作业帮-sin x+√(1+sin^2 x) =1/[sin x+√(1+sin^2 x)] 怎么算的怎么算的?
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/12 01:16:21
![-sin x+√(1+sin^2 x) =1/[sin x+√(1+sin^2 x)] 怎么算的怎么算的?](/uploads/image/z/3022089-33-9.jpg?t=-sin+x%2B%E2%88%9A%281%2Bsin%5E2+x%29+%3D1%2F%5Bsin+x%2B%E2%88%9A%281%2Bsin%5E2+x%29%5D+%E6%80%8E%E4%B9%88%E7%AE%97%E7%9A%84%E6%80%8E%E4%B9%88%E7%AE%97%E7%9A%84%3F)
-sin x+√(1+sin^2 x) =1/[sin x+√(1+sin^2 x)] 怎么算的怎么算的?
-sin x+√(1+sin^2 x) =1/[sin x+√(1+sin^2 x)] 怎么算的
怎么算的?
-sin x+√(1+sin^2 x) =1/[sin x+√(1+sin^2 x)] 怎么算的怎么算的?
利用分子有理化啊
把方程左边看成分式,分子是-sin x+√(1+sin^2 x）
分母是1
然后,分子分母同乘以[sin x+√(1+sin^2 x)] ,得到：
分子：平方差公式化简得1
分母：[sin x+√(1+sin^2 x)]
x*（1+sin^2 x )/sin^2x 不定积分
傅里叶级数作图f（x）=2sin[x] - sin[2x] + 2/3sin[3x] - 1/2sin[4x]我用mathematica输入程序Plot[{2sin[x],-2sin[x],2sin[x] - sin[2x],-2sin[x] + sin[2x],2sin[x] - sin[2x] + 2/3sin[3x],-2sin[x] + sin[2x] - 2/3sin[3x],2sin[x] - sin[2x] + 2/3si
化简(sin x+cos x-1)(sin x-cos x+1)/sin 2x
(sin 1/x) /x
化简√1+sin x /1-sin x -√1-sin x /1+sin x ,其中x为第2象限角
(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x
化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3sin^2 x
化简(sin^2 x/sin x-cosx)-(sin x+cosx/tan^2 x-1)
∫sin^2x/(1+sin^2x )dx求解,
求不定积分【sin(x)^5 + sin(x)^3】^(1/2)
1-2sinx*sin(60+x)sin(60-x)
sin(x^2)/sin^2x 求导
三角函数√1-sin^2x=?
cosx/√1+sin^2x dx
-sin x+√(1+sin^2 x) =1/[sin x+√(1+sin^2 x)] 怎么算的怎么算的?
化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
请教一道函数极限问题：lim[sin√（x^2+1)-sin(x)] (x→+∞)
1/sin^2x的不定积分谢谢!（是1/sin x * sin x）