task/30201919 Доказать тождество sinα+sin(α+14π/3)+sin(α -8π/3) =0
решение
sinα+sin(α+14π/3)+sin(α -8π/3) =sinα+2sin(α+π)*cos(11π/3) =
sinα+2sin(π+α)*cos(4π -π/3) =sinα -2sinα*cos(π/3) =
sinα -2sinα*(1/2) = sinα -sinα = 0 ч.т.д.
* * * * * * * * * * P.S. * * * * * * * * * * *
sinα +sinβ =2sin[(α +β)/2]*cos [(α -β)/2] сумма → произведения
[(α+14π/3)+(α -8π/3] /2 =[(2α +6π/3)] /2 =(2α +2π) /2 =2(α +π) /2 =π+α
[(α+14π/3)- (α -8π/3] /2 =[14π/3+8π/3] /2 =(22π/3) /2 =11π/3 =4π-π/3
формула приведение:
sin(π+α) = - sinα ; cos(2πk ±α) =cosα
cos(4π -π/3)= cos(2π*2 -π/3)= cos(π/3) = 1/2 .
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task/30201919 Доказать тождество sinα+sin(α+14π/3)+sin(α -8π/3) =0
решение
sinα+sin(α+14π/3)+sin(α -8π/3) =sinα+2sin(α+π)*cos(11π/3) =
sinα+2sin(π+α)*cos(4π -π/3) =sinα -2sinα*cos(π/3) =
sinα -2sinα*(1/2) = sinα -sinα = 0 ч.т.д.
* * * * * * * * * * P.S. * * * * * * * * * * *
sinα +sinβ =2sin[(α +β)/2]*cos [(α -β)/2] сумма → произведения
[(α+14π/3)+(α -8π/3] /2 =[(2α +6π/3)] /2 =(2α +2π) /2 =2(α +π) /2 =π+α
[(α+14π/3)- (α -8π/3] /2 =[14π/3+8π/3] /2 =(22π/3) /2 =11π/3 =4π-π/3
формула приведение:
sin(π+α) = - sinα ; cos(2πk ±α) =cosα
cos(4π -π/3)= cos(2π*2 -π/3)= cos(π/3) = 1/2 .