高一变态数学题
发布网友
发布时间：2023-07-30 10:39
共4个回答
热心网友
时间：2024-10-30 06:18
∵cos3A=cos(2A+A)
=cos2AcosA-sin2AsinA
=[2(cosA)^2-1]cosA-2sinAcosAsinA
=[2(cosA)^2-1-2(sinA)^2]cosA
=[2(cosA)^2+2-2(sinA)^2-3]cosA
=[2(cosA)^2+2(cosA)^2-3]cosA
=4(cosA)^3-3cosA
∴2cosAcos2A=2cosA[2(cosA)^2-1]=4(cosA)^3-2cosA=cos3A+cosA
∵cos2A=2(cosA)^2-1
∴(cosA)^2=(1+cos2A)/2
∵cos2A+cos4A
=2(cosA)^2-1+2(cos2A)^2-1
=2(cosA)^2-1+2[2(cosA)^2-1]^2-1
=2(cosA)^2+2[4(cosA)^4-4(cosA)^2+1]-2
=8(cosA)^4-6(cosA)^2
=2cosA[4(cosA)^3-3cosA]
=2cosA*cos3A
∴cosA*cos3A=(cos2A+cos4A)/2
∴原式=(cos5°)^2+(cos10°)^2-2cos5°cos10°cos15°
=(1+cos10°)/2+(1+cos20°)/2-(cos15°+cos5°)cos15°
=1+（cos10°+cos20°）/2-(cos15°)^2-cos5°*cos15°
=1+（cos10°+cos20°）/2-(1+cos30°)/2-（cos10°+cos20°）/2
=1-1/2-cos30°/2
=1/2-[(根号3)/2]/2
=2/4-(根号3)/4
=[2-(根号3)]/4
2.证明：
∵cos3A=4(cosA)^3-3cosA
sin3A=sin(2A+A)
=sin2AcosA+cos2AsinA
=2sinAcosAcosA+[1-2(sinA)^2]sinA
=sinA[2(cosA)^2+1-2(sinA)^2]sinA
=sinA[2-2(sinA)^2+1-2(sinA)^2]
=sinA[-4(sinA)^2+3]
=-4(sinA)^3+3sinA
(cosA)^4-(sinA)^4=[(cosA)^2+(sinA)^2][(cosA)^2-(sinA)^2]=cos2A
(cosA)^3+(sinA)^3
=(cosA+sinA)[(cosA)^2-cosAsinA+(sinA)^2]
=(cosA+sinA)(1-1/2sin2A)
(cosA)^3-(sinA)^3
=(cosA-sinA)[(cosA)^2+cosAsinA+(sinA)^2]
=(cosA-sinA)(1+1/2sin2A)
∴sin3Asin^3A+cos3A*cos^3A
=[-4(sinA)^3+3sinA](sinA)^3+[4(cosA)^3-3cosA](cosA)^3
=-4(sinA)^6+3(sinA)^4+4(cos)^6-3(cosA)^4
=4(cos)^6-4(sinA)^6-[3(cosA)^4-3(sinA)^4]
=4{[(cosA)^3]^2-[(sinA)^3]^2}-3cos2A
=4[(cosA)^3+(sinA)^3][(cosA)^3-(sinA)^3]-3cos2A
=4(cosA-sinA)(1-1/2sin2A)(cosA+sinA)(1+1/2sin2A)-3cos2A
=4[(cosA)^2-(sin)^2][1-1/4(sin2A)^2]-3cos2A
=4cos2A[1-1/4(sin2A)^2]-3cos2A
=4cos2A-cos2A(sin2A)^2-3cos2A
=cos2A-cos2A(sin2A)^2
=cos2A[1-(sin2A)^2]
=cos2A*(cos2A)^2
3.解：
原式=-(cos48°+cos24°-cos12°-cos84°)
=-[cos(36°+12°)+cos(36°-12°)-cos(48°-36°)-cos(48°+36°)]
=-[cos36°cos12°-sin36°sin12°+cos36°cos12°+sin36°sin12°-
(cos48°cos36°+sin48°sin36°)-(cos48°cos36°-sin48°sin36°)]
=-[2cos36°cos12°-2cos48°cos36°]
=-2cos36°(cos12°-cos48°)
=-2cos36°[cos(30°-18°)-cos(30°+18°)]
=-2cos36°[(cos30°cos18°+sin30°sin18°)-(cos30°cos18°-sin30°sin18°)]
=-2cos36°(2sin30°sin18°)
=-2cos36°sin18°
=-4cos36°sin18°cos18°/(2cos18°)
=-2cos36°sin36°/(2cos18°)
=-sin72°/(2cos18°)
=-cos18°/(2cos18°)
=-1/2
声明：本人数学水平一般，经常犯低级错误，如果有错，请指出，谢谢!
热心网友
时间：2024-10-30 06:18
热心网友
时间：2024-10-30 06:19
热心网友
时间：2024-10-30 06:20
