(5π+a)=-2且cosa>0则sin(a-π)=sin(-π/3)+2sin4/3π+3sin2/3π=
sin(a-π)= (根号5)分之2
sin(-π/3)+2sin4/3π+3sin2/3π= 自己算吧,太难写了,应该不难