Odpowiedzi

2009-11-14T20:55:03+01:00
A) cos(π/2 +x) = -sin x
cos (2π -x) = =-cos x
ctg(2π +x ) =ctg x
ctg(3*π/2 -x) =ctg(π/2 - x) = 1/ctg x
Zatem mamy
= -sin x*(-cos x)(ctg x + 1/ctg x) =sin X*cosX*(cosx/sinx +sinx/cosx)=
= (cos x)^2 + ( (sin x)^2 = 1

b)
(sin x)/(1 +cos x) + (sin x)/(1- cos x) =
= [sin x*(1-cos x) +sin x*(1+cos x)/(1 +cos x)*(1-cos x) =
= [sin x -sin xcos x + sinx +sinxcosx]/[(1 -(cos x)^2 ] =
= [2*sin x]/(sin x)^2 = 2/ sin x
c)
Obliczam pod pierwiastkiem
(sin x)^2 *(1+ctg x) + (cosx)^2*(1 +tg x) =
= (sinx)^2*+(sinx)^2*(cos x/sin x) +(cosx)^2+(cosx)^2 *(sin x/cos x =1 + sin x*cos x + sin x*cos x = 1 + 2*sin x*cos x =
= 1 + sin 2x
Odp. √(1 +sin 2x)
Pierwiastek kwadratowy z (1 +sin 2x)
1 5 1