sin(540-x)cos(360-x) / tan(900-x)tan(450-x)tan(810-x)sin(-x)
= sin(180-x)cos(-x) / tan(180-x)tan(90-x)tan(90-x)sin(-x)
= - sin(x)cos(x)tan(x)tan(x)cos(180-x) / sin(180-x)sin(x)
= cos^2(x)tan^2(x) / sin(x)
= sinx
[sin(540-x)/tan(900-x)]*[1/tan(450-x)*tan(810-x)]*[cos(360-x
3个回答
相关问题
-
化简：（1）sin(540°−x)tan(900°−x)•[1tan(450°−x)tan(810°−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
-
sin(540°+α)cos(360°-α)/sin(450°+α)tan(900°-α)
-
求证 （1—tan^2X)/(1+tan^2X)=cos^2X—sin^2X
-
(2sin^2x-sinxcosx+cos^2x)/sin^2x+cos^2x=(2tan^x-tanx+1)/(tan
-
求证：1-2sinxcosx/cos方x-sin方x=1-tan/+tan
-
求证：tan x/2=sin x/(1+cos x)
-
1/cos(x) - tan(x) = 3 求sin(x)
-
tan(x+y)tan(x-y)=sin^2x-sin^2y/cos^2x-sin^2y 顺便问一下. tan,sin,
-
已知（1-tan x）/(2+tan x )=1,求证3 sin 2x =-4cos 2x