诱导公式(二)-高一上学期数学课时作业人教版A版（含解析）

PAGE诱导公式(二)(时间:45分钟分值:90分)【基础全面练】1.(5分)(2026·长沙高一检测)已知角α终边上一点P(3,4),则sin(π2+α)=()A.35 B.45 C.-35 2.(5分)(2025·哈尔滨高一检测)sin(π2+α)+cos(π+α)=()A.1 B.-1 C.0 D.-tanα3.(5分)化简:cos(α-3π)A.1tan2α B.-1tan2α C.tan4.(5分)(2026·衡水高一检测)已知tanα=-2,则cos(π2+αA.-23 B.23 C.-2 5.(5分)已知α是第四象限角,且3sin2α=8cosα,则cos(α+2025π2)=(A.-223 B.-13 C.2236.(5分)(多选)已知cosα=-55,α∈(π2,π),则下列说法正确的是(A.sin(α-π2)=-55 B.tan(πC.sin(π2+α)=-55 D.cos(3π27.(5分)已知sin(π2+α)=35,α∈(0,π2),则sin(π+α8.(5分)已知角α的终边经过点(-5,2),则sin(49.(5分)sin21°+sin22°+sin23°+…+sin289°=.

10.(10分)已知sin(53°-α)=15,且-270°<α<-90°(1)求sin(127°+α)的值;(2)求sin(37°+α)的值.【综合应用练】11.(5分)已知cos(α+π3)=-513,则sin(7π6-α)-2cos(2π3-α)=A.-513 B.513 C.-1513 12.(5分)(多选)下列结论正确的有()A.sin(π6+α)=cos(π3B.cos(5π6+θ)+sin(2π3C.sin2(15°-α)+cos2(75°+α)=1D.sin2(15°-α)+sin2(75°+α)=113.(5分)(2025·徐州高一检测)已知cos(75°+α)=513,α是第三象限角,则sin(195°-α)+cos(105°-α.

14.(10分)已知tan(3π+α)=2,求sin(α15.(10分)已知f(α)=sin((1)化简f(α);(2)若α=-16π3,求f(α(3)若α为第三象限角,且cos(α-π2)=-35,求f(α

答案解析【基础全面练】1.(5分)(2026·长沙高一检测)已知角α终边上一点P(3,4),则sin(π2+α)=()A.35 B.45 C.-35 【解析】选A.由诱导公式和三角函数的定义可知sin(π2+α)=cosα=32.(5分)(2025·哈尔滨高一检测)sin(π2+α)+cos(π+α)=()A.1 B.-1 C.0 D.-tanα【解析】选C.sin(π2+α)+cos(π+α)=cosα-cosα=03.(5分)化简:cos(α-3π)A.1tan2α B.-1tan2α C.tan【解析】选C.cos(α-3π4.(5分)(2026·衡水高一检测)已知tanα=-2,则cos(π2+αA.-23 B.23 C.-2 【解析】选C.因为tanα=-2,则cos(π2+α)sin(π-5.(5分)已知α是第四象限角,且3sin2α=8cosα,则cos(α+2025π2)=(A.-223 B.-13 C.223【解析】选C.因为3sin2α=8cosα,所以sin2α+(3sin2α8)2=1,整理可得9sin4α+64sin2α-64=0,解得sin2α=89或sin2α=-8(舍去).又因为α是第四象限角,所以sin所以cos(α+2025π2)=cos(α+1012π+π2)=cos(α+π2)=-6.(5分)(多选)已知cosα=-55,α∈(π2,π),则下列说法正确的是(A.sin(α-π2)=-55 B.tan(πC.sin(π2+α)=-55 D.cos(3π2【解析】选BC.由题意,sinα=1-cos2α=2对于A,sin(α-π2)=-cosα=5对于B,tan(π-α)=tan(-α)=-tanα=2,故B正确;对于C,sin(π2+α)=cosα=-5对于D,cos(3π2-α)=-sinα=-257.(5分)已知sin(π2+α)=35,α∈(0,π2),则sin(π+α)=-【解析】由诱导公式可得sin(π2+α)=cosα=35,且α∈(0,所以sinα=1-cos2α=1-(358.(5分)已知角α的终边经过点(-5,2),则sin(4π+α【解析】由角α的终边经过点(-5,2),则tanα=-2sin(4π+α)·cos(9.(5分)sin21°+sin22°+sin23°+…+sin289°=44.5.

【解析】设S=sin21°+sin22°+sin23°+…+sin289°,因为sin21°=cos289°,sin22°=cos288°,sin23°=cos287°,…,sin289°=cos21°,所以S=cos289°+cos288°+cos287°+…+cos21°,两式相加得:2S=1×89,所以S=44.5.10.(10分)已知sin(53°-α)=15,且-270°<α<-90°(1)求sin(127°+α)的值;【解析】(1)因为sin(53°-α)=15,所以sin(127°+α)=sin[180°-(53°-α)]=sin(53°-α)=1(2)求sin(37°+α)的值.【解析】(2)因为sin(53°-α)=15,且-270°<α<-所以143°<53°-α<323°,又sin(53°-α)=15>0,所以143°<53°-α所以cos(53°-α)=-1-sin2(53°-α所以sin(37°+α)=cos(53°-α)=-26【综合应用练】11.(5分)已知cos(α+π3)=-513,则sin(7π6-α)-2cos(2π3-α)=A.-513 B.513 C.-1513 【解析】选A.sin(7π6-α)=sin[π+(π6-α)]=-sin(π6-α)=-cos(π3cos(2π3-α)=-cos[π-(2π3-α)]=-cos(π3+α所以sin(7π6-α)-2cos(2π3-α)=513-2×51312.(5分)(多选)下列结论正确的有()A.sin(π6+α)=cos(π3B.cos(5π6+θ)+sin(2π3C.sin2(15°-α)+cos2(75°+α)=1D.sin2(15°-α)+sin2(75°+α)=1【解析】选ABD.对于A,sin(π6+α)=sin(π2+α-π3)=cos(α-π3对于B,因为cos(5π6+θ)=cos(π2+π3+θ)=-sin(π3+θ)=-sin[π-(2π3-θ)]=所以cos(5π6+θ)+sin(2π3对于C,sin(15°-α)=sin[90°-(75°+α)]=cos(75°+α),所以sin2(15°-α)+cos2(75°+α)=2cos2(75°+α)≠1,故C错误;对于D,sin2(15°-α)+sin2(75°+α)=cos2(75°+α)+sin2(75°+α)=1,故D正确.13.(5分)(2025·徐州高一检测)已知cos(75°+α)=513,α是第三象限角,则sin(195°-α)+cos(105°-α-1013【解析】因为cos(75°+α)=513所以sin(195°-α)+cos(105°-α)=sin[270°-(75°+α)]+cos[180°-(75°+α)]=-cos(75°+α)-cos(75°+α)=-2×513=-1014.(10分)已知tan(3π+α)=2,求sin(α【解析】因为tan(3π+α)=2,所以tanα=2,所以sin=sinαsin15.(10分)已知f(α)=sin((1)化简f(α);【解析】(1)f(α)