倍角公式

Question

1.求证：tan(x/2)=(sinx)/(1+cosx)=(1-cosx)/(sinx)
2.求证：cosx-cosy=-2sin((x+y)/2)*sin((x-y)/2)
要详细的解答过程

Answer 1

1.sinx=2sin(x/2)cos(x/2)
cosx=2cos^2(x/2)-1=1-sin^2(x/2)
所以sinx/(1+cosx)=[2sin(x/2)cos(x/2)]/[2cos^2(x/2)]=sin(x/2)/cos(x/2)=tan(x/2)
同理(1-cosx)/(sinx)=2sin^2(x/2)/[2sin(x/2)cos(x/2)]=sin(x/2)/cos(x/2)=tan(x/2)
得证
2.x=(x+y)/2+(x-y)/2
y=(x+y)/2-(x-y)/2
所以cosx-cosy=cos((x+y)/2)*cos((x-y)/2)-sin((x+y)/2)*sin((x-y)/2)-cos((x+y)/2)*cos((x-y)/2)-sin((x+y)/2)*sin((x-y)/2)=-2sin((x+y)/2)*sin((x-y)/2)
得证

Answer 2

1.tanx/2=(sinx/2)/cos(x/2)
上下同乘以2cosx/2
上面变成sinx因为cosx=2(cosx/2)^2-1
1+cosx=2(cosx/2)^2
下面变成1+cosx (因为sinx=2sinx/2 * cosx/2 )
所以tan(x/2)=(sinx)/(1+cosx)

上下同乘以2sinx/2
上面变成1-cosx 因为cosx=1-2(sinx/2)^2
1-cosx=2(sinx/2)^2
下面变成sinx (因为sinx=2sinx/2 * cosx/2 )
所以tan(x/2)=(1-cosx)/(sinx)

2.利用和角公式
cos(a-b)=cosa*cosb+sina*sinb (1)
cos(a+b)=cosa*cosb-sina*sinb (2)

令x=a+b，y=a-b,那么a=（x+y）/2,b=（x-y）/2
由（2）-(1),得
cos(a+b)-cos(a-b)=-2sina*sinb
将上述假设代入可得
cosx-cosy=-2sin((x+y)/2)*sin((x-y)/2)，得证

Answer 3

(1)sinx=2sin(x/2)cos(x/2)
cosx=2cos^2(x/2)-1＝1-2sin^2(x/2)
代入原式可得证
（2）cosx=cos((x+y)/2+(x-y)/2)
cosy=cos((x+y)/2-(x-y)/2)
代入原式，再用cos(a+b)=cosacosb-sinasinb 即得证

Answer 4

1.
sinx=2sin(x/2)cos(x/2)
cosx=1-2(sinx)^2=2(cosx)^2-1
第一个题就是考你这两个公式
sinx/(1+cosx)=2sin(x/2)cos(x/2)/[1+(2(cosx)^-1)]=tan(x/2)
(1-cosx)/(sinx)=[1-(1-2(sinx)^2)]/2sin(x/2)cos(x/2)=tan(x/2)

2.cosx=cos[(x+y)/2+(x-y)/2]
cosy=cos[(x+y)/2-(x-y)/2]
带入化简后得右面结果

Answer 5