wyznacz.. olk: Wiadomo, ze cosx=12/13 i x nalezy (3π/2, 2π) wyznacz sin4x

M:

K700: sin(x+y)=sinx*cosx+cosx*siny sin(4x)=sin(2x+2x)= sin(2x)*cos(2x)+cos(2x)*sin(2x)= sin(2x)=2sin(x)*cos(x) cos(2x)= 2cos²x−1 sin(4x)= 2sinx*cos(x)(2cos²x−1)+2(2cos²x−1)*sin(x)cos(x)= =4sin(x)cos³(x)−2sin(x)cos(x)+4cos³(x)sin(x)−2sin(x)cos(x)= 4cos³(x)sin(x)+4cos³(x)sin(x)−4sin(x)cos(x)=8cos³(x)sin(x)−4sin(x)cos(x)= 4sin(x)cos(x)[2cos²(x)−1]

	12		144		25		5
cos(x)=		sin(x)=−√1−		= −√		=−
	13		169		169		13

	5		12		12		240		288
4*(−		)*		[2*(		)2−1]=−		[		−1]=
	13		13		13		169		169

	240		119		28560
=−		*		=−
	169		169		28561