jwbf.net
当前位置:首页 >> 若θ属于[0,π/2],且sinθ=4/5,则tAnθ/2= >>

若θ属于[0,π/2],且sinθ=4/5,则tAnθ/2=

解由sin2θ=-4/5,且θ是第二象限角, 知2θ是第三象限角 故cos2θ=-3/5 故tanθ =sinθ/cosθ =2cosθsinθ/2cos^2θ =sin2θ/(1+cos2θ) =(-4/5)/(1+(-3/5)) =(-4/5)/(2/5) =-2

-4/3 解: ∵ θ是第四象限角 ∴ θ+π/4是第四象限或第一象限角 又∵ sin(θ+π/4)=3/5>0 ∴ θ+π/4是第一象限角 ∴ cos(θ+π/4)=√[1-sin²(θ+π/4)]=4/5 ∴ cot(θ+π/4)=4/3 tan(θ-π/4) =tan[(θ+π/4)-π/2] =-tan[π/2-(θ+π/4)] =-cot(θ+π/4) =-4/3

解:∵tan(π/4-θ)=1/2 ==>(tan(π/4)-tanθ)/(1+tan(π/4)*tanθ)=1/2 (应用差角公式) ==>(1-tanθ)/(1+tanθ)=1/2 ∴tanθ=1/3 ∵θ属于(π,2π),tanθ=1/3>0 ==>θ属于(π,3π/2),即θ在第三象限 ==>secθ>0 ∴secθ=√(1+(tanθ)^2)=√10/3 故sinθ+cosθ=(sinθ/cosθ+...

tan2x=4/3 sin2x=4/5,cos2x=3/5 sin(2x+pie/4)=7sqt(2)/10

∵ sinθ= 3 5 ,∴cos 2 θ=1-sin 2 θ= 16 25 ∵sin2θ=2sinθcosθ<0,∴cosθ=- 4 5 (舍正)因此, tan θ 2 = sin θ 2 cos θ 2 = 2si n 2 θ 2 2sin θ 2 cos θ 2 = 1-cosθ sinθ =3故答案为:3

sinθ-2丨cosθ丨=0 sinθ=2丨cosθ丨 sinθ/丨cosθ丨=2 因为θ为第二象限角,所以sinθ>0,cosθ

-4/3 解析: tan(θ-π/4) =-tan(π/4-θ) =-cot[π/2-(π/4-θ)] =-cot(π/4+θ) =-cos(π/4+θ)/sin(π/4+θ) =-4/3

sinθ/2-2cosθ/2=0 tanθ/2=2 根据倍角公式,tanθ =2*2/(1-2²)= -4/3

设x 1 ,x 2 为关于x的方程x 2 -2xtanθ-3=0的两个根,∴x 1 +x 2 =2tanθ,x 1 ?x 2 =-3.又∵x 1 2 +x 2 2 =10,∴(x 1 +x 2 ) 2 -2x 1 ?x 2 =10,∴4tan 2 θ+6=10,∴tanθ=±1.∵0°<θ<90°,∴tanθ>0,∴tanθ=1,∴θ=45度.当θ=45°时,△=4tan 2 θ+1...

网站首页 | 网站地图
All rights reserved Powered by www.jwbf.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com