题目内容
若sin(+α)=,则cos2α=________.
若|sin(4π-α)|=sin(π+α),则角α的取值范围是________.
[答案] [2kπ-π,2kπ],(k∈Z)
[解析] ∵|sin(4π-α)|=sin(π+α),
∴|sinα|=-sinα,∴sinα≤0,
∴2kπ-π≤α≤2kπ,k∈Z.
若sinα+cosα=-,则tanα+=( )
A.1 B.2 C.-1 D.-2
若sin α+cos α=,则sin 2α= .
若sin α+cos α=,则sin 2α=