新高考數(shù)學(xué)一輪復(fù)習(xí)精講精練3.3 誘導(dǎo)公式及恒等變化（基礎(chǔ)版）（解析版）

3.3誘導(dǎo)公式及恒等變化（精講）（基礎(chǔ)版）思維導(dǎo)圖思維導(dǎo)圖考點呈現(xiàn)考點呈現(xiàn)例題剖析例題剖析考點一誘導(dǎo)公式基本運用【例1-1】（2022·寧夏）已知SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0SKIPIF1<0，故選：D【例1-2】（2022·廣西南寧）化簡：SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0,故選：D總結(jié)：總結(jié)：“負化正，大化小，化到銳角就好了”溫馨提示【一隅三反】1．（2022·北京）計算SKIPIF1<0的結(jié)果是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0.故選：D.2．（2022·寧夏中衛(wèi)·一模）已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0故SKIPIF1<0故選：D3．（2022·江西省臨川）化簡SKIPIF1<0（）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0故選：B考點二兩角和差與二倍角公式基本運用【例2-1】（2022·四川省岳池中學(xué)）SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0故選：A【例2-2】（2022·四川省瀘縣第一中學(xué)）SKIPIF1<0的值等于(

)A．0 B．1 C．－1 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0.故選：A.【例2-3】（2022·貴州·模擬預(yù)測（理））SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0.故選：A【例2-4】（2022·重慶八中）（多選）下列選項中，值為SKIPIF1<0的是（

）A．SKIPIF1<0 B．SKIPIF1<0SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】AC【解析】A.SKIPIF1<0；B.SKIPIF1<0；C.SKIPIF1<0；D.SKIPIF1<0；故選：AC【一隅三反】1．（2022·江蘇省響水中學(xué)）SKIPIF1<0=（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0.故選：A2．（2022·廣東·佛山一中）（多選）下列等式成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ACD【解析】對于A：SKIPIF1<0，故A正確；對于B：SKIPIF1<0SKIPIF1<0SKIPIF1<0，故B錯誤；對于C：SKIPIF1<0，故C正確；對于D：SKIPIF1<0，故D正確；故選：ACD3．（2022·河南焦作·二模）已知SKIPIF1<0，則x的值可以是（

）A．0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因為SKIPIF1<0，所以SKIPIF1<0，對于A，若SKIPIF1<0，則SKIPIF1<0，所以A不正確，對于B，若SKIPIF1<0，則SKIPIF1<0，所以B不正確，對于C，若SKIPIF1<0，則SKIPIF1<0，所以C正確，對于D，若SKIPIF1<0，則SKIPIF1<0，所以D不正確，故選：C4．（2022·甘肅）SKIPIF1<0_______.【答案】SKIPIF1<0【解析】由題，SKIPIF1<0，SKIPIF1<0，故原式可化為SKIPIF1<0，故答案為：SKIPIF1<0考點三公式的綜合基礎(chǔ)運用【例3-1】（2022·北京·一模）已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因為SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0；故選：C【例3-2】（2022·陜西·二模）已知SKIPIF1<0為銳角，若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0，故SKIPIF1<0，又SKIPIF1<0為銳角，則SKIPIF1<0，SKIPIF1<0.故選：A.【例3-3】（2022·河南）已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0______．【答案】SKIPIF1<0【解析】由題意，SKIPIF1<0,即SKIPIF1<0，所以SKIPIF1<0．故答案為：SKIPIF1<0．1.1.弦切互化，異名化同名，異角化同角，降冪或升冪．2.在三角函數(shù)式的化簡中“次降角升”和“次升角降”是基本的規(guī)律，根號中含有三角函數(shù)式時，一般需要升次．方法總結(jié)【一隅三反】1．（2022·寧夏六盤山高級中學(xué)二模）已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由已知可得SKIPIF1<0，等式兩邊平方得SKIPIF1<0，解得SKIPIF1<0.故選：B.2．（2022·安徽蚌埠·三模）已知SKIPIF1<0，則SKIPIF1<0的值為（

）A．3 B．-3 C．SKIPIF1<0 D．-1【答案】A【解析】原式SKIPIF1<0.故選：A3．（2022·山東·模擬預(yù)測）若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0.故選：D4．（2022·陜西·二模）角SKIPIF1<0頂點與原點重合，始邊與x軸非負半軸重合，終邊在直線SKIPIF1<0上，則SKIPIF1<0___________.【答案】SKIPIF1<0【解析】因為角SKIPIF1<0終邊在直線SKIPIF1<0上，所以SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.故答案為：SKIPIF1<05（2022·遼寧·撫順縣高級中學(xué)校）已知SKIPIF1<0是第三象限角，且SKIPIF1<0.(1)化簡SKIPIF1<0；(2)若SKIPIF1<0，求SKIPIF1<0的值.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】（1）SKIPIF1<0為第三象限角，則SKIPIF1<0.(2)SKIPIF1<0，所以，SKIPIF1<0，由已知可得SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0.考點四角的拼湊【例4-1】（2022·廣東·一模）已知SKIPIF1<0為銳角，且SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由SKIPIF1<0為銳角得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0.故選：C.【例4-2】（2022·四川·眉山市彭山區(qū)第一中學(xué)）已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由題SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0．故選：B【例4-3】．（2022·遼寧撫順·一模）已知SKIPIF1<0，則SKIPIF1<0的值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0.故選：B【例4-4】（2022·廣西·桂林十八中）若SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0________.【答案】SKIPIF1<0【解析】由SKIPIF1<0，且SKIPIF1<0，可得SKIPIF1<0，而SKIPIF1<0，由SKIPIF1<0可知：SKIPIF1<0，故SKIPIF1<0SKIPIF1<0，故答案為：SKIPIF1<0【一隅三反】1．（2022·江西九江·二模）已知SKIPIF1<0，則SKIPIF1<0___________.【答案】SKIPIF1<0【解析】SKIPIF1<0.故答案為：SKIPIF1<02．（2022·北京市房山區(qū)房山中學(xué)）已知SKIPIF1<0，則SKIPIF1<0______.【答案】SKIPIF1<0【解析】SKIPIF1<0.故答案為：SKIPIF1<03．（2022·河南·模擬預(yù)測）已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0.故選：B.4．（2022·全國·模擬預(yù)測）若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0，可得SKIPIF1<0SKIPIF1<0.故選：D.5．（2022·山西晉中·二模）已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0等于（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．2或6【答案】C【解析】因為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0.故選：C.3.3誘導(dǎo)公式及恒等變化（精練）（基礎(chǔ)版）題組一題組一誘導(dǎo)公式基礎(chǔ)運用1．（2022·北京·高一階段練習(xí)）化簡SKIPIF1<0為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0SKIPIF1<0.故選：A2．（2022·山西·懷仁市第一中學(xué)校二模）若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0,故選:D3．（2022·云南保山）已知角SKIPIF1<0的終邊過點SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因為SKIPIF1<0，由已知角SKIPIF1<0的終邊過點SKIPIF1<0可得SKIPIF1<0∵SKIPIF1<0，解得SKIPIF1<0所以SKIPIF1<0故選：C．4．（2022·北京·人大附中）已知角SKIPIF1<0的終邊經(jīng)過點SKIPIF1<0，則SKIPIF1<0的值等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】依題意SKIPIF1<0，所以SKIPIF1<0.故選：C5．（2022·山東·煙臺二中）已知角SKIPIF1<0的頂點在原點，始邊與x軸的非負半軸重合，終邊經(jīng)過點SKIPIF1<0，則SKIPIF1<0等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0,又SKIPIF1<0,故SKIPIF1<0.故SKIPIF1<0故選：B6．（2022·山東·淄博中學(xué)）（多選）下列式子中正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【解析】對于A選項：SKIPIF1<0，故A錯誤；對于B選項：SKIPIF1<0SKIPIF1<0，故B正確；對于C選項：因為SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，故C正確；對于D選項：SKIPIF1<0，即SKIPIF1<0，故D正確.故選：BCD.題組二題組二兩角和差與二倍角的基礎(chǔ)運用1．（2022·山西太原·一模）SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0；故選：D.2．（2022·四川雅安·二模）已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故選：C.3．（2022·全國·模擬預(yù)測（理））SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】C【解析】由題得SKIPIF1<0.故選：C4．（2022·陜西咸陽·二模（文））已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因為SKIPIF1<0所以SKIPIF1<0SKIPIF1<0故選：A5．（2022·全國·高三專題練習(xí)）SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0SKIPIF1<0，故選：D6．（2022·全國·哈師大附中模擬預(yù)測）若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，綜上所述，SKIPIF1<0.故選：C.7．（2022·云南·昆明一中高三階段練習(xí)（文））已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．1【答案】A【解析】由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故選：A.8．（2022·重慶一中高三階段練習(xí)）式子SKIPIF1<0的值為___________【答案】SKIPIF1<0【解析】原式SKIPIF1<0SKIPIF1<0SKIPIF1<0，故答案：SKIPIF1<0題組三題組三公式的基礎(chǔ)綜合運用1．（2022·寧夏吳忠）已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】D【解析】由誘導(dǎo)公式可得SKIPIF1<0，所以，SKIPIF1<0.因此，SKIPIF1<0.故選：D.2．（2022·山東濰坊·一模）已知SKIPIF1<0，且SKIPIF1<0，則（

）.A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0或SKIPIF1<0（舍去），∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故選：A.3．（2022·河南省杞縣高中）已知SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因為SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0．故選：D4．（2022·廣西南寧·一模（理））已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0．故選：C．5．（2022·河北邯鄲·一模）已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因為SKIPIF1<0，則SKIPIF1<0SKIPIF1<0.故選：C.6．（2022·河北·模擬預(yù)測）（多選）已知SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】ACD【解析】對于A選項，SKIPIF1<0，故A選項正確；對于B選項，SKIPIF1<0，故B選項錯誤；對于C選項，SKIPIF1<0，故C選項正確；對于D選項，SKIPIF1<0，故D選項正確.故選：ACD7．（2022·全國·模擬預(yù)測）已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，=SKIPIF1<0，SKIPIF1<0．故選：D．8．（2022·江西·高三階段練習(xí)（理））若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】依題意，SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，故選：A9．（2022·北京·人大附中）若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0，解得SKIPIF1<0,則SKIPIF1<0SKIPIF1<0.故選：C.10（2022·陜西·西安中學(xué)）已知SKIPIF1<0，則SKIPIF1<0_______．【答案】SKIPIF1<0【解析】SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<0題組四題組四角的拼湊1．（2022·福建漳州·二模）已知SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SK