sin3π/8cos3π/8=1/3,则tan3π/8+tanπ/8sin3π/8cos3π/8=1/3,则tan3π/8+tanπ/8=?

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/16 09:14:48

sin3π/8cos3π/8=1/3,则tan3π/8+tanπ/8sin3π/8cos3π/8=1/3,则tan3π/8+tanπ/8=?
sin3π/8cos3π/8=1/3,则tan3π/8+tanπ/8
sin3π/8cos3π/8=1/3,则tan3π/8+tanπ/8=?

sin3π/8cos3π/8=1/3,则tan3π/8+tanπ/8sin3π/8cos3π/8=1/3,则tan3π/8+tanπ/8=?
sin3π/8cos3π/8=1/3
2*sin3π/8cos3π/8=2/3
sin²(3π/8)=2/3
sin(3π/8)=√6/3
cos²(3π/8)=1-sin²(3π/8)=1-2/3=1/3
cos(3π/8)=√3/3
tan(3π/8)=sin(3π/8)/cos(3π/8)=(√6/3)/(√3/3)=(√6/3)x(3/√3)=√2
tan(3π/8)=tan[3(π/8)]=[3tan(π/8)-tan³(π/8)]/[1-3tan²(π/8)]=√2
tan(π/8)=tan(45°/2)=sin45°/(1+cos45°)=(√2/2)/(1+√2/2)=(√2)/(2+√2)=(√2)(2-√2)/2=√2-1
∴tan3π/8+tanπ/8 = √2+√2-1 = 2√2-1