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为什么sin2x+2sinx平方=1啊???

解: y=1/2sin2x+sin²x =1/2·sin2x+(1-cos2x)/2 =1/2·sin2x-1/2·cos2x -1/2 =1/2·(sin2x- cos2x)-1/2 =√2/2·sin(2x-π/4)-1/2 ∵ -1 ≤ sin(2x-π/4)≤1 -√2/2 ≤√2/2· sin(2x-π/4)≤√2/2 -√2/2 -1/2≤√2/2· sin(2x-π/4)-1/2≤√2/2-1/2 -(√2+1)/2 ≤√...

y=1/2sin2x+sinx^2 =1/2sin2x+(1-cos2x)/2 =√2/2sin(2x-π/4)+1/2 因为sin(2x-π/4)的值域是[-1,1] 所以函数的值域是[(1-√2)/2,(1+√2)/2]

(1+x)^a-1~ax,这就是他的等价无穷小,证明的时候比一下洛必达就可以了,或者泰勒

y=√3cos²x+1/2sin2x =√3/2(cos2x+1)+1/2sin2x =√3/2cos2x+1/2sin2x+√3/2 =sin(π/3)cos2x+cos(π/3)sin2x+√3/2 ....//注:sin(π/3)=√3/2,cos(π/3)=1/2 =sin(2x+π/3)+√3/2

不是有两个重要极限之一:lim(x→0)sinx/x=1这就说明x→0时sinx与x是等价无穷小,因此可以代换啊.情愫Bo60 2014-11-131-cosx=2sin^2(x/2)~2*(x/2)^2=1/2x^2sinx~x,sin(x/2)~x/2,2sin^2(x/2)~2*(x/2)^2=1/2x^2 其实就是等量代换

f'(x)=1/2cos2x*(2x)]+cosx =cos2x+cosx 显然这是偶函数 cos2x+cosx =2cos²x-1+cosx =2(cosx+1/4)²-9/8 -1

倍角公式 cos2x=1-2sin²x 所以sin²x=1/2(1-cos2x)

y= (sinx)^2-2sinx +3 =(sinx-1)^2 +2 值域= [2, 3]

其实就是把sinx看做整体,进行配方。应该是初二的知识吧。 解: a=-sin²x+4sinx+1 =-sin²x+4sinx-4+5 =-(sin²x-4sinx+4)+5 =-(sinx-2)²+5

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