![已知cos2α=4/5,且2α∈[π,2π],求sinα cos2α](https://img.wumaow.org/upload/tu/441037115.jpg)
已知cos2α=4/5,且2α∈[π,2π],求sinα

2cosα-√2sinα 化简

1-2cos(α/2的n次方 化简

cos(2a-π/2) =cos(π/2 -2a) =sin2a 请问这是怎么转换的呢

cos(π-α)=1/2,则cos2α
已知cos2α=4/5,且2α∈[π,2π],求sinα
2cosα-√2sinα 化简
1-2cos(α/2的n次方 化简
cos(2a-π/2) =cos(π/2 -2a) =sin2a 请问这是怎么转换的呢
cos(π-α)=1/2,则cos2α