Question

已知0<a<π/2,tan(a/2)+1/tan(a/2)=5/2,则sin(a-π/3)的值为？

1.已知0<a<π/2,tan(a/2)+1/tan(a/2)=5/2,则sin(a-π/3)的值为？ 2.已知cosa+cosa=1,则sina+sin六次方a+sin8次方=？ 3.设cos[a-(b/2)]=-1/9,sin[(a/2)-b]=2/3,且π/2<a<π，0<b<π/2,求cos(a+b). 4.已知6sina+sinacosa-2cosa=0,a∈[π/2,π],求sin(2a+π/3)。 请写明详细的解题过程

Answer 1

1、解: 由tan(a/2)+1/tan(a/2)=tan(x/2)+cot(x/2)=5/2 则: sin(x/2)/cos(x/2)+cos(x/2)/sin(x/2)=5/2 [sin^2(x/2)+cos^2(x/2)]/[sin(x/2)cos(x/2)]=5/2 2/[2sin(x/2)cos(x/2)=5/2 2/sin(x)=5/2 5sin(x)=4 sin(x)=4/5 由于0 <x< Л/2 则: cos(x)>0 则: cos(x)=根号[1-sin^2(x)] =3/5 则:sin(x-Л/3) =sinxcospi/3-sinpi/3cosx =4/5*1/2-根号3/2*3/5 =(4-3根号3)/10 2、因为cosa+cos^2a=1 又因为 sin^2a+cos^2a=1 所以得:sin^2a=cosa 则: sin^2a+sin^6a+sin^8a =sin^2a+(sin^2a)^3+(sin^2a)^4 =cosa+cos^3a+cos^4a =cosa+cos^2a(cosa+cos^2a) =cosa+cos^2a(sin^2a+cos^2a) =cosa+cos^2a =sin^2a+cos^2a =1 3、