Trigonometry

1.呢題我用左兩邊一齊計既方法計因為呢條數originally from sin^2+cos^2=1

sin^2 x+cos^2 x= 1 呢度左右都要除以cosxsinx

sinx/cosx + cosx/sinx =1/cosxsinx

tanx + cotx =1/(1/2)sin2x

=2/(sin2x)

tanx + cotx =2cosec2x

2. 呢條我只係用左邊計

cos3x = cos(2x+x) = cos2x cosx - sin2x sinx

= (cos^2 x- sin^2 x)cosx - 2sin^2 xcosx

= cos^3 x- sin^2 xcosx - 2sin^2 xcosx

= cos^3 x- 3sin^2 xcosx 呢度用sin^2 x =1-cos^2 x

= cos^3 x- (3cosx - 3cos^3 x)

= 4cos^3 x - 3cosx

3.呢條都係用左邊計

cosec2x - cot2x

= (1/sin2x) - (cos2x/sin2x)

= (1-cos2x)/sin2x 呢度再用cos2x = cos^2 x - sin^2 x

= (1-[cos^2 x - sin^2 x]) / sin2x 再用1= cos^2 x + sin^2 x

= ([sin^2 x + cos^2 x] - [cos^2 x - sin^2 x] ) / sin2x

= 2sin^2 x / sin2x

= 2sin^2 x / 2sinxcosx

= sinx/cosx

=tanx

以上幾個prove因為我怕解得唔清晰so每一step都寫左出黎所以有d長=.=

答問題我估4個step倒就夠...

2007-09-09 07:05:22 補充：

IVANCYP 第一題答案你打少左野= ( sin^X + cos^X ) / (sin X cos X) <-----(通份母)呢度你唔記得打sin^2 同cos^2

1) LHS = tan X + cot X

= sin X/cos X + cos X/sin X

= ( sin^X + cos^X ) / (sin X cos X) <-----(通份母)

= 1 / (sin X cos X)

= 1 / 0.5 sin 2X <-----(因為 sin 2X = 2 sin X cos X)

= 2cosec 2X

= RHS

2) LHS = cos3X

= cos (2X + X)

= [cos2X]cosX - (sin2X)sinX

= [2(cosX)^2 - 1]cosX - (2sinXcosX)sinX <-----(對返括號就會明)

= 2cosX(cosX)^2 - cosX - 2cosX(sinX)^2

= 2cosX(cosX)^2 - cosX - 2cosX[1- (cosX)^2]

= 2cosX(cosX)^2 - cosX - 2cosX + 2cosX(cosX)^2

= 4cosX(cosX)^2 - 3cosX

= 4(cosX)^3 - 3cosX

= RHS

3) LHS = cosec2X - cot2X

= 1/(sin2X) - cos2X/sin2X

= (1 - cos2X)/sin2X <------(通份母)

= [2(sinX)^2] / sin2X <------{ 用 cos2X = 1 - 2(sinX)^2 }

= [2(sinX)^2] / 2sinXcosX

= sinX / cosX

= tanX

= RHS