1.已知0
網友廻答:
- 1、解: 由tan(a/2) 1/tan(a/2)=tan(x/2) cot(x/2)=5/2 則: 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 由於0
0 則: cos(x)=根號[1-sin^2(x)] =3/5 則:sin(x-Л/3) =sinxcospi/3-sinpi/3cosx =4/5*1/2-根號3/2*3/5 =(4-3根號3)/10 2、因爲cosa cos^2a=1 又因爲 sin^2a cos^2a=1 所以得:sin^2a=cosa 則: 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、
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