cos α-sin α=-=-=-=-.
[答案] -
8.已知2cos2x+sin 2x=Asin(ωx+φ)+b(A>0),则A=________,b=________.
【导学号:79402131】
[解析] ∵2cos2x+sin 2x=1+cos 2x+sin 2x=1+sin,
∴1+sin=Asin(ωx+φ)+b,∴A=,b=1.
[答案] 1
三、解答题
9.化简.
[解] 原式=
====1.
10.(1)已知sin α+cos α=,求cos 2α,tan 2α的值.
(2)已知sin sin =,求sin 2α的值.
[解] (1)因为(sin α+cos α)2=,所以1+2sin αcos α=,所以2sin αcos α=sin 2α=-,所以(sin α-cos α)2=1-2sin αcos α=1+=.
又<α<π,所以sin α>0,cos α<0,所以sin α-cos α=,所以cos 2α=cos2α-sin2α=(cos α+sin α)·(cos α-sin α)=×=-,
所以tan 2α===.