Question

tg(α+$frac{ pi }{4}$), если sin2α=$frac{4}{5}$

Answer 1

 sin2α= .
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tg(α+)   -?

tq(α+π/4) =tq(π/4+ α)= (tqπ/4+tqα)/(1-tqπ/4*tqα)  =(1+tqα)/(1-tqα) =
(cosα +sinα)/(cosα -sinα)=(cosα +sinα)(cosα -sinα)/(cosα -sinα)²=
(cos²α -sin²α)/(cos²α -2sinα*cosα+sin²α) = cos2α/(1 -sin2α) = ±  √(1-sin²2α)/(1-sin2α) =±√(1 -(4/5)²) / (1 -4/5) =   ± 3/25.
* * *    или по другому  * * *
π/4+α = (π/2 +2α)/2  =β/2  ;  β= π/2 +2α
tqβ/2 =(sinβ/2) /(cosβ/2) = 2(sinβ/2)*(cosβ/2) /2(cosβ/2)*(cosβ/2) =sinβ/(1+cosβ). 

tq(π/4+α )=tq((π/2+2α )/2) =sin(π/2+2α)(/(1+cos(π/2 +2α) )= cos2α/(1-sin2α).