新高考数学一轮复习精讲精练3.3 诱导公式及恒等变化（基础版）（解析版）

3.3诱导公式及恒等变化（精讲）（基础版）思维导图思维导图考点呈现考点呈现例题剖析例题剖析考点一诱导公式基本运用【例1-1】（2022·宁夏）已知SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0SKIPIF1<0，故选：D【例1-2】（2022·广西南宁）化简：SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0,故选：D总结：总结：“负化正，大化小，化到锐角就好了”温馨提示【一隅三反】1．（2022·北京）计算SKIPIF1<0的结果是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0.故选：D.2．（2022·宁夏中卫·一模）已知SKIPIF1<0，则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，则SKIPIF1<0故SKIPIF1<0故选：D3．（2022·江西省临川）化简SKIPIF1<0（）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0故选：B考点二两角和差与二倍角公式基本运用【例2-1】（2022·四川省岳池中学）SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0故选：A【例2-2】（2022·四川省泸县第一中学）SKIPIF1<0的值等于(

)A．0 B．1 C．－1 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0.故选：A.【例2-3】（2022·贵州·模拟预测（理））SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0.故选：A【例2-4】（2022·重庆八中）（多选）下列选项中，值为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；故选：AC【一隅三反】1．（2022·江苏省响水中学）SKIPIF1<0=（

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

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ACD【解析】对于A：SKIPIF1<0，故A正确；对于B：SKIPIF1<0SKIPIF1<0SKIPIF1<0，故B错误；对于C：SKIPIF1<0，故C正确；对于D：SKIPIF1<0，故D正确；故选：ACD3．（2022·河南焦作·二模）已知SKIPIF1<0，则x的值可以是（

）A．0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因为SKIPIF1<0，所以SKIPIF1<0，对于A，若SKIPIF1<0，则SKIPIF1<0，所以A不正确，对于B，若SKIPIF1<0，则SKIPIF1<0，所以B不正确，对于C，若SKIPIF1<0，则SKIPIF1<0，所以C正确，对于D，若SKIPIF1<0，则SKIPIF1<0，所以D不正确，故选：C4．（2022·甘肃）SKIPIF1<0_______.【答案】SKIPIF1<0【解析】由题，SKIPIF1<0，SKIPIF1<0，故原式可化为SKIPIF1<0，故答案为：SKIPIF1<0考点三公式的综合基础运用【例3-1】（2022·北京·一模）已知SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因为SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0；故选：C【例3-2】（2022·陕西·二模）已知SKIPIF1<0为锐角，若SKIPIF1<0，则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为锐角，则SKIPIF1<0，SKIPIF1<0.故选：A.【例3-3】（2022·河南）已知SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0______．【答案】SKIPIF1<0【解析】由题意，SKIPIF1<0,即SKIPIF1<0，所以SKIPIF1<0．故答案为：SKIPIF1<0．1.1.弦切互化，异名化同名，异角化同角，降幂或升幂．2.在三角函数式的化简中“次降角升”和“次升角降”是基本的规律，根号中含有三角函数式时，一般需要升次．方法总结【一隅三反】1．（2022·宁夏六盘山高级中学二模）已知SKIPIF1<0，则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，等式两边平方得SKIPIF1<0，解得SKIPIF1<0.故选：B.2．（2022·安徽蚌埠·三模）已知SKIPIF1<0，则SKIPIF1<0的值为（

）A．3 B．-3 C．SKIPIF1<0 D．-1【答案】A【解析】原式SKIPIF1<0.故选：A3．（2022·山东·模拟预测）若SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因为SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0.故选：D4．（2022·陕西·二模）角SKIPIF1<0顶点与原点重合，始边与x轴非负半轴重合，终边在直线SKIPIF1<0上，则SKIPIF1<0___________.【答案】SKIPIF1<0【解析】因为角SKIPIF1<0终边在直线SKIPIF1<0上，所以SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.故答案为：SKIPIF1<05（2022·辽宁·抚顺县高级中学校）已知SKIPIF1<0是第三象限角，且SKIPIF1<0.(1)化简SKIPIF1<0；(2)若SKIPIF1<0，求SKIPIF1<0的值.【答案】(1)SKIPIF1<0(2)SKIPIF1<0【解析】（1）SKIPIF1<0为第三象限角，则SKIPIF1<0.(2)SKIPIF1<0，所以，SKIPIF1<0，由已知可得SKIPIF1<0，解得SKIPIF1<0，则SKIPIF1<0.考点四角的拼凑【例4-1】（2022·广东·一模）已知SKIPIF1<0为锐角，且SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由SKIPIF1<0为锐角得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0.故选：C.【例4-2】（2022·四川·眉山市彭山区第一中学）已知SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由题SKIPIF1<0，则SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0．故选：B【例4-3】．（2022·辽宁抚顺·一模）已知SKIPIF1<0，则SKIPIF1<0的值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0.故选：B【例4-4】（2022·广西·桂林十八中）若SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，则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，故答案为：SKIPIF1<0【一隅三反】1．（2022·江西九江·二模）已知SKIPIF1<0，则SKIPIF1<0___________.【答案】SKIPIF1<0【解析】SKIPIF1<0.故答案为：SKIPIF1<02．（2022·北京市房山区房山中学）已知SKIPIF1<0，则SKIPIF1<0______.【答案】SKIPIF1<0【解析】SKIPIF1<0.故答案为：SKIPIF1<03．（2022·河南·模拟预测）已知SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】因为SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0.故选：B.4．（2022·全国·模拟预测）若SKIPIF1<0，则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.故选：D.5．（2022·山西晋中·二模）已知SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0等于（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．2或6【答案】C【解析】因为SKIPIF1<0，则SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0.故选：C.3.3诱导公式及恒等变化（精练）（基础版）题组一题组一诱导公式基础运用1．（2022·北京·高一阶段练习）化简SKIPIF1<0为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0SKIPIF1<0.故选：A2．（2022·山西·怀仁市第一中学校二模）若SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因为SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0,故选:D3．（2022·云南保山）已知角SKIPIF1<0的终边过点SKIPIF1<0，且SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因为SKIPIF1<0，由已知角SKIPIF1<0的终边过点SKIPIF1<0可得SKIPIF1<0∵SKIPIF1<0，解得SKIPIF1<0所以SKIPIF1<0故选：C．4．（2022·北京·人大附中）已知角SKIPIF1<0的终边经过点SKIPIF1<0，则SKIPIF1<0的值等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】依题意SKIPIF1<0，所以SKIPIF1<0.故选：C5．（2022·山东·烟台二中）已知角SKIPIF1<0的顶点在原点，始边与x轴的非负半轴重合，终边经过点SKIPIF1<0，则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故选：B6．（2022·山东·淄博中学）（多选）下列式子中正确的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【解析】对于A选项：SKIPIF1<0，故A错误；对于B选项：SKIPIF1<0SKIPIF1<0，故B正确；对于C选项：因为SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，故C正确；对于D选项：SKIPIF1<0，即SKIPIF1<0，故D正确.故选：BCD.题组二题组二两角和差与二倍角的基础运用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；故选：D.2．（2022·四川雅安·二模）已知SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故选：C.3．（2022·全国·模拟预测（理））SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】C【解析】由题得SKIPIF1<0.故选：C4．（2022·陕西咸阳·二模（文））已知SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因为SKIPIF1<0所以SKIPIF1<0SKIPIF1<0故选：A5．（2022·全国·高三专题练习）SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0SKIPIF1<0，故选：D6．（2022·全国·哈师大附中模拟预测）若SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因为SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，当SKIPIF1<0时，SKIPIF1<0，当SKIPIF1<0时，SKIPIF1<0，综上所述，SKIPIF1<0.故选：C.7．（2022·云南·昆明一中高三阶段练习（文））已知SKIPIF1<0，则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，故选：A.8．（2022·重庆一中高三阶段练习）式子SKIPIF1<0的值为___________【答案】SKIPIF1<0【解析】原式SKIPIF1<0SKIPIF1<0SKIPIF1<0，故答案：SKIPIF1<0题组三题组三公式的基础综合运用1．（2022·宁夏吴忠）已知SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．2【答案】D【解析】由诱导公式可得SKIPIF1<0，所以，SKIPIF1<0.因此，SKIPIF1<0.故选：D.2．（2022·山东潍坊·一模）已知SKIPIF1<0，且SKIPIF1<0，则（

）.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.故选：A.3．（2022·河南省杞县高中）已知SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0的值为（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因为SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0，SKIPIF1<0．故选：D4．（2022·广西南宁·一模（理））已知SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因为SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0．故选：C．5．（2022·河北邯郸·一模）已知SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】因为SKIPIF1<0，则SKIPIF1<0SKIPIF1<0.故选：C.6．（2022·河北·模拟预测）（多选）已知SKIPIF1<0，则下列结论正确的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】ACD【解析】对于A选项，SKIPIF1<0，故A选项正确；对于B选项，SKIPIF1<0，故B选项错误；对于C选项，SKIPIF1<0，故C选项正确；对于D选项，SKIPIF1<0，故D选项正确.故选：ACD7．（2022·全国·模拟预测）已知SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因为SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，=SKIPIF1<0，SKIPIF1<0．故选：D．8．（2022·江西·高三阶段练习（理））若SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】依题意，SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，故选：A9．（2022·北京·人大附中）若SKIPIF1<0，则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,则SKIPIF1<0SKIPIF1<0.故选：C.10（2022·陕西·西安中学）已知SKIPIF1<0，则SKIPIF1<0_______．【答案】SKIPIF1<0【解析】SKIPIF1<0，则SKIPIF1<0，所以SKIPIF1<0，因为SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，则SKIPIF1<0.故答案为：SKIPIF1<0题组四题组四角的拼凑1．（2022·福建漳州·二模）已知SKIPIF1<0，则SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SK