Question

Докажите тождество
cos(3/2π+4α)+sin(3π-8α)-sin(4π-12α)=4cos2α cos4α sin 6α

Answer 1

cos(3/2π+4α)+sin(3π-8α)-sin(4π-12α)=4cos2α cos4α sin 6α
Упростим левую часть
cos(3/2π + 4α) + sin(3π - 8α) - sin(4π - 12α) = sin4α + sin8α + sin12α = 
= (sin4α + sin8α) + sin12α = 2sin(4α + 8α)/2*cos(4α - 8α)/2 + sin12α =
= 2sin6α*cos2α + sin12α =  2sin6α*cos2α + sin[2*(6α)] = 
2sin6α*cos2α + 2sin6αcos6α  = 2sin6α(cos2α + cos6α) = 
=  [2sin6α]*[2cos(2α + 6α)/2*cos(2α - 6α)/2] = 4cos2αcos4αsin6α
4cos2αcos4αsin6α = 4cos2αcos4αsin6α
Тождество доказано