Odpowiedzi

2010-02-20T23:17:44+01:00
Zad. 6
Obliczam c
c² = 2² ++ 4²
c² = 4 + 16
c² = 20
c = √20
c = √4*√5
c = 2√5

a) 1 + sinx *cosx =
=1 + 2*2:2√5 *4: 2√5 =
= 1 + 16: 4*5 =
= 1 + 4/5=
= 5/5 + 4/5 =
= 9/5

b) (tg x *cosx + ctgx*sinx)²=
= (2/4*4/2√5 + 4/2√5 *2/2√5 )² =
= (1/√5 + 2/5)² =
= (5 + 2√5):(5√5)²=
= (25 + 2*5*2√5 +4*5): (25*5)=
= (45 + 20√5) : 75 =
= 5(9 + 4√5): 75=
= (9 + 4√5): 15

Zad. 7
Obliczam x
x² + (√3)² = (√7)²
x² = 7 -3
x² = 4
x = √4
x = 2
a ) 4*sin²x *cos²x + tg²x * ctg²x=
= 4*(2/√7)²*(√3/√7)² + (2/√7)² *(√3/√7)²=
= 4*4/7 *3/7 + 4/7*3/7=
= 48/49 + 12/49=
= 60/49

b) (sinx + tgx)² =
= (2/√7)² + (2/√3)²=
=4/7 + 4/3 =
= 12/21 + 28/21=
= 40/21

Zad.8

Obliczam x
x² + 1² = (√3)²
x² = 3 -1
x² = 2
x = √2

a) (cosx + sinx): (sinx - cosx) =
= (1/√3 + √2/√3 ) : ( √2/√3 -1/√3)=
= [(1 + √2): √3] *[ √3 : (√2 -1)] =
= (1 + √2) : ( √2 -1)=
= [(1 + √2) : ( √2 -1)]*[(√2 +1) : (√2 +1)] usuwam niewyierność mianownika
= [(1 + √2)(√2 +1)] : [ (√2)² -1²]=
= [(1 + √2)(√2 +1)] : (2 -1)=
= (√2 +1)² : 1
= (√2 +1)²

b) (tgx + 2ctgx) : (sinx *cosx)=
= (√2/1 + 2*1/√2) : (√2/√3 *1/√3) =
= [(√2)² +2):√2] : [ √2/3]=
= [(2 + 2):√2]*(3/√2]=
= (4 : √2)*(3/√2)=
= 12 : 2= 6

Wszystkich przykładów z zad 5 nie zrobię. zbyt dużo pisania a już północ, moz jutro !

zad5
h : 10 = sin 60°
h =10*sin 60°
h = 10*(1/2)*√3
h = 5√3

b) h : 10 = sin 62°
h = 10*sin 62°
h = 10*0,8829
h = 8,829