sin3π/8cos3π/8=1/3,则tan3π/8+tanπ/8
问题描述：
sin3π/8cos3π/8=1/3,则tan3π/8+tanπ/8
sin3π/8cos3π/8=1/3，则tan3π/8+tanπ/8=？
答
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为什么sin2(3π/8)=2tan(3π/8)/[1+tan²(3π/8)]=2/3呢？就是说sin2α=tan2α？万能公式：sina=2tan(a/2)/[1+tan²(a/2) ]