已知tanα求cos 已知tan(π/4+α)=2,求1/(2sinαcosα+cos^2α)的值

已知tan(π/4+α)=2,求1/(2sinαcosα+cos^2α)的值  

已知tan(π/4+α)=2,求1/(2sinαcosα+cos^2α)的值

tan(π/4+α)=sin(π/4+α)/cos(π/4+α)
=[sin(π/4)cosα+cos(π/4)sinα]/[cos(π/4)cosα-sin(π/4)sinα]
=[cosα+sinα]/[cosα-sinα]
=2
cosα+sinα=2cosα-2sinα
cosα=3sinα
tanα=1/3
1/(2sinαcosα+cos^2α)=1/[6(sinα)^2+9(sinα)^2]
=1/[15(sinα)^2]
2(sinα)^2=1-cos(2α)
cos(2α)=[1-(tanα)^2]/[1+(tanα)^2]
=[1-1/9]/[1+1/9]
=4/5
(sinα)^2=[1-cos(2α)]/2=1/10
1/(2sinαcosα+cos^2α)=1/[15(sinα)^2]=1/[15/10]=2/3

已知tan(π/4+α)=2,求1/(2sinα*cosα+cos平方α)的值

1/(2sinα*cosα+cos平方α)

= (sina*sina+cosa*cosa)/(2sinα*cosα+cosα*cosa )
对式子上下同时除以 cosa*cosa就有
=（tana*tana + 1）/（2tana + 1 ）
有tan(π/4+α)=2, 就有（1+tana）/(1-tana) =2 ,就有 tana=1/3
就有 （1/9+1）/(1+2/3)=2/3

已知tan(π/4+α）=2，求1/2sinαcosα+（cosα）^2

tan(π/4+α）=2
(tanπ/4+tanα)/(1-tanπ/4tanα)=2
(1+tanα)/(1-tanα)=2
1+tanα=2(1-tanα)
1+tanα=2-2tanα
3tanα=1
tanα=1/3
1/2sinαcosα+（cosα）^2
=1/4sin2α+(1+cos2α)/2
=1/4sin2α+1/2*cos2α+1/2
=1/4*2tanα/(1+tan^2α)+1/2*(1-tan^2α)/(1+tan^2α)+1/2
=1/2*tanα/(1+tan^2α)+1/2*(1-tan^2α)/(1+tan^2α)+1/2
=(1/2*1/3)/[1+(1/3)^2]+1/2*[1-(1/3)^2]/[1+(1/3)^2]+1/2
=(1/6)/(10/9)+(4/9)/(10/9)+1/2
=3/20+4/10+1/2
=3/20+8/20+10/20
=21/20

求解：已知tan(派/4+α）=2，求1/（2sinαcosα+cos平方α）的值。

tan(α+π/4)=(tanα+tanπ/4)/(1-tanαtanπ/4).
=(tanα+1)/(1-tanα)=2.
tanα+1=2-2tanα.
3tanα=1
tanα=1/3.
1/(2sinαcosα+cos^2α)=(1/cos^2α)/(2tanα+1).
=sec^2α/(2tanα+1).
=(1+tan^2α)/(2tanα+1)
=[1+(1/3)^2]/[2*(1/3)+1].
=(10/9)/(5/3)
=2/3. ----即为所求。

已知tanα=3,求1/(sin^2 α+2sinαcosα)的值

1=（sina）^2+(cosa)^2
所以原式=[（sina）^2+(cosa)^2]/(sin^2α+2sinαcosα)
分子、分母同时除以(cosa)^2
得原式=[（tana）^2+1]/[(tanα)^2+2tanα]
=(9+1)/(9+6)
=2/3

已知tan（π/4+θ）+tan（π/4-θ)=4,且-π＜θ＜-π/2，求sin²θ-2sinθcosθ-cos²θ的值

tan( π/4 + θ ) + tan( π/4 - θ ) = 4
[ sin( π/4 + θ )cos( π/4 - θ ) + cos( π/4 + θ )sin( π/4 - θ )] / [cos( π/4 + θ )cos( π/4 - θ )] = 4
sin( π/2 ) / [cos( π/4 + θ )cos( π/4 - θ )] = 4
cos( π/4 + θ )cos( π/4 - θ ) = 1 / 4
( cos π/2 + cos 2θ ) / 2= 1 / 4
cos 2θ = 1 / 2
sin 2θ = √( 1 - cos² 2θ ) = √3 / 2
sin² θ - 2sin θ cos θ - cos² θ = - ( sin 2θ + cos 2θ ) = - (√3 + 1) / 2

已知tanθ=1/2,求2sin^2θ+3sinθcosθ﹣cos^2θ的值

2sin^2θ+3sinθcosθ﹣cos^2θ
=(2sin^2θ+3sinθcosθ﹣cos^2θ)/(sin^2θ+cos^2θ) 分子分母同时除以cos^2θ
=(2tan^2θ+3tanθ﹣1)/(tan^2θ+1) 代入tanθ=1/2
=(1/2+3/2-1)/(1/4+1)
=4/5

已知tanα=1/2，求2sinα-cos²α+4sinαcosα的值

tanα=1/2
sina=(1/2)cosa
2sinα-cos²α+4sinαcosα （代入：sina=(1/2)cosa）
=cosa-cos^2a+2cos^2a
=cosa+cos^2a
∵tanα=1/2
∴cosa=±2√5/5
原式=(4/5)±2√5/5

已知tanθ/(tanθ-6)=1,求(2sinθ-cosθ)/(sinθ+2cosθ)的值

将此式子：(2sinθ-cosθ)/(sinθ+2cosθ)分子分母同时除以cosθ
得：(2tanθ-1)/(tanθ+2)~~~~~~~~~~①
又∵tanθ/(tanθ-6)=1，分子分母同时平方，
得：tan²θ/(tanθ-6)²=1，把分母开启乘到等号右边
tan²θ=tan²θ-12tanθ+36
解得：tanθ=3~~~~~~~~~②
把②带入①中，解得：原式=1