人教A版高中數(shù)學(xué)（必修第一冊(cè)）同步講義 5.2三角函數(shù)概念及同角三角函數(shù)（教師版）

第第頁(yè)5.2三角函數(shù)概念及同角三角函數(shù)5.2.1三角函數(shù)的概念課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①理解結(jié)合單位圓定義三角函數(shù)的意義。②結(jié)合任意角終邊與單位圓的交點(diǎn)會(huì)求任意角的正弦、余弦、正切值。③根據(jù)任意角終邊所在象限的位置，會(huì)判斷任意角三角函數(shù)值的符號(hào)。1.掌握三角函數(shù)的定義；2會(huì)求任意角的三個(gè)三角函數(shù)值；3.能準(zhǔn)確判斷任意角的三角函數(shù)值的符號(hào)；知識(shí)點(diǎn)01：任意角的三角函數(shù)定義1、單位圓定義法：如圖,設(shè)SKIPIF1<0是一個(gè)任意角,SKIPIF1<0,它的終邊SKIPIF1<0與單位圓相交于點(diǎn)SKIPIF1<0①正弦函數(shù)：把點(diǎn)SKIPIF1<0的縱坐標(biāo)SKIPIF1<0叫做SKIPIF1<0的正弦函數(shù),記作SKIPIF1<0,即SKIPIF1<0②余弦函數(shù)：把點(diǎn)SKIPIF1<0的橫坐標(biāo)SKIPIF1<0叫做SKIPIF1<0的余弦函數(shù),記作SKIPIF1<0,即SKIPIF1<0

③正切函數(shù)：把點(diǎn)SKIPIF1<0的縱坐標(biāo)與橫坐標(biāo)的比值SKIPIF1<0叫做SKIPIF1<0的正切,記作SKIPIF1<0,即SKIPIF1<0(SKIPIF1<0)

我們將正弦函數(shù)、余弦函數(shù)和正切函數(shù)統(tǒng)稱為三角函數(shù)【即學(xué)即練1】已知角θ的終邊經(jīng)過點(diǎn)SKIPIF1<0，則SKIPIF1<0等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0，故SKIPIF1<0在單位圓上，根據(jù)三角函數(shù)值的定義，SKIPIF1<0的橫坐標(biāo)的值即為SKIPIF1<0，故SKIPIF1<0.故選：B2、終邊上任意一點(diǎn)定義法：在角SKIPIF1<0終邊上任取一點(diǎn)SKIPIF1<0，設(shè)原點(diǎn)到SKIPIF1<0點(diǎn)的距離為SKIPIF1<0①正弦函數(shù)：SKIPIF1<0②余弦函數(shù)：SKIPIF1<0

③正切函數(shù)：SKIPIF1<0(SKIPIF1<0)

【即學(xué)即練2】若點(diǎn)SKIPIF1<0在角SKIPIF1<0的終邊上，則SKIPIF1<0.【答案】SKIPIF1<0/0.8【詳解】點(diǎn)SKIPIF1<0在角SKIPIF1<0的終邊上，所以SKIPIF1<0.故答案為：SKIPIF1<0.知識(shí)點(diǎn)02：三角函數(shù)值在各象限的符號(hào)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0在各象限的符號(hào)如下:（口訣“一全正,二正弦,三正切,四余弦”）知識(shí)點(diǎn)03：特殊的三角函數(shù)值角度SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0弧度SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0正弦值SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0余弦值SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0正切值SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0知識(shí)點(diǎn)04：誘導(dǎo)公式一(1)語(yǔ)言表示:終邊相同的角的同一三角函數(shù)的值相等.

(2)式子表示:①SKIPIF1<0②SKIPIF1<0③SKIPIF1<0其中SKIPIF1<0.

知識(shí)點(diǎn)05：三角函數(shù)線設(shè)SKIPIF1<0角的終邊與單位圓相交點(diǎn)SKIPIF1<0；④由點(diǎn)SKIPIF1<0向SKIPIF1<0軸做垂線，垂足為點(diǎn)SKIPIF1<0；⑤由點(diǎn)SKIPIF1<0作單位圓的切線與終邊相交于點(diǎn)SKIPIF1<0。如下圖所示：在SKIPIF1<0中：SKIPIF1<0SKIPIF1<0為正弦線，長(zhǎng)度為正弦值。SKIPIF1<0SKIPIF1<0為余弦線，長(zhǎng)度為余弦值。在SKIPIF1<0中：SKIPIF1<0。SKIPIF1<0為正切線，長(zhǎng)度為正切值。題型01利用三角函數(shù)的定義求三角函數(shù)值【典例1】）已知角SKIPIF1<0的頂點(diǎn)在原點(diǎn)，始邊與SKIPIF1<0軸的非負(fù)半軸重合，終邊在第三象限且與單位圓交于點(diǎn)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】SKIPIF1<0在單位圓上即SKIPIF1<0終邊在第三象限所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0所以SKIPIF1<0.故選：C【典例2】已知角SKIPIF1<0的終邊與單位圓交于點(diǎn)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0的終邊與單位圓交于點(diǎn)SKIPIF1<0,故SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，故選：B.【典例3】在平面直角坐標(biāo)系SKIPIF1<0中，角SKIPIF1<0的頂點(diǎn)與原點(diǎn)SKIPIF1<0重合，它的始邊與SKIPIF1<0軸的非負(fù)半軸重合，終邊SKIPIF1<0交單位圓SKIPIF1<0于點(diǎn)SKIPIF1<0，則SKIPIF1<0的值為A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由題意，角SKIPIF1<0的頂點(diǎn)與原點(diǎn)SKIPIF1<0重合，它的始邊與SKIPIF1<0軸的非負(fù)半軸重合，終邊SKIPIF1<0交單位圓SKIPIF1<0于點(diǎn)SKIPIF1<0，根據(jù)三角函數(shù)的定義可得SKIPIF1<0.故選：C.【變式1】設(shè)角SKIPIF1<0的終邊與單位圓的交點(diǎn)坐標(biāo)為SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．1【答案】C【詳解】由題意得SKIPIF1<0，故選：C【變式2】已知角SKIPIF1<0的頂點(diǎn)與原點(diǎn)重合，始邊與x軸的非負(fù)半軸重合，若SKIPIF1<0的終邊與圓心在原點(diǎn)的單位圓交于SKIPIF1<0，且SKIPIF1<0為第四象限角，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】SKIPIF1<0在單位圓上，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0為第四象限角，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，故選：B.題型02由終邊或終邊上點(diǎn)求三角函數(shù)值【典例1】已知角SKIPIF1<0的終邊落在直線SKIPIF1<0上，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】設(shè)直線SKIPIF1<0上任意一點(diǎn)P的坐標(biāo)為SKIPIF1<0（SKIPIF1<0），則SKIPIF1<0（O為坐標(biāo)原點(diǎn)），根據(jù)正弦函數(shù)的定義得：SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)，SKIPIF1<0，所以選項(xiàng)D正確，選項(xiàng)A，B，C錯(cuò)誤，故選：D.【典例2】（多選）已知角SKIPIF1<0的終邊經(jīng)過點(diǎn)SKIPIF1<0，則SKIPIF1<0的值可能為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】CD【詳解】已知角SKIPIF1<0的終邊經(jīng)過點(diǎn)SKIPIF1<0所以SKIPIF1<0，SKIPIF1<0則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0；所以SKIPIF1<0的值可能為SKIPIF1<0或SKIPIF1<0.故選：CD.【典例3】）已知角SKIPIF1<0的終邊過點(diǎn)SKIPIF1<0，且SKIPIF1<0，求SKIPIF1<0及SKIPIF1<0的值．【答案】SKIPIF1<0，SKIPIF1<0【詳解】由角SKIPIF1<0的終邊過點(diǎn)SKIPIF1<0，可知SKIPIF1<0，又SKIPIF1<0，得SKIPIF1<0．所以SKIPIF1<0，SKIPIF1<0.【變式1】若角SKIPIF1<0的終邊經(jīng)過點(diǎn)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】設(shè)SKIPIF1<0，則點(diǎn)SKIPIF1<0到原點(diǎn)的距離為SKIPIF1<0，則SKIPIF1<0.故選：D.【變式2】（多選）已知角SKIPIF1<0的終邊上有一點(diǎn)SKIPIF1<0，若SKIPIF1<0，則（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】BD【詳解】由題知，因?yàn)镾KIPIF1<0，所以點(diǎn)SKIPIF1<0在第三象限，所以SKIPIF1<0，SKIPIF1<0，故選：BD.【變式3】若SKIPIF1<0的終邊所在射線經(jīng)過點(diǎn)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0.【答案】SKIPIF1<0/SKIPIF1<0SKIPIF1<0【詳解】由于SKIPIF1<0的終邊所在射線經(jīng)過點(diǎn)SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<0題型03由三角函數(shù)值求終邊上的點(diǎn)或參數(shù)【典例1】角SKIPIF1<0的終邊經(jīng)過點(diǎn)SKIPIF1<0且SKIPIF1<0，則b的值為（

）A．3 B．SKIPIF1<0 C．SKIPIF1<0 D．5【答案】B【詳解】根據(jù)三角函數(shù)定義可得SKIPIF1<0，且SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0．故選：B．【典例2】已知角SKIPIF1<0的終邊上有一點(diǎn)SKIPIF1<0，且SKIPIF1<0，則實(shí)數(shù)m取值為．【答案】0或SKIPIF1<0【詳解】因?yàn)榻荢KIPIF1<0的終邊上有一點(diǎn)SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.故答案為：0或SKIPIF1<0.【典例3】已知角SKIPIF1<0的終邊上一點(diǎn)SKIPIF1<0，且SKIPIF1<0，求SKIPIF1<0值.【答案】SKIPIF1<0或SKIPIF1<0.【詳解】解：依題意有：SKIPIF1<0即：SKIPIF1<0解得：SKIPIF1<0或SKIPIF1<0即SKIPIF1<0或SKIPIF1<0【變式1】已知角SKIPIF1<0的終邊經(jīng)過點(diǎn)SKIPIF1<0，且SKIPIF1<0，則實(shí)數(shù)m的值是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0D．SKIPIF1<0或SKIPIF1<0【答案】A【詳解】由三角函數(shù)的定義得SKIPIF1<0，SKIPIF1<0解得SKIPIF1<0故選：A【變式2】已知角SKIPIF1<0的終邊經(jīng)過點(diǎn)SKIPIF1<0，且SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0.【答案】SKIPIF1<0【詳解】由題意，根據(jù)余弦函數(shù)的定義，可得SKIPIF1<0.整理得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.【變式3】已知點(diǎn)SKIPIF1<0角SKIPIF1<0的終邊上，且SKIPIF1<0，求m，SKIPIF1<0，SKIPIF1<0．【答案】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．【詳解】根據(jù)三角函數(shù)定義SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0．題型04三角函數(shù)值符號(hào)的運(yùn)用【典例1】）求SKIPIF1<0（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】C【詳解】由SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.故選：C【典例2】若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0是（

）A．第一象限角 B．第二象限角 C．第三象限角 D．第四象限角【答案】D【詳解】由SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0是第四象限角.故選:D.【典例3】SKIPIF1<0所有可能取值的集合為．【答案】SKIPIF1<0【詳解】解：因?yàn)镾KIPIF1<0，由已知可得角SKIPIF1<0的終邊不在坐標(biāo)軸上，當(dāng)角SKIPIF1<0的終邊在第一象限，則原式SKIPIF1<0，當(dāng)角SKIPIF1<0的終邊在第二象限，則原式SKIPIF1<0，當(dāng)角SKIPIF1<0的終邊在第三象限，則原式SKIPIF1<0，當(dāng)角SKIPIF1<0的終邊在第四象限，則原式SKIPIF1<0，故SKIPIF1<0所有可能取值的集合為SKIPIF1<0，故答案為：SKIPIF1<0【變式1】已知SKIPIF1<0，則點(diǎn)P所在象限為（

）A．第一象限 B．第二象限 C．第三象限 D．第四象限【答案】D【詳解】因?yàn)?（rad）是第一象限角，2（rad）是第二象限角，所以SKIPIF1<0，所以點(diǎn)P所在象限為第四象限.故選：D.【變式2】）已知點(diǎn)SKIPIF1<0是第三象限的點(diǎn)，則SKIPIF1<0的終邊位于（

）A．第一象限 B．第二象限 C．第三象限 D．第四象限【答案】C【詳解】∵點(diǎn)SKIPIF1<0是第三象限的點(diǎn)，∴SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0可得，SKIPIF1<0的終邊位于第二象限或第三象限或x軸的非正半軸；由SKIPIF1<0可得，SKIPIF1<0的終邊位于第一象限或第三象限，綜上所述，SKIPIF1<0的終邊位于第三象限．故選:C【變式3】點(diǎn)SKIPIF1<0位于第象限．【答案】四【詳解】SKIPIF1<0，∴SKIPIF1<0是第三象限角，則SKIPIF1<0.則點(diǎn)SKIPIF1<0位于第四象限．故答案為：四題型05畫三角函數(shù)線【典例1】不等式SKIPIF1<0在區(qū)間SKIPIF1<0上的解集為．【答案】SKIPIF1<0【詳解】如圖所示，由于SKIPIF1<0，所以在SKIPIF1<0上SKIPIF1<0的解集為SKIPIF1<0．故答案為：SKIPIF1<0【典例2】利用三角函數(shù)線，寫出滿足下列條件的角x的集合：(1)SKIPIF1<0且SKIPIF1<0；(2)SKIPIF1<0．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【詳解】（1）分別作出三角函數(shù)線圖象如下所示：

由圖（1）知當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，角SKIPIF1<0滿足的集合SKIPIF1<0.（2）由圖（2）知：當(dāng)SKIPIF1<0時(shí)，角SKIPIF1<0滿足的集合SKIPIF1<0，即SKIPIF1<0；所以SKIPIF1<0的解集為SKIPIF1<0.【變式1】使SKIPIF1<0成立的x的一個(gè)變化區(qū)間是（）A．[SKIPIF1<0，SKIPIF1<0]B．[SKIPIF1<0，SKIPIF1<0]C．[SKIPIF1<0，SKIPIF1<0]D．[0，SKIPIF1<0]【答案】A【詳解】當(dāng)x的終邊落在如圖所示的陰影部分時(shí)，滿足SKIPIF1<0．

故選：A【變式2】已知SKIPIF1<0，則SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0【詳解】如圖，作出單位圓中的三角函數(shù)線，則有SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，SKIPIF1<0，∴SKIPIF1<0,又SKIPIF1<0，∴SKIPIF1<0即SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0取等號(hào)，∴SKIPIF1<0，故答案為：SKIPIF1<0.【變式3】利用單位圓分別寫出符合下列條件的角α的集合：（1）SKIPIF1<0；（2）SKIPIF1<0；（3）SKIPIF1<0.【答案】（1）SKIPIF1<0或SKIPIF1<0；（2）SKIPIF1<0或SKIPIF1<0；（3）SKIPIF1<0．【詳解】解（1）作出如圖所示的圖形，則根據(jù)圖形可得SKIPIF1<0或SKIPIF1<0；（2）作出如圖所示的圖形，則根據(jù)圖形可得SKIPIF1<0或SKIPIF1<0；（3）作出如圖所示的圖形，則根據(jù)圖形可得SKIPIF1<0．題型06三角函數(shù)線的應(yīng)用【典例1】）已知SKIPIF1<0是SKIPIF1<0的一個(gè)內(nèi)角，且SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】解：SKIPIF1<0，SKIPIF1<0令SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，作角SKIPIF1<0的正切線SKIPIF1<0，如圖所示.由圖可得，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)，SKIPIF1<0，即SKIPIF1<0的取值范圍是SKIPIF1<0.故選：SKIPIF1<0.【典例2】（1）設(shè)SKIPIF1<0，試證明：SKIPIF1<0；（2）若SKIPIF1<0，試比較SKIPIF1<0與SKIPIF1<0的大小.【答案】（1）證明見解析；（2）SKIPIF1<0【詳解】（1）如下單位圓中，若SKIPIF1<0，SKIPIF1<0軸，SKIPIF1<0與單位圓切于SKIPIF1<0點(diǎn)，所以SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0扇形SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0.

（2）作單位圓如下圖，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，過SKIPIF1<0作SKIPIF1<0于SKIPIF1<0，連接SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，由SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0.

【變式1】如圖，已知點(diǎn)SKIPIF1<0是單位圓與SKIPIF1<0軸的交點(diǎn)，角SKIPIF1<0的終邊與單位圓的交點(diǎn)為SKIPIF1<0，SKIPIF1<0軸于SKIPIF1<0，過點(diǎn)SKIPIF1<0作單位圓的切線交角SKIPIF1<0的終邊于SKIPIF1<0，則角SKIPIF1<0的正弦線、余弦線、正切線分別是（

）A．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0B．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0D．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0【答案】D【詳解】由題圖，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0，所以角SKIPIF1<0的正弦線、余弦線、正切線分別是SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故選：D

【變式2】利用三角函數(shù)線,確定滿足不等式SKIPIF1<0的SKIPIF1<0取值范圍.【答案】SKIPIF1<0,SKIPIF1<0或SKIPIF1<0,SKIPIF1<0.【詳解】解:作出以坐標(biāo)原點(diǎn)為圓心的單位圓,分別作直線SKIPIF1<0,SKIPIF1<0,直線SKIPIF1<0與單位圓交于點(diǎn)SKIPIF1<0,SKIPIF1<0與x軸交于點(diǎn)M,直線SKIPIF1<0與單位圓交于點(diǎn)SKIPIF1<0,SKIPIF1<0,與x軸交于點(diǎn)SKIPIF1<0,連接SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.在SKIPIF1<0范圍內(nèi),SKIPIF1<0,SKIPIF1<0,則點(diǎn)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0分別在角SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0的終邊上.又SKIPIF1<0,結(jié)合圖形可知,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0或SKIPIF1<0,故SKIPIF1<0的取值范圍為SKIPIF1<0,SKIPIF1<0或SKIPIF1<0,SKIPIF1<0.A夯實(shí)基礎(chǔ)一、單選題1．若角SKIPIF1<0的終邊經(jīng)過點(diǎn)SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)題意，由三角函數(shù)的定義，即可得到結(jié)果.【詳解】因?yàn)榻荢KIPIF1<0的終邊經(jīng)過點(diǎn)SKIPIF1<0，則SKIPIF1<0.故選：D2．角SKIPIF1<0的終邊上一點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】借助三角函數(shù)定義求出SKIPIF1<0，然后利用定義SKIPIF1<0可求答案.【詳解】SKIPIF1<0，解得：SKIPIF1<0，所以SKIPIF1<0.故選：A.3．已知角SKIPIF1<0的頂點(diǎn)與原點(diǎn)SKIPIF1<0重合，始邊與SKIPIF1<0軸的非負(fù)半軸重合，它的終邊過點(diǎn)SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)三角函數(shù)的定義求解即可.【詳解】由題意，SKIPIF1<0.故選：D.4．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的終邊在（

）A．第一、三象限B．第二、四象限C．第一、三象限或在x軸的非負(fù)半軸上D．第二、四象限或在x軸上【答案】D【分析】根據(jù)題意得到SKIPIF1<0是第四象限或x軸正半軸，結(jié)合角的表示方法，進(jìn)求得SKIPIF1<0所在的象限，得到答案.【詳解】因?yàn)镾KIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0是第一、四象限或x軸正半軸，又因?yàn)镾KIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0是二、四象限或x軸，所以SKIPIF1<0是第四象限或x軸正半軸，所以SKIPIF1<0，可得SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0在二象限或x軸負(fù)半軸；令SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0在四象限或x軸正半軸，綜上可得，SKIPIF1<0的終邊在第二、四象限或在x軸上.故選：D.5．SKIPIF1<0的符號(hào)為（

）A．正 B．0 C．負(fù) D．無法確定【答案】C【分析】先判斷所給角位于的象限，進(jìn)而判斷正負(fù)即可．【詳解】由1弧度為第一象限角，2弧度為第二象限角，3弧度為第二象限角，4弧度為第三象限角，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．故選：C．6．在平面直角坐標(biāo)系中，SKIPIF1<0是圓SKIPIF1<0上的四段?。ㄈ鐖D），點(diǎn)P在其中一段上，角SKIPIF1<0以SKIPIF1<0為始邊，OP為終邊，若SKIPIF1<0，則P所在的圓弧是A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】C【詳解】分析：逐個(gè)分析A、B、C、D四個(gè)選項(xiàng)，利用三角函數(shù)的三角函數(shù)線可得正確結(jié)論.詳解：由下圖可得：有向線段SKIPIF1<0為余弦線，有向線段SKIPIF1<0為正弦線，有向線段SKIPIF1<0為正切線.A選項(xiàng)：當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0上時(shí)，SKIPIF1<0，SKIPIF1<0，故A選項(xiàng)錯(cuò)誤；B選項(xiàng)：當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0上時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故B選項(xiàng)錯(cuò)誤；C選項(xiàng)：當(dāng)點(diǎn)SKIPIF1<0在SKIPIF1<0上時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故C選項(xiàng)正確；D選項(xiàng)：點(diǎn)SKIPIF1<0在SKIPIF1<0上且SKIPIF1<0在第三象限，SKIPIF1<0，故D選項(xiàng)錯(cuò)誤.綜上，故選C.點(diǎn)睛：此題考查三角函數(shù)的定義，解題的關(guān)鍵是能夠利用數(shù)形結(jié)合思想，作出圖形，找到SKIPIF1<0所對(duì)應(yīng)的三角函數(shù)線進(jìn)行比較.7．質(zhì)點(diǎn)P和Q在以坐標(biāo)原點(diǎn)O為圓心，半徑為1的圓周上順時(shí)針作勻速圓周運(yùn)動(dòng)，同時(shí)出發(fā).P的角速度為3rad/s，起點(diǎn)為射線SKIPIF1<0與圓的交點(diǎn)；Q的角速度為5rad/s，起點(diǎn)為圓與x軸正半軸交點(diǎn)，則當(dāng)質(zhì)點(diǎn)Q與P第二次相遇時(shí)，Q的坐標(biāo)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】當(dāng)質(zhì)點(diǎn)Q與P第二次相遇時(shí)，質(zhì)點(diǎn)Q比P多旋轉(zhuǎn)SKIPIF1<0，解方程確定質(zhì)點(diǎn)Q所在終邊，求坐標(biāo).【詳解】設(shè)當(dāng)質(zhì)點(diǎn)Q與P第二次相遇時(shí)，用了時(shí)間SKIPIF1<0，依題意有SKIPIF1<0，解得SKIPIF1<0，此時(shí)質(zhì)點(diǎn)Q轉(zhuǎn)過角度為SKIPIF1<0，因?yàn)槭琼槙r(shí)針作勻速圓周運(yùn)動(dòng)，質(zhì)點(diǎn)Q轉(zhuǎn)在SKIPIF1<0角的終邊上，圓的半徑為1，Q的坐標(biāo)為SKIPIF1<0.故選：C8．已知函數(shù)SKIPIF1<0（SKIPIF1<0，且SKIPIF1<0）的圖像恒過點(diǎn)P，若點(diǎn)SKIPIF1<0是角SKIPIF1<0終邊上的一點(diǎn)，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)對(duì)數(shù)型函數(shù)過定點(diǎn)求得SKIPIF1<0，利用三角函數(shù)的定義求解即可.【詳解】解：∵SKIPIF1<0，∴函數(shù)SKIPIF1<0（SKIPIF1<0，且SKIPIF1<0）的圖像恒過點(diǎn)SKIPIF1<0，∴由三角函數(shù)定義得SKIPIF1<0故選：D二、多選題9．已知SKIPIF1<0，則SKIPIF1<0可能是（

）A．第一象限角 B．第二象限角 C．第三象限角 D．第四象限角【答案】BC【分析】根據(jù)余弦值的正負(fù)確定角SKIPIF1<0所在象限即可.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0可能是第二、三象限角.故選：BC.10．若SKIPIF1<0是第四象限角，則點(diǎn)SKIPIF1<0在第（

）象限.A．第一象限 B．第二象限 C．第三象限 D．第四象限【答案】CD【分析】根據(jù)給定條件確定SKIPIF1<0角的范圍，再確定SKIPIF1<0與SKIPIF1<0符號(hào)，即可判斷作答.【詳解】因SKIPIF1<0是第四象限角，即SKIPIF1<0，則SKIPIF1<0，當(dāng)k是奇數(shù)時(shí)，SKIPIF1<0是第二象限角，SKIPIF1<0，點(diǎn)SKIPIF1<0在第三象限，當(dāng)k是偶數(shù)時(shí)，SKIPIF1<0是第四象限角，SKIPIF1<0，點(diǎn)SKIPIF1<0在第四象限，所以點(diǎn)SKIPIF1<0在第三或四象限.故選：CD三、填空題11．已知角SKIPIF1<0的終邊上一點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0，則角SKIPIF1<0的最小正值為【答案】SKIPIF1<0【分析】根據(jù)坐標(biāo)值符號(hào)確定SKIPIF1<0所在象限，由三角函數(shù)定義求SKIPIF1<0，最后確定其對(duì)應(yīng)的最小正角.【詳解】因?yàn)镾KIPIF1<0，所以角α的終邊在第四象限，根據(jù)三角函數(shù)的定義，知SKIPIF1<0，故角α的最小正值為SKIPIF1<0．故答案為：SKIPIF1<012．如果SKIPIF1<0是第二象限角，則SKIPIF1<0的符號(hào)為（選填“正”或“負(fù)”）．【答案】負(fù)【分析】由SKIPIF1<0是第二象限角，可得SKIPIF1<0，SKIPIF1<0，進(jìn)而可得SKIPIF1<0，即可得答案.【詳解】解：因?yàn)镾KIPIF1<0是第二象限角，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故答案為：負(fù).四、解答題13．已知SKIPIF1<0，試判斷SKIPIF1<0的符號(hào)．【答案】負(fù)【分析】分SKIPIF1<0、SKIPIF1<0、SKIPIF1<0與SKIPIF1<0討論即可求解.【詳解】由題意可得SKIPIF1<0，若SKIPIF1<0,則SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0，符合題意；若SKIPIF1<0，則SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0，符合題意.故SKIPIF1<0或SKIPIF1<0.所以SKIPIF1<0.所以SKIPIF1<0的符號(hào)為負(fù).14．）已知SKIPIF1<0,,試判斷角SKIPIF1<0所在的象限.【答案】第三象限【解析】化簡(jiǎn)得到SKIPIF1<0故SKIPIF1<0且SKIPIF1<0，討論得到答案.【詳解】SKIPIF1<0SKIPIF1<0且SKIPIF1<0.由SKIPIF1<0，得SKIPIF1<0為第二象限或第三象限或終邊在x軸負(fù)半軸上的角由SKIPIF1<0，得SKIPIF1<0為第三象限或第四象限或終邊在y軸負(fù)半軸上的角綜上所述：SKIPIF1<0為第三象限的角.【點(diǎn)睛】本題考查了角的象限問題，意在考查學(xué)生的計(jì)算能力.15．已知SKIPIF1<0，SKIPIF1<0可能為第幾象限的角？分別求出SKIPIF1<0，SKIPIF1<0的值.【答案】答案見解析【分析】根據(jù)余弦值的符號(hào)判斷角所在象限，討論SKIPIF1<0所在象限，應(yīng)用同角三角函數(shù)的關(guān)系求正弦、正切值即可.【詳解】由SKIPIF1<0，則SKIPIF1<0可能是第二象限或者第三象限角，若SKIPIF1<0是第二象限角，則SKIPIF1<0，故SKIPIF1<0；若SKIPIF1<0是第三象限角，則SKIPIF1<0，故SKIPIF1<0.綜上，SKIPIF1<0可能是第二象限或者第三象限角，若是第二象限角，SKIPIF1<0，SKIPIF1<0；若是第三象限角，SKIPIF1<0，SKIPIF1<0.16．利用三角函數(shù)線，說明當(dāng)SKIPIF1<0時(shí)，求證：SKIPIF1<0．【答案】證明見詳解【分析】由正弦線及SKIPIF1<0面積即可求證.【詳解】證明：如圖：

在直角坐標(biāo)系中作出單位圓，SKIPIF1<0的終邊與單位圓交于P，SKIPIF1<0的正弦線為有向線段MP，則SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，又SKIPIF1<0.所以SKIPIF1<0，即SKIPIF1<0．B能力提升1．已知角SKIPIF1<0，則SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】結(jié)合三角函數(shù)定義與單位圓數(shù)形結(jié)合求解即可；【詳解】如下圖示，在單位圓中SKIPIF1<0，SKIPIF1<0軸，SKIPIF1<0軸，且SKIPIF1<0，

所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的面積SKIPIF1<0，扇形SKIPIF1<0的面積SKIPIF1<0，SKIPIF1<0的面積SKIPIF1<0，由圖知：SKIPIF1<0，故SKIPIF1<0.故選：A.2．在直角坐標(biāo)系SKIPIF1<0中，角SKIPIF1<0的始邊為SKIPIF1<0軸的正半軸，頂點(diǎn)為坐標(biāo)原點(diǎn)SKIPIF1<0，已知角SKIPIF1<0的終邊SKIPIF1<0與單位圓交于點(diǎn)SKIPIF1<0，將SKIPIF1<0繞原點(diǎn)逆時(shí)針旋轉(zhuǎn)SKIPIF1<0與單位圓交于點(diǎn)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．0.6 B．0.8 C．-0.6 D．-0.8【答案】B【詳解】解：已知角SKIPIF1<0的終邊SKIPIF1<0與單位圓交于點(diǎn)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0，所以SKIPIF1<0在第四象限，角SKIPIF1<0為第四象限角，SKIPIF1<0繞原點(diǎn)逆時(shí)針旋轉(zhuǎn)SKIPIF1<0與單位圓交于點(diǎn)SKIPIF1<0，可知點(diǎn)SKIPIF1<0在第一象限，則SKIPIF1<0，所以SKIPIF1<0，即：SKIPIF1<0，解得：SKIPIF1<0.故選：B.【點(diǎn)睛】本題考查單位圓中任意角的三角函數(shù)的定義的應(yīng)用以及運(yùn)用誘導(dǎo)公式化簡(jiǎn)，考查計(jì)算能力.3．（多選）已知函數(shù)SKIPIF1<0且SKIPIF1<0的圖象經(jīng)過定點(diǎn)SKIPIF1<0，且點(diǎn)SKIPIF1<0在角SKIPIF1<0的終邊上，則SKIPIF1<0的值可能是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】BD【詳解】根據(jù)題意可知函數(shù)SKIPIF1<0的圖象經(jīng)過定點(diǎn)SKIPIF1<0,則SKIPIF1<0或SKIPIF1<0,當(dāng)點(diǎn)SKIPIF1<0在角SKIPIF1<0的終邊上，則SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，當(dāng)點(diǎn)SKIPIF1<0在角SKIPIF1<0的終邊上，則SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，故選：SKIPIF1<0.4．已知角SKIPIF1<0（SKIPIF1<0）的終邊過點(diǎn)SKIPIF1<0，則SKIPIF1<0.【答案】SKIPIF1<0【詳解】SKIPIF1<0，SKIPIF1<0，因此SKIPIF1<0在第四象限，又SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0．故答案為：SKIPIF1<0．

5.2.2同角三角函數(shù)的基本關(guān)系課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①掌握同角三角函數(shù)的基本關(guān)系式。②能正確運(yùn)用同角三角函數(shù)的基本關(guān)系式進(jìn)行求值、化簡(jiǎn)和證明。會(huì)用同角三角函數(shù)的基本關(guān)系進(jìn)行求值、化簡(jiǎn)、證明知識(shí)點(diǎn)01：同角三角函數(shù)的基本關(guān)系1、平方關(guān)系：SKIPIF1<02、商數(shù)關(guān)系:SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）【即學(xué)即練1】已知SKIPIF1<0，SKIPIF1<0，求SKIPIF1<0的值．【答案】SKIPIF1<0【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.知識(shí)點(diǎn)02：關(guān)系式的常用等價(jià)變形1、SKIPIF1<0SKIPIF1<0SKIPIF1<02、SKIPIF1<0SKIPIF1<0SKIPIF1<0【即學(xué)即練2】已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0.【答案】SKIPIF1<0/0.75【詳解】由同角三角函數(shù)的平方關(guān)系及已知條件可知：SKIPIF1<0，當(dāng)SKIPIF1<0，此時(shí)SKIPIF1<0，不合題意；當(dāng)SKIPIF1<0，符合題意；所以SKIPIF1<0.故答案為：SKIPIF1<0題型01同角三角函數(shù)的基本關(guān)系【典例1】已知SKIPIF1<0是第二象限角，且SKIPIF