2025高考數(shù)學(xué)一輪復(fù)習(xí)-4.2-同角三角函數(shù)的基本關(guān)系與誘導(dǎo)公式-專項(xiàng)訓(xùn)練【含答案】

2025高考數(shù)學(xué)一輪復(fù)習(xí)-4.2-同角三角函數(shù)的基本關(guān)系與誘導(dǎo)公式-專項(xiàng)訓(xùn)練【A級(jí)基礎(chǔ)鞏固】1.已知sin(α-π4)=13,則cos(π4A.223 B.-223 C.2.已知3sin(θ+π2)+sin(θ+π)=0,θ∈(-π,0),則sinθ等(A.-31010 B.-1010 C.33.已知α∈(0,π),且2cos2α-sin2α-4cosα=3,則sinα等于()A.53 B.23 C.134.(多選題)在△ABC中,下列結(jié)論正確的是()A.sin(A+B)=sinCB.sinB+CC.tan(A+B)=-tanC(C≠π2D.cos(A+B)=cosC5.已知角α的頂點(diǎn)在原點(diǎn),始邊與x軸非負(fù)半軸重合,終邊與直線2x+y+3=0平行,則sinαA.-2 B.-14 C.2 6.已知tanθ=2,則1+sinθcosθ的值為.

7.已知角θ的終邊過點(diǎn)A(4,a),且sin(θ-π)=35,則tanθ等于8.若sinx=3sin(x-π2),則cosxcos(x+π2)=9.已知角θ的終邊與單位圓x2+y2=1在第四象限交于點(diǎn)P,且點(diǎn)P的坐標(biāo)為(12(1)求tanθ的值;(2)求cos(πINCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級(jí)能力提升】10.已知tanα=-3,則sinA.-34 B.34 C.31011.(多選題)已知3sin(π+θ)=cos(2π-θ),θ∈(-π3A.-π6 B.-π3 C.π312.已知sin(3π+θ)=13,則cos(π+.

13.已知sin(π-α)-cos(π+α)=23(1)sinα-cosα;(2)sin3(π+α)+cos3(-α).INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級(jí)應(yīng)用創(chuàng)新練】14.已知f(α)=1+sinα1-(1)若f(α)=3,求43sin2α+cos2(2)若cos2αf(α)=12,求cos(2023π+α)+cos(3π參考答案【A級(jí)基礎(chǔ)鞏固】1.解析:因?yàn)棣?+α-(α-π4)=所以cos(π4+α)=cos[π2+(α-π4)]=-sin(α-π2.解析:因?yàn)?sin(θ+π2)+sin(θ+π)=0,所以3cosθ-sinθ=0,因?yàn)棣取?-π,0),sin2θ+cos2θ=1,所以sinθ=-33.解析:因?yàn)?cos2α-sin2α-4cosα=3,所以3cos2α-4cosα-4=0,解得cosα=-231-cos2α4.解析:在△ABC中,有A+B+C=π,則sin(A+B)=sin(π-C)=sinC,A正確;sinB+C2=sin(π2-(C≠π25.解析:因?yàn)榻铅恋慕K邊與直線2x+y+3=0平行,即角α的終邊在直線y=-2x上,所以tanα=-2,所以sinα-cos6.解析:因?yàn)閠anθ=2,所以1+sinθcosθ=sintan2θ+tanθ答案:77.解析:sin(θ-π)=-sinθ=35,sinθ=-3由于角θ的終邊過點(diǎn)A(4,a),所以θ在第四象限,所以cosθ=1-sin所以tanθ=sinθcosθ答案:-38.解析:因?yàn)閟inx=3sin(x-π2)=-3cosx,所以sin2x=9cos2x=1-cos2所以cos2x=110所以cosxcos(x+π2)=-cosxsinx=(-cosx)·=3cos2x=3×110=3答案:39.解:(1)由θ為第四象限角,終邊與單位圓交于點(diǎn)P(12得(12)2+y2解得y=-32所以tanθ=-321(2)因?yàn)閠anθ=-3,所以cos(π2-θ)+cos(θ-2πINCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級(jí)能力提升】10.解析:因?yàn)閠anα=-3,所以sin3α-sin-sinαcosα=-sinαcosαco11.解析:因?yàn)?sin(π+θ)=cos(2π-θ),所以-3·sinθ=cosθ,所以tanθ=-33,因?yàn)棣取?-π3,π),所以θ=-π612.解析:由sin(3π+θ)=13,可得sinθ=-1所以cos(π+=-cosθ=11+cosθ=2=21-=18.答案:1813.解:(1)由sin(π-α)-cos(π+α)=23,得sinα+cosα=2所以1+2sinαcosα=29,故2sinαcosα=-7所以sinα>0,cosα<0,所以sinα-cosα>0,所以sinα-cosα=(sinα-cosα)2(2)sin3(π+α)+cos3(-α)=cos3α-sin3α=(cosα-sinα)(cos2α+cosαsinα+sin2α)=(-43)×(1-718)=-INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級(jí)應(yīng)用創(chuàng)新練】14.解:(1)因?yàn)棣翞榈诙笙藿?所以|cosα|=-cosα,f(α)=1+sinα1=(1+sinα=(1+sinα=1+sinα|cosα|若f(α)=3,則-2tanα=3,所以tanα=-32所以43sin2α+cos2α=43sin2α+co(2)cos2αf(α)=cos2α×(-2tanα)=-cos2α×2sinα因?yàn)?/p>