Ответ(ы) на вопрос:
Гость1)
sinα= - 3/5 ; π <α<3π/2 || 2π < 2α < 3π ||
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sin2α -?
sin2α =2sinα*cosα
cosα= - √(1-(-3/5)²) = -4/5.
sin2α =2sinα*cosα =2*(-3/5)*(-4/5) =24/25 .
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2)
cosα=5/13 ; 3π/2 < α <2π
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cos2α -?
cos2α = 2cos²α -1 = 2*(5/13)² =2*25/169 = 50/169.
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3)
tqα/2 =2
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cosα -?
tqα -?
sinα -?
cosα =(1-tq²α/2) / (1+tq² α/2) = =(1-2²) / (1+2²) = - 3/5.
tqα =2tqα/2 / (1- tq²α/2) = 2*2/(1-2²) = -4/3;
sinα= 2tqα/2 / (1+ tq²α/2) = 2*2/(1+2²) =4/5.
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4)
cos7x*cos5x =(cos(7x-5x)+ cos(7x+5x))/2 =(cos2x+ cos12x)/2.
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5)
sin11x*sinx =(cos(11x -x) -cos((11x+x) )/2 =(cos10x -cos12x)/2 .
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6)
sin5x * cos2x =( sin(5x-2x) +sin(5x+2x) )/2 =(sin3x+sin7x) /2.
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