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F.4 a.maths....

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1a) Prove that (cosx-sinx)^2 = 1 - sin2x.b) Find the maximum and minimum values of cosx-sinx.Hence find the range of the values of 1/(cox-sinx-2) for all possible values of x.2 Solve the following equations for 0<=x<=360a) sin2xsin6x = sin^2(4x)b) cos(2x-45)sin(2x+45) = 1/43) suppose m>0 and... 顯示更多 1a) Prove that (cosx-sinx)^2 = 1 - sin2x. b) Find the maximum and minimum values of cosx-sinx. Hence find the range of the values of 1/(cox-sinx-2) for all possible values of x. 2 Solve the following equations for 0<=x<=360 a) sin2xsin6x = sin^2(4x) b) cos(2x-45)sin(2x+45) = 1/4 3) suppose m>0 and n>=4. In the expansion of (x-2/x)^2 (1+mx)^n in ascending powers of x, the constant term is 20 and there is no x term. Find the values of m and n. 更新: 1b) cosx-sinx既MAX.係√2 , MIN.係 -√2

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但(b)中係二次方後先係1-sin2x 所以應該係0<1-sin2x<1~~~~ 是嗎???? 2010-07-14 13:27:13 補充： 1a) Prove that (cosx-sinx)^2 = 1 - sin2x. LHS =(cosx-sinx)^2 =cos^2x-2cosxsinx+sin^2x =1-2cosxsinx =1-(cosxsinx+cosxsinx) =1-sin2x b) Find the maximum and minimum values of cosx-sinx. Hence find the range of the values of 1/(cox-sinx-2) for all possible values of x. -1-sin2x>-1 2>1-sin2x>0 0<(cosx-sinx)^2<2 (cosx-sinx)^2<2 -sqrt21/(cosx-sinx-2)>1/(-2+sqrt2) 2 Solve the following equations for 0<=x<=360 a) sin2xsin6x = sin^2(4x) [cos(2x-6x)-cos(2x+6x)]/2=sin^2(4x) [cos(-4x)-cos(8x)]/2=sin^2(4x) (cos4x-cos8x)/2=sin^2(4x) [(cos4x-cos(4x+4x)]/2=sin^2(4x) [cos4x-(cos^2(4x)-sin^2(4x)]/2=sin^2(4x) [cos4x-cos^2(4x)+sin^2(4x)]/2=sin^2(4x) [cos4x-cos^2(4x)+1-cos^2(4x)]/2=1-cos^2(4x) Let y=cos4x (y-y^2+1-y^2)/2=1-y^2 (y-2y^2+1)=2-2y^2 y+1=2 y=1 cos4x=1 cos4x=cos0 4x=0 or 360 or 720 or 1080 or 1440 x=0 or 90 or 180 or 270 or 360 b) cos(2x-45)sin(2x+45) = 1/4 [sin(2x+45+2x-45)+sin(2x+45-2x+45)]/2=1/4 [sin(4x)+sin90]/2=1/4 sin4x+1=1/2 sin4x=-1/2 sin4x=-sin30 4x=180+30 or 360-30 or 540+30 or 720-30 or 900+30 or 1080-30 or 1260+30 or 1440-30 4x=210 or 330 or 570 or 690 or 930 or 1050 or 1290 or 1410 x=52.5 or 82.5 or 142.5 or 172.5 or 232.5 or 262.5 or 322.5 or 352.5 3)suppose m>0 and n>=4. In the expansion of (x-2/x)^2 (1+mx)^n inascending powers of x, the constant term is 20 and there is no x term.Find the values of m and n. (x-2/x)^2(1+mx)^n =(x^2-2(x)(2/x)+(2/x)^2)(1+nmx+n(n-1)(mx)^2/2!+...) =(x^2-4+4/x^2)(1+(nm)x+n(n-1)m^2x^2/2!+n(n-1)(n-2)(m^3)x^3/3!...) constant term=-4+4n(n-1)m^2/2!=20 4n(n-1)m^2/2=24 4n(n-1)m^2=48 n(n-1)m^2=12----(1) x term=-4nm+4n(n-1)(n-2)m^3/6=0 4n(n-1)(n-2)m^3/6=4nm n(n-1)(n-2)m^3/6=nm n(n-1)(n-2)m^3=6nm (n-1)(n-2)m^2=6----(2) (1)/(2) n(n-1)/(n-1)(n-2)=2 n/(n-2)=2 n=2n-4 4=n when n=4 4(4-1)m^2=12 4x3xm^2=12 m^2=1 m=1

其他解答:

1a) Prove that (cosx-sinx)^2 = 1 - sin2x. L.H.S. = (cosx-sinx)^2 = cos^2x - 2cosxsinx + sin^2x = 1 - 2cosxsinx = 1-sin2x = R.H.S. 2010-07-14 12:23:50 補充： b)Given (cosx-sinx)^2 = 1 - sin2x., For Max: sin2x = -1, (cosx-sinx)^2 = 1+1 = 2 For Min: sin2x = 1, (cosx-sinx)^2 = 1-1 = 0 Max for cosx-sinx = 2^(1/2) Min for cosx - sinx = 0 0<=(cosx - sinx)<=2^(1/2) -2<=(cosx - sinx-2)<= 2^(1/2)-2 -1/2>=1/(cosx - sinx-2)>=1/[2^(1/2)-2]E2A5F59BAA12C031