sqrt(2)/2*sin2x-sqrt(2)/2*cos2x=sqrt(2)*sqrt(2)/2*sin2x
Ответ :
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F'(x) = (cos5x)'(cos3x)+(cos3x)'(cos5x) + (sin5x)'(sin3x)+(sin3x)'(sin5x)-x' =
-5sin5xcos3x-3sin3xcos5x +5cos5xsin3x+3cos3xsin5x-1 = 2sin3xcos5x -2sin5xcos3x - 1=
-2(sin5xcos3x- sin3xcos5x) -1 =
-2sin2x-1