考點(diǎn):命題的真假判斷與應(yīng)用
專題:三角函數(shù)的圖像與性質(zhì)
分析:對(duì)于A,當(dāng)x∈(0,
)時(shí),由y=sinx,y=cosx的性質(zhì)可判斷故A錯(cuò)誤;
對(duì)于B,令x+
=kπ+
,k∈Z,當(dāng)x=
π時(shí),找不到整數(shù)k使上式成立,可判斷B錯(cuò)誤;
對(duì)于C,由tan
2x≥0,可得1+tan
2x≥1,y=
≤π,從而可判斷C正確;
對(duì)于D,y=sin(2x-
),利用三角函數(shù)的圖象變換可判斷D錯(cuò)誤.
解答:
解:對(duì)于A,當(dāng)x∈(0,
)時(shí),由y=sinx,y=cosx的性質(zhì)得:
當(dāng)x∈(0,
)時(shí),cosx>sinx,x=
時(shí),sinx=cosx,x∈(
,
)時(shí),sinx>cosx,故A錯(cuò)誤;
對(duì)于B,令x+
=kπ+
,k∈Z,顯然當(dāng)x=
π時(shí),找不到整數(shù)k使上式成立,故B錯(cuò)誤;
對(duì)于C,由于tan
2x≥0,∴1+tan
2x≥1.
∴y=
≤π.
∴函數(shù)y=
的最大值為π,C正確;
對(duì)于D,y=sin(2x-
)的圖象向右平移
個(gè)單位得到:y=sin[2(x-
)-
]=sin(2x-
)=-cos2x,故D錯(cuò)誤.
故選:C.
點(diǎn)評(píng):本題考查三角函數(shù)的圖象與性質(zhì),著重考查正弦函數(shù)與余弦函數(shù)及正切函數(shù)的單調(diào)性與最值,考查三角函數(shù)的平移變換,屬于中檔題.