和角差、倍角與半角計算

(1). sinX = 6/5cosX

tanX = 6/5


圖片參考：http://imgcld.yimg.com/8/n/AC06686869/o/20131201102545.jpg


所以sinX = 6/√61= (6√61)/61


(2). sinX-cosX = 1/3

(sinX-cosX)^2 = (1/3)^2

1 - 2sinXcosX = 1/9

sinXcosX = 4/9


sin3X + cos3X = 3sinX-4*(sinX)^3 + 4(cosX)^3-3cosX

=3(sinX-cosX)-4(sinX-cosX)(sin^2X+sinXcosX+cos^2X)

=3*(1/3)-4(1/3)[1+(4/9)]

=1-52/27= -25/27


(3). sinX-cosX = t

(sinX-cosX)^2 = t^2

1 - 2sinXcosX = t^2

sinXcosX= (1-t^2)/2


sin3X + cos3X = 3sinX-4*(sinX)^3 + 4(cosX)^3-3cosX

=3(sinX-cosX)-4(sinX-cosX)(sin^2X+sinXcosX+cos^2X)

=3t-4t[1+(1-t^2)/2]

=3t-6t+2t^3 = 2t^3-3t


2013-12-01 10:57:18 補充：
a^3-b^3 = (a-b)(a^2+ab+b^2)

2013-12-01 23:09:08 補充：
第一題

sinX = 6/5cos1/2X, 求cosX

5sinX = 6cosX/2

5sinX用二倍角公式~

5*2sinX/2cosX/2 = 6cosX/2

5sinX/2cosX/2 = 3cosX/2

5sinX/2cosX/2-3cosX/2 = 0

cosX/2(5sinX/2-3) = 0

cosX/2 = 0 或 5sinX/2-3 = 0

2013-12-01 23:09:15 補充：
先看cosX/2 = 0

cosX/2 = ±√(1+cosX)/2

0^2 = 1+cosX

cos = -1

再看5sinX/2-3 = 0

5sinX/2 = 3

sinX/2 = 3/5

sinX/2 = ±√(1-cosX)/2

(3/5)^2 = (1-cosX)/2

9/25 = (1-cosX)/2

18/25 = 1-cosX

cosX = 7/25

2013-12-01 23:37:05 補充：
過程中~哪裡不了解的

歡迎補充發問^_____^

1 設Sinx=6Cosx/5則 Cosx值?
Sol
Sinx=6Cosx/5
5Sinx=6Cosx
25Sin^2 x=36Cos^2 x
25Sin^2 x+25Cos^2 x=61Cos^2 x
Cos^2 x=25/61
Cosx=+/-5/√61=+/-5√61/61

2若Sinx－Cosx=1/3則Sin3x+Cos3x=?
Sol
Sinx－Cosx=1/3
Sin^2 x－2SinxCosx+Cos^2x=1/9
SinxCosx=4/9
Sin3x=3Sinx－4Sin^3x
Cos3x=4Cos^3 x－3Cosx
Sin3x+Cos3x
=3(Sinx－Cosx)－4(Sin^3 x－Cos^3 x)
=3*(1/3)－4(Sinx－Cosx)(Sin^2x+SinxCosx+Cos^2 x)
=1－4*(1/3)*(1+4/9)
=－25/27

3設Sinx－Cosx=t，則t表示Sin3x+Cos3x=?
Sol
Sinx－Cosx=t
Sin^2 x－2SinxCosx+Cos^2x=1/9
1－2SinxCosx=t^2
2SinxCosx=1－t^2
Sin3x=3Sinx－4Sin^3x
Cos3x=4Cos^3 x－3Cosx
Sin3x+Cos3x
=3(Sinx-Cosx)－4(Sin^3x－Cos^3x)
=3*t－4(Sinx－Cosx)(Sin^2x+SinxCosx+Cos^2 x)
=3t－4*t*[1+(1－t^2)/2]
=3t－(4t+2t+2t^3)
=2t^3－3t

4 Sinx=6/5Cos(x/2)求Cosx
Sol
Sinx=(6/5)Sin(x/2)
5Sinx=6Sin(x/2)
10Sin(x/2)Cos(x/2)=6Sin(x/2)
Sin(x/2)*[5Cos(x/2)-3]=0
(1) Sin(x/2)=0
CosX=1-2Sin^2 (x/2)=1
(2) 5Cos(x/2)－3=0
Cos(x/2)=3/5
Cosx=2cos^ (x/2)－1=2*9/25－1=－7/25




2013-12-01 12:04:44 補充：
修正
Sinx=6/5Cos(x/2)求Cosx
Sol
Sinx=(6/5)Cos(x/2)
5Sinx=6Cos(x/2)
10Sin(x/2)Cos(x/2)=6Cos(x/2)
Cos(x/2)*[5Sin(x/2)－3]=0
(1) Cos(x/2)=0
Cosx=2Cos^2 (x/2)－1=－1
(2) 5Sin(x/2)－3=0
Sin(x/2)=3/5
Cosx=1－2Sin^2 (x/2)=1－2*9/25==7/25