人教A版高中数学必修四新课标优秀教案备课资料任意角的三角函数

备课资料

1 一、一个三角不等式的证明

已知θ∈(0,6d1a6127d3610e7b68659478ed0c2ae2.png),求证:sinθ<θ

图13

证明:如图13,设锐角θ的终边交单位圆于点P,过单位圆与x轴正半轴的交点A作圆的切线交OP于点T,过点P作PM⊥x轴于点M,则MP=sinθ,AT=tanθ,的长为θ,连结PA.

∵S△OPA扇形OPA△OAT,

∴93b05c90d14a117ba52da1d743a43ab1.png·｜OA｜·｜MP｜<93b05c90d14a117ba52da1d743a43ab1.png｜OA｜2·θ<93b05c90d14a117ba52da1d743a43ab1.png｜OA｜·｜AT｜.

∴|MP|<θ<|AT|,则MP<θ即sinθ<θ

二、备用习题

1.若cd3ba7dbe6650fbf0b092d3d3c833d5d.png<θ<6d1a6127d3610e7b68659478ed0c2ae2.png,则sinθ,cosθ,tanθ的大小关系是( )

A.tanθ

C.cosθ

2.若0<α<2π,则使sinα<aed430fdf4c64058b58e05bf9ccbbbde.png和cosα>93b05c90d14a117ba52da1d743a43ab1.png同时成立的α的取值范围是( )

A.(3a67e99b91e2ae0567dad7fb4bd2c07a.png,c87d41c12d441153b97f3593f330c121.png) B.(0,c87d41c12d441153b97f3593f330c121.png)

C.(a98ce72d61a0534e6ac1234f9e05fedc.png,2π) D.(0,c87d41c12d441153b97f3593f330c121.png)∪(a98ce72d61a0534e6ac1234f9e05fedc.png,2π)

3.在(0,2π)内,使sinx>cosx成立的x的取值范围是_______.

4.如图14,点B、C在x轴的负半轴上,且BC=CO,角α的顶点重合于坐标原点O,始边重合于x轴的正半轴,终边落在第二象限,点A在角α的终边上,且有∠BAC=45°,∠CAO=90°,求sinα,cosα,tanα.

图14

5.求函数y=b4beb82c8a39a71b449a11e0bacb0289.png+lg(25-x2)的定义域.

6.设0<β<α<6d1a6127d3610e7b68659478ed0c2ae2.png,求证:α-β>sinα-sinβ.

7.当α∈［0,2π)时,试比较sinα与cosα的大小.

参考答案:

1.D 2.D

3.(cd3ba7dbe6650fbf0b092d3d3c833d5d.png,526d66ca8c25ffc560316305ea6beb1e.png)

4.解:∵AB是∠CAO的外角的平分线,∴cd259a91c594807e48ccbab13a25baca.png=1fc8e0a1c6a82d65ef36e621432acc88.png=93b05c90d14a117ba52da1d743a43ab1.png.

在Rt△ACO中,设AC=a,则AO=2a,CO=ea7fda6d43521d1e34cd01f7875374ab.png,∴sin∠CAO=da9dd4d9820ae70db880039c2369cb44.png=2caf8383a51dc5a80a30f35520848a35.png.

∵角α的终边与OA重合,而OA落在第二象限,

∴sinα=2caf8383a51dc5a80a30f35520848a35.png,cosα=b8c3e01d8154c321a4e304428f05120c.png,tanα=7ddf743a44884d9bbc8c1789b835ac41.png.

5.x∈(-5,20e384c830c88788969e4e189603b7ee.png］∪［112cd3a4bd7251e8821eb20289c812c0.png,7153b98388afacd1997b7ce07d65cdaa.png］∪［526d66ca8c25ffc560316305ea6beb1e.png,5).

6.解:如图15,设单位圆与角α,β的终边分别交于P1,P2,作P1M1⊥x轴于M1,作P2M2⊥x轴于M2,

图15

作P2C⊥P1M于C,连结P1P2,则

sinα=M1P1,sinβ=M2P2,α-β=,

∴α-β=>P1P2>CP1=M1P1-M1C=M1P1-M2P2=sinα-sinβ,

即α-β>sinα-sinβ.

图16

7.解:如图16.

(1)当0≤α<cd3ba7dbe6650fbf0b092d3d3c833d5d.png时,设角α的终边与单位圆交于点P1(x1,y1),此时x1>y1,而sinα=y1,

cosα=x1,

∴cosα>sinα.

(2)当α=cd3ba7dbe6650fbf0b092d3d3c833d5d.png时,x1=y1,此时sinα=cosα.

(3)当cd3ba7dbe6650fbf0b092d3d3c833d5d.png<α≤6d1a6127d3610e7b68659478ed0c2ae2.png时,设角α的终边与单位圆交于点P2(x2,y2),此时y2>x2,而sinα=y2,

cosα=x2,

∴sinα>cosα.

(4)当6d1a6127d3610e7b68659478ed0c2ae2.png<α≤π时,sinα≥0,cosα<0,∴sinα>cosα.

(5)当π<α<526d66ca8c25ffc560316305ea6beb1e.png时,设角α的终边与单位圆交于点P3(x3,y3),此时x33<0,而sinα=y3,

cosα=x3,

∴sinα>cosα.

(6)当α=526d66ca8c25ffc560316305ea6beb1e.png时,有sinα=cosα.

(7)当526d66ca8c25ffc560316305ea6beb1e.png<α≤e6e9794ee82bcf3d55007ff942283739.png时,设角α的终边与单位圆交于点P4(x4,y4),此时y44<0,而sinα=y4,cosα=x4,∴sinα

(8)当e6e9794ee82bcf3d55007ff942283739.png<α<2π时,cosα≥0,sinα<0,

∴cosα>sinα.

综上所述,当α∈(cd3ba7dbe6650fbf0b092d3d3c833d5d.png,526d66ca8c25ffc560316305ea6beb1e.png)时,sinα>cosα;

当α=cd3ba7dbe6650fbf0b092d3d3c833d5d.png或526d66ca8c25ffc560316305ea6beb1e.png时,sinα=cosα;

当α∈［0,cd3ba7dbe6650fbf0b092d3d3c833d5d.png)∪(526d66ca8c25ffc560316305ea6beb1e.png,2π)时,sinα

(设计者：房增凤)

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