早教吧作业答案频道 -->数学-->
0-90°时,1+cos2x+8sin平方x/sin2x的最小值为?
题目详情
0-90°时,1+cos2x+8sin平方x/sin2x的最小值为?
▼优质解答
答案和解析
cos2x=1-2sin^2x
2sin^2x=1-cos2x
8sin^2x=4-4cos2x
y=f(x)=(1+cos2x+8sin^2x)/sin2x
=(1+cos2x+4-4cos2x)/sin2x
=(5-3cos2x)/sin2x
=(5-3cos2x)/√[1-(cos2x)^2]
已知00
∴y>0
y*√[1-(cos2x)^2]=5-3cos2x
y^2*[1-(cos2x)^2]=(5-3cos2x)^2
(9+y^2)*(cos2x)^2-30(cos2x)+25-y^2=0
上方程未知数为(cos2x)的判别式△≥0,即
(-30)^2-4*(9+y^2)*(25-y^2)≥0
y^4-16y^2≥0
y^2*(y+4)*(y-4))≥0
y≥4(另一解y≤-4舍去)
y的最小值=4
y=4
(9+y^2)*(cos2x)^2-30(cos2x)+25-y^2=0
(5cos2x-3)^2=0
cos2x=3/5,sin2x=4/5
y=f(x)=(1+cos2x+8sin^2x)/sin2x
=(1+cos2x+4-4cos2x)/sin2x
=(5-3cos2x)/sin2x
=(5-3*3/5)/(4/5)
=4
答:当0
2sin^2x=1-cos2x
8sin^2x=4-4cos2x
y=f(x)=(1+cos2x+8sin^2x)/sin2x
=(1+cos2x+4-4cos2x)/sin2x
=(5-3cos2x)/sin2x
=(5-3cos2x)/√[1-(cos2x)^2]
已知00
∴y>0
y*√[1-(cos2x)^2]=5-3cos2x
y^2*[1-(cos2x)^2]=(5-3cos2x)^2
(9+y^2)*(cos2x)^2-30(cos2x)+25-y^2=0
上方程未知数为(cos2x)的判别式△≥0,即
(-30)^2-4*(9+y^2)*(25-y^2)≥0
y^4-16y^2≥0
y^2*(y+4)*(y-4))≥0
y≥4(另一解y≤-4舍去)
y的最小值=4
y=4
(9+y^2)*(cos2x)^2-30(cos2x)+25-y^2=0
(5cos2x-3)^2=0
cos2x=3/5,sin2x=4/5
y=f(x)=(1+cos2x+8sin^2x)/sin2x
=(1+cos2x+4-4cos2x)/sin2x
=(5-3cos2x)/sin2x
=(5-3*3/5)/(4/5)
=4
答:当0
看了0-90°时,1+cos2x+...的网友还看了以下：
分解因式：（1）4a2b-6ab2+2ab（2）6（a-b）2-12（a-b）（3）x（x+y）2 2020-04-08 …
先化简,再求值 （1）[(x-y)的平方+（x+y)(x-y)]÷2x 其中X=2010,y=20 2020-05-16 …
已知3f(x)+2f(x)=x,求f(x)怎么算我自己算了一半因为3f(x)+2f(x)=x3f( 2020-06-03 …
用[x]表示不超过x的最大整数,记｛x｝=x-[x],其中x∈R,设f（x）=[x]•｛x｝.用[ 2020-06-04 …
设f（x）=ln10x，g（x）=x，h（x）=ex10，则当x充分大时有（）A．g（x）＜h（x 2020-06-18 …
设总体X~N(μ,σ^2),已知样本容量n=24,样本方差s^2=12.5227,求总体标准差σ大 2020-06-30 …
导数乘法证明中h是什么意思?(f(x)g(x))'=lim(h→0)[f(x+h)g(x+h)-f 2020-07-22 …
,关于集合的..设集合M={x|m-4/5≤x≤m},N={x|n≤x≤n+1/4},且M,N都是 2020-07-29 …
,关于集合的..设集合M={x|m-4/5≤x≤m},N={x|n≤x≤n+1/4},且M,N都是 2020-07-29 …
设总体X~N(μ,σ^2),已知样本容量n=24,样本方差s^2=12.5227,求总体标准差σ大 2020-08-02 …