求证:tan(x/2+派/4)+tan(x/2-派/4)=2tanx{[tan(x/2)+1]^2-[1-tan(x/2)]^2}=4tan(x/2)具体怎么得到

问题描述:

求证:tan(x/2+派/4)+tan(x/2-派/4)=2tanx
{[tan(x/2)+1]^2-[1-tan(x/2)]^2}=4tan(x/2)具体怎么得到

tan(x/2+派/4)+tan(x/2-派/4)=[sin(x/2+派/4)*cos(x/2-派/4)+cos(x/2+派/4)*sin(x/2-派/4)]/[cos(x/2-派/4)cos(x/2+派/4)]=sinx / [1/2*(cosx+cos派/2)]=sinx/[1/2*cosx]=2tanx

直接按完全平方公式展开就得到了

设,tan(x/2)=a
(a+1)^2-(1-a)^2=a^2+2a+1-(a^2-2a+1)=4a=4tan(x/2)