证明题:1-cosx=2乘以sin (x/2)的平方
证明题:1-cosx=2乘以sin (x/2)的平方
因为: cosx=cos(2*x/2)=[cosx/2]^2-[sinx/2]^2
=1-[sinx/2]^2-[sinx/2]^2
所以:1-cosx=2[sinx/2]^2
因为:
cos(a+b)=cos(a)cos(b)-sin(a)sin(b) //和角公式
所以:
cos(x)=[cos(x/2)]^2 - [sin(x/2)]^2
又因为:
[cos(x/2)]^2 + [sin(x/2)]^2 =1
所以:
cos(x/2)]^2 = 1 - [sin(x/2)]^2
cos(x)=1 - 2[sin(x/2)]^2
1- cos(x) = 2[sin(x/2)]^2
这是一个基本三角公式:1-cosx=1-[(cos(x/2))^2-(sin(x/2))^2]=1-[1-2*(sin(x/2))^2]=2*(sin(x/2))^2,不知道这样行不行,这确实是最基本的公式了,要是再不会,我也无能为力了。谢谢!
1-cosx=1-cos(2*x/2)=1-(2cosx/2^2-1)=2-2cosx/2^2=2(1-cosx/2^2)=2sinx/2^2
证明完毕。
这是二倍角公式
1-2sin^2(x/2)=cosx
证明
cosx=cos(x/2+x/2)
=cos^2(x/2)-sin^2(x/2)
=1-sin^2(x/2)-sin^2(x/2)
=1-2sin^2(x/2)
1-cosx=2sin^2(x/2)
1-cosx
=1-cos(x/2+x/2)
=1-{cos(x/2)*cos(x/2)-sin(x/2)*sin(x/2)}
=1-[cos(x/2)]^2+[sin(x/2)]^2
=[cos(x/2)]^2+[sin(x/2)]^2-[cos(x/2)]^2+[sin(x/2)]^2
=2[sin(x/2)]^2
即证!