Соотношение между функциями

Date: 2015-10-07; view: 523.

Формулы преобр. произв. в сумму

Ф-лы преобразования суммы в произв.

Ф-лы половинного аргумента.

sin² a /2 = (1 - cos a )/2

cos² a /2 = (1 + cosa )/2

tg a /2 = sina /(1 + cosa ) = (1-cos a )/sin a

a ¹ p + 2p n, n Î Z

sin x + sin y = 2 sin ((x+y)/2) cos ((x-y)/2)

sin x - sin y = 2 cos ((x+y)/2) sin ((x-y)/2)

cos x + cos y = 2cos (x+y)/2 cos (x-y)/2

cos x - cos y = -2sin (x+y)/2 sin (x-y)/2

sin x sin y = ½ (cos (x-y) - cos (x+y))

cos x cos y = ½ (cos (x-y)+ cos (x+y))

sin x cos y = ½ (sin (x-y)+ sin (x+y))

sin x = (2 tg x/2)/(1+tg²x/2)

cos x = (1-tg² 2/x)/ (1+ tg² x/2)

sin2x = (2tgx)/(1+tg²x)

sin² a = 1/(1+ctg² a ) = tg² a /(1+tg² a )

cos² a = 1/(1+tg² a ) = ctg² a / (1+ctg² a )

ctg2a = (ctg² a -1)/ 2ctga

sin3a = 3sina -4sin³ a = 3cos² a sina -sin³ a

cos3a = 4cos³ a -3 cosa= cos³ a -3cosa sin² a

tg3a = (3tga -tg³ a )/(1-3tg² a )

ctg3a = (ctg³ a -3ctga )/(3ctg² a -1)

sin a /2 = ± Ö ((1-cosa )/2)

cos a /2 = ± Ö ((1+cosa )/2)

tga /2 = ± Ö ((1-cosa )/(1+cosa ))=

sina /(1+cosa )=(1-cosa )/sina

ctga /2 = ± Ö ((1+cosa )/(1-cosa ))=

sina /(1-cosa )= (1+cosa )/sina

sin(arcsin a ) = a

cos( arccos a ) = a

tg ( arctg a ) = a

ctg ( arcctg a ) = a

arcsin (sina ) = a ; a Î [-p /2 ; p /2]

arccos(cos a ) = a ; a Î [0 ; p ]

arctg (tg a ) = a ; a Î [-p /2 ; p /2]

arcctg (ctg a ) = a ; a Î [ 0 ; p ]

arcsin(sina )=

1)a - 2p k; a Î [-p /2 +2p k;p /2+2p k]

2) (2k+1)p - a ; a Î [p /2+2p k;3p /2+2p k]

arccos (cosa ) =

1) a -2p k ; a Î [2p k;(2k+1)p ]

2) 2p k-a ; a Î [(2k-1)p ; 2p k]

arctg(tga )= a -p k

a Î (-p /2 +p k;p /2+p k)

arcctg(ctga ) = a -p k

a Î (p k; (k+1)p )

arcsina = -arcsin (-a )= p /2-arccosa =

= arctg a /Ö (1-a ² )

arccosa = p -arccos(-a )=p /2-arcsin a =

= arc ctga /Ö (1-a ² )

arctga =-arctg(-a ) = p /2 -arcctga =

= arcsin a /Ö (1+a ² )

arc ctg a = p -arc cctg(-a ) =

= arc cos a /Ö (1-a ² )

arctg a = arc ctg1/a =

= arcsin a /Ö (1+a ² )= arccos1/Ö (1+a ² )

arcsin a + arccos = p /2

arcctg a + arctga = p /2