Question

高一数学，谢谢

已知tg^a=2tg^b+1,求证sin^b=2sin^a-1

Answer 1

很简单嘛 你学过倒推在正推没
tg^a=2tg^b+1
推出sin^a 2sin^b
--- = ------- +1
cos^a cos^b
推出
sin^a 2sin^b+cos^b
--- = -------------
cos^a cos^b
推出
sin^a sin^b+1
--- = --------
cos^a cos^b

推出sin^acos^b=sin^bcos^a+cos^a
推出 -cos^a=sin^bcos^a-sin^acos^b(这是正推）
sin^b=2sin^a-1
推出sin^b=sin^a-cos^a 又上面可知

-cos^a=sin^bcos^a-sin^acos^b

所以sin^b=sin^a＋sin^bcos^a-sin^acos^b

变形sin^b（1－cos^a ）＝sin^a（1－cos^b）

所以sin^bsin^a＝sin^asin^b显然成立 所以 sin^b=2sin^a-1