化简:(1)1-tanˆ2(x+y)/2tan(x+y) (2)sin4a/1+cos4a
1个回答
1)
=1/[2tan(x+y)/1-tan²(x+y)]
=1/[tan2(x+y)]
2)
=2sin2acos2a/(2cos²2a-1+1)
=sina2a/cos2a
=tan2a
相关问题
化简{(cos x*tan x +1-2 sin平方x/2)[根号2(x-π/4)] }/1+sin2x
化简题及证明题请帮帮忙!1.化简:[1-tan²(x+y)]/2tan(x+y)2.证明 ①.(2sinα-sin2α)
化简(cos2x-sin2x)/(1-cos2x)(1-tan2x)
化简:(2cos²a-1)/【2tan(π/4-a)sin²(π/46+a)】
化简2sin2a^2+√3sin4a-4tan2a/sin8a×(1-tan2a^2)/(1+tan2a^2) ^2 急
化简1-2cos2a/2tan(π/4+a)sin2(π/4-a)得
若xˆ2-yˆ2=1,化简(x+y)ˆ2010 • (x-y) ˆ2
化简:sin(3丌-a)/tan(a-5丌)x tan(-a+2丌)/cot(丌-a)x cos(4丌-a)/sin(-
求证:(1)tan(x/2+π/4)+tan(x/2 - π/4)=2tan x(2)(1+sin 2φ)/(cos φ
tan(x+y)tan(x-y)=sin^2x-sin^2y/cos^2x-sin^2y 顺便问一下. tan,sin,