[latex] \\ [/latex]tg(α+[latex] \frac{ \pi }{4} [/latex]), если sin2α=[latex]\frac{4}{5} [/latex]

 sin2α= . --------------- 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).