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若,则sin2θ+cos2θ= .
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高一上学期数学《》真题及答案
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下列公式是正确的
sin 2α+cos2α=1
sin 2α-cos2α=1
sin 2α+cos2α=-1
sin 2α-cos2α=-1
已知fx=x2+sinθ﹣cosθx+sinθθ∈R.的图象关于y轴对称则sin2θ+cos2θ的值
2
1
cos2–sin2=
下列公式是正确的
sin 2α= 2sin α cos α
sin 2α= cos2α-sin 2α
sin 2α= 21(1+ cos2 α)
sin 2α= 21(1- cos2 α)
已知α为第四象限角且cosα﹣|sinα﹣cosα|=﹣求tanαsin2αcos2α的值.
化简得到
sin2α
﹣sin2α
cos2α
﹣cos2α
若α∈0π且cos2α=sin+α则sin2α的值为.
已知sinα=+cosα且α∈0则sin2α=cos2α=.
定义在R.上的偶函数fx满足fx=fx+2当x∈[35]时fx=2-|x-4|则
f(sin
)
)
f(sin1)>f(cos1)
f(cos
)
)
f(cos2)>f(sin2)
要证明sin4θ-cos4θ=2sin2θ-1过程为sin4θ-cos4θ=sin2θ+cos2θs
分析法
反证法
综合法
间接证明法
下列公式是正确的
sin2 α= 2sin α? cos α
sin2 α= 21(1- cos2 α)
sin2 α= 21(1+ cos2 α)
sin2 α= 21sin αcosα
已知tanα=2求下列代数式的值.2sin2α+sinαcosα+cos2α.
设α为第四象限的角若=则cos2α﹣sin2α=.
已知sinα=<α<π求sin2αcos2α的值.
若cosα﹣=cos2α则sin2α=
﹣1
如图所示小球由细线AB.AC拉住静止AB保持水平AC与竖直方向成α角此时AC对球的拉力为T.1.现将
)1:1 (
)1:cos2α (
)cos2α:1 (
)sin2α:cos2α
用tanα表示sin2αcos2α和tan2α.
已知abc为正数且满足acos2θ+bsin2θ
定义在R.上的函数fx满足fx=fx+2当x∈[35]时fx=2-|x-4|则
f(sin
)
)
f(sin1)>f(cos1)
f(cos
)
)
f(cos2)>f(sin2)
点Axy在单位圆x2+y2=1上沿逆时针方向匀速旋转每秒旋转ω弧度己知1秒时点A的坐标为10则3秒
(cos2ω,sin2ω)
(cos3ω,sin3ω)
(
cos2ω,
sin2ω)
(2cos
,2sin
)
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