[sin(540-x)/tan(900-x)]*[1/tan(450-x)*tan(810-x)]*[cos(360-x)/sin(-x)]
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/28 12:48:10
x).051Э/IӰ4 1c
\SWĴ0$ku4me~
U݅0 BA_a*@yP*x0#!bu!z!`$Hb@%P@5qF`m6yvаs#4V([
5hUh%Thi ͭKBLR [.H!8)v<>/,cY-/?_l_ `!
[sin(540-x)/tan(900-x)]*[1/tan(450-x)*tan(810-x)]*[cos(360-x)/sin(-x)]
化简:sin(540度-X)/tan(900度-X)*1/tan(450度-x)tan(810度-x)*cos(360度-x)/sin(-x)
化简 (sin(540°-x))/(tan(900°-x))*(1/(tan(450°-x)tan(810°-x))*((cos(360°-x))/(sin(-x))
一道高一三角函数化简.1.sin(540-x)/tan(900-x)2.1/tan(450-x)tan(810-x)3.cos(360-x)/sin(-x)将上面这三个乘起来(450,900等都是度数)
lim (x->0) [tan(tan x)-sin(sin x)]/(tan x -sin x)
帮帮哦,原题是这样的:sin(540-x)/tan(900-x) 乘以 1/tan(450-x)tan(810-x) 乘以 cos(360-x)/sin(-x)条件是tanx=2我怎么化简都是得sinx,真是邪门了,
化简①sin(π+α)×cos(3π/α)+sin(π/2+α)×cos(π+α)②sin(540°-x)×cos(x-360°)/tan(900°-x)×tan(450°-x)×tan(810°-x)×sin(-x)
lim x->0 (sin x-tan x)/sin (x^3)
(1+tan(x))/sin(2x)不定积分
为什么cos(sin(tan x)))=1
求证tan^x-sin^2x=tan^2x*sin^2x
tan(x)-sin(x)等价无穷小tan(x)-sin(x)的等价无穷小
证明:(tan^2)x-(sin^2)x=(tan^2)x(sin^2)x
sin(540-α)/tan(α-180)*cot(-α-360)/tan(900-α)*cos(720-α)/sin(-α-360)
sin(540-α)/tan(α-180)*cot(-α-360)/tan(900-α)*cos(720-α)/sin(-α-360)
证明tan^2x-sin^2x=tan^2 sin^2x
已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x
若x属于零到六分之派,比较tan(tan x),tan(cos x),tan(sin x)的大小并证明