三角恒等变形11函数y=sin^4+cos^2的最小正周期是2已知在三角形ABC中,3sinA+4cosB=6,4sinB+3cosA=1,则角C的大小为3计算：（sin65°+sin15°sin10°）／（sin25°-cos15°cos80°）4计算sin6°sin42°sin66°sin78°5计算sin^2

Question

三角恒等变形1
1函数y=sin ^4+cos ^2的最小正周期是
2已知在三角形ABC中,3sinA+4cosB=6,4sinB+3cosA=1,则角C的大小为
3计算：（sin65°+sin15°sin10°）／（sin25°-cos15°cos80°）
4计算sin6°sin42°sin66°sin78°
5计算sin^2(20°)+cos^2 (50°)+sin20°cos50°
6计算log2 cos∏/9+ log2 cos2∏/9+ log2 cos4∏/9
7已知A+B=∏/4,求证：(1+tanA)(1+tanB)=2
刷分的不要来

Answer 1

y=sin^4+cos^2=sin^4+(1-2sin^2)=(sin^2-1)^2
sinx的周期是π,有曲线可知sin^2的最小周期为π/2,
所以函数y=sin ^4+cos ^2的最小正周期是π/2；
(1)3sinA+4cosB=6
(2)4sinB+3cosA=1
(1)式两边平方可得
9sinA*sinA++24sinAcosB+16cosB*cosB=36 (3)
(2)式两边平方可得
16sinB*sinB+24sinBcosA+9cosA*cosA=1 (4)
(3)式+(4)式得
9sinA*sinA+24sinAcosB+16cosB*cosB+16sinB*sinB+24sinBcosA+9cosA*cosA=37
化简可得
9(sinA*sinA+cosA*cosA)+16(sinB*sinB+cosB*cosB)+24sinAcosB+24sinBcosA=37
可得
9+16+24(sinAcosB+sinBcosA)=37
sinAcosB＋cosAsinB =1/2
sinC=sin(180-(A+B))=sin(A+B)=sinAcosB＋cosAsinB=1/2
原式＝[sin(90°-25°)+sin15°sin10°]/[sin(15°+10°)-cos15°cos(90°-10°)]
＝(cos25°+sin15°sin10°)/(sin15°cos10°＋cos15°sin10°-cos15°sin10°)
＝[cos(15°+10°)+sin15°sin10°]/sin15°cos10
＝(cos15°cos10°-sin15°sin10°+sin15°sin10°)/sin15°cos10
＝cos15°/sin15°
＝cot15°=cot(30°/2)
＝(1+cos30°)/sin30°
原式=sin6°sin(90°-48°)sin(90°-24°)sin(90°-12°)
=(sin6°cos24°)(cos48°cos12°)
={1/2*[sin(6°＋24°)＋sin(6°－24°)]}*{1/2*[cos(48°＋12°)＋cos(48°－12°)]}
=1/4*(1/2-sin18°)(1/2+cos36)
=1/16-1/8*sin18°+1/8*cos36°-1/4*sin18°cos36°
=1/16-1/8*sin18°+1/8*cos36°-1/4*{1/2[sin(18°+36°)+sin(18°-36°)]}
=1/16-1/8*sin18°+1/8*cos36°-1/4*1/2(sin54°-sin18°)
=1/16-1/8*sin18°+1/8*cos36°-1/8[(sin(90°-36°)-sin18°]
=1/16-1/8*sin18°+1/8*cos36°-1/8cos36°+1/8sin18°
=1/16
原式=sin^2(20°)+cos^2(90°-40°)+sin20°cos(90°-40°)
=sin^2(20°)+sin^2(40°)+sin(20°)sin(40°)
=(sin20°+sin40°)^2-sin20°sin40°
=[2sin((20°+40°)/2)cos((20°-40°)/2)]^2-(-1/2)[cos(20°+40°)-cos(20°-40°)]
=[2*(1/2)cos10°]^2+1/2(1/2-cos20°)
=cos^2(10°)+1/4-1/2cos20°
=cos^2(10°)+1/4-1/2[1-2sin^2(10°)]
=cos^2(10°)-1/4+sin^2(10°)
=-1/4
原式=log2[cos(π/9)cos(2π/9)cos(4π/9)]
=log2{1/2[cos(π/9+2π/9)＋cos(π/9-2π/9)]*cos(4π/9)}
=log2{1/2[1/2+cos(π/9)]*cos(4π/9)}
=log2{1/2[1/2cos(4π/9)+cos(4π/9)cos(π/9)]}
=log2{1/2[1/2cos(4π/9)+1/2(cos(4π/9+π/9)+cos(4π/9-π/9))]}
=log2{1/2[1/2cos(4π/9)+1/2(cos(5π/9)+cos(3π/9))]}
=log2{1/2[1/2cos(4π/9)+1/2cos(5π/9)+1/4]}
=log2[1/4cos(4π/9)+1/4cos(5π/9)+1/8]
=log2[1/4*2*cos((4π/9+5π/9)/2)*cos((4π/9-5π/9)/2)+1/8]
=log2[1/2cos(π/2)cos(π/9)+1/8]
=log2(1/8)
tan(A+B) = (tanA+tanB)/(1-tanAtanB)=tan(π/4)
可得 tanA+tanB=1-tanAtanB;
tanA+tanB+tanAtanB=1;
tanA+tanB(1+tanA)=1;
方程两边+1
(1+tanA)+tanB(1+tanA)=1+1;
抽取公因式1+tanA
(1+tanA)(1+tanB)=2;