Question

TgA=1/2. Найдите sin(2A+пи/4)

Answer 1

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 ||

Answer 2

Для начала найдем значение sin(2A):

tg(A) = 1/2 Применим теорему Пифагора: tg^2(A) + 1 = sec^2(A) 1/4 + 1 = sec^2(A) 5/4 = sec^2(A) sec(A) = √(5/4) sec(A) = √5 / 2

Используя определение тангенса и секанса: sin(A) = 1 / sec(A) sin(A) = 2 / √5

Теперь найдем sin(2A): sin(2A) = 2 * sin(A) * cos(A) sin(2A) = 2 * (2 / √5) * (√(1 - sin^2(A))) sin(2A) = 4 / √5 * √(1 - (4/5)) sin(2A) = 4 / √5 * √(1/5) sin(2A) = 4 / √25 sin(2A) = 4 / 5

Теперь найдем sin(2A + π/4): sin(2A + π/4) = sin(2A) * cos(π/4) + cos(2A) * sin(π/4) sin(2A + π/4) = (4/5) * (√2/2) + (4/5) * (√2/2) sin(2A + π/4) = 4√2 / 10 + 4√2 / 10 sin(2A + π/4) = 8√2 / 10 sin(2A + π/4) = 4√2 / 5

Итак, sin(2A + π/4) = 4√2 / 5.