Дано: tqα=1/2.
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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 ||