Question

谢谢啊！求帮忙！高中数学

Answer 1

有1+(tanα)^2=(secα)^2=(1/cosα)^2
故(tanα)^2=2(tanβ)^2+1可得(1/cosα)^2=2(1/cosβ)^2可得(cosβ)^2=2(cosα)^2
cos2β-2cos2α=2(cosβ)^2-1-2[2(cosα)^2-1]=1

追问

谢谢

Answer 2

(tana)^2=2(tanb)^2+1;
(sina/cosa)^2=2(sinb/cosb)^2+1
(sina)^2/cosa)^2+1=2(sinb/cosb)^2+2
[(sina)^2+(cosa)^2]/(cosa)^2=2[(sinb)^2+(cosb)^2]/(cosb)^2
1/(cosa)^2=2/(cosb)^2
(cosb)^2=2(cosa)^2
(1+cos2b)/2=2(1+cos2a)/2
cos2b-2cos2a=1

追问

谢谢！