TgA=1/2. Найдите sin(2A+пи/4)

Дано: tqα=1/2. ------- sin(2α +π/4) -? B = sin(2α +π/4) =sin2α*cosπ/4 + cos2α* sinπ/4=sin2α*1/√2 +cos2α*1/√2= (1/√2)*(sin2α +cos2α). но  sin2α=2sinα*cosα =2sinα*cosα/(cos²α+sin²α) =2tqα /(1+tq²α) ; cos2α =cos²α-sin²α =(cos²α - sin²α)/(cos²α+sin²α)=(1-tq²α)/(1+tq²α),  поэтому B=(1/√2)*(sin2α +cos2α)=(1/√2)*(2tqα/(1+tq²α)  +(1-tq²α)/(1+tq²α) )= 1/√2(1+tq²α) *(2tqα  +1-tq²α) =1/√2(1+1/4) *(2*1/2  +1-1/4) = (4/5√2)*(7/4) =7/5√2  =7√2 / 10 .    || 0,7√2 ||