已知tan2α=2tan2β+1,求证sin2β=2sin2α-1
问题描述：
已知tan2α=2tan2β+1,求证sin2β=2sin2α-1
函数后面的2都是代表平方
答
因为(tanα)^2=2*(tanβ)^2+1所以(sinα)^2/(cosα)^2=2*(sinβ)^2/(cosβ)^2+1两边同时乘以(cosα)^2*(cosβ)^2,有(sinα)^2*(cosβ)^2=2*(sinβ)^2*(cosα)^2+(cosα)^2*(cosβ)^2又有(cosα)^2=1-(sinα)^2,(cos...