541 … dokaz 541 dokažite ako je p prost broj razli čit od ... · 3 dokaz 543 dokažite ako...

1

541 …

Dokaz 541

Dokažite ako je p prost broj različit od 2 i 3 tada je p + 1 ili p - 1 djeljivo sa 6.

Teorija

Za prirodni broj a kažemo da je djeljiv s prirodnim brojem b ako postoji prirodan broj q tako da vrijedi .a b q= ⋅

Broj q zovemo količnikom brojeva a i b i pišemo

ili : .a

q a b qb

= =

Prosti brojevi (prim – brojevi) su prirodni brojevi djeljivi bez ostatka samo s brojem 1 i sami sa sobom, a veći od broja 1. Prirodni brojevi koji su veći od broja 1, a nisu prosti brojevi nazivaju se složenim brojevima. Složen broj je prirodan broj veći od jedan koji je djeljiv brojem 1, samim sobom i barem još jednim brojem. Svaki se složeni broj može rastaviti na proste faktore. Broj 1 nije ni prost, ni složen broj. Prostih brojeva ima beskonačno mnogo. Prosti brojevi su: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, …

��

Svaki se broj koji nije djeljiv ni sa 2 ni sa 3 pa i prost broj može napisati u jednom obliku 6 ∙ n – 1 ili 6 ∙ n + 1. Ako je p = 6 ∙ n – 1 tada je

1 6 1 1 6 1 1 6 .p n n n+ = ⋅ − + = =+ ⋅−⋅

Ako je p = 6 ∙ n + 1 tada je 1 6 1 1 6 1 1 6 .p n n n− = ⋅ + − = =− ⋅+⋅ ■

2

Dokaz 542

Dokažite ako je neki broj zbroj kvadrata dvaju cijelih brojeva njegov je kvadrat također zbroj

kvadrata dvaju cijelih brojeva.

Teorija

Kvadrat zbroja:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b+ = + ⋅ ⋅ + + ⋅ ⋅ + = +

Kvadrat razlike:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b− = − ⋅ ⋅ + − ⋅ ⋅ + = −

Množenje potencija jednakih eksponenata:

( ) ( ), .n nn n n n

a b a b a b a b⋅ = ⋅ ⋅ = ⋅

Potenciranje potencija:

( ) ( ) ( ) ( ), .m n m nn m n m n m n m

a a a a a a⋅ ⋅

= = = =

��

Neka je n = a2 + b2. Tada slijedi:

( ) ( ) ( )2 2 22 2 2 2 2 2 2 2 2 2 2 22/ 2n a b n a b n a b n a a b b= + = + = + = + ⋅ ⋅ +

2 4 2 2 4 2 4 2 2 4 2 22 2 4n a a b b n a a b b a b = + ⋅ ⋅ + = − ⋅ ⋅ + + ⋅ ⋅

( ) ( )2 22 2 2 2 .n a b a b = − + ⋅ ⋅ ■

3

Dokaz 543

Dokažite ako brojevi a1, a2, a3, … čine geometrijski niz, onda je i ovaj niz geometrijski:

a1 ∙ b, a2 ∙ b, a3 ∙ b, …

Teorija

Niz (an) je geometrijski niz ako je svaki član niza, počevši od drugog, jednak prethodnom članu pomnoženom s konstantom q ≠ 0, tj.

1 .a a qnn = ⋅+

Broj q naziva se količnik geometrijskog niza. Niz je geometrijski ako je omjer svakog člana i člana ispred njega stalan:

.1

an qa

n

=

−

Skratiti razlomak znači brojnik i nazivnik tog razlomka podijeliti istim brojem različitim od nule i jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

��

Ako brojevi a1, a2, a3, … čine geometrijski niz, onda je kvocijent između člana i člana pred njim stalan. Vrijedi:

3 12 4 ... ... .1 2 3

a aa an q

a a a an

+= = = = = =

Treba pokazati da isto vrijedi i za navedeni niz.

♥ 2 2 2 .1 1 1

b

b

a b a aq

a b a a

⋅ ⋅= = =

⋅ ⋅

♥ 3 3 3 .2 2 2

b

b

a b a aq

a b a a

⋅ ⋅= = =

⋅ ⋅

♥ 4 4 4 .3 3 3

b

b

a b a aq

a b a a

⋅ ⋅= = =

⋅ ⋅

…

♥ 1 1 1 .a b a a

n n n qa b an

b

an nb

⋅ ⋅+ + += = =

⋅ ⋅ itd. ■

4

Dokaz 544


, , , ...1 2 3

b b b

a a a

Teorija


1 .a a qnn = ⋅+


.1

an qa

n

=

−


, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

Dvojni razlomak:

.

a

a db

c b c

d

⋅=

⋅

��


3 12 4 ... ... .1 2 3

a aa an q

a a a an

+= = = = = =


♥ 1 12 2 1 .22

1 1 1

b

a a a

b

b aa q

a a a

b= = = =

♥ 1 13 3 2 .33

2 2 2

b

a a a

b

b aa q

a a a

b= = = =

♥ 1 134 4 .44

3 3 3

b

aa a

b aa q

a a a

b

b= = = =

…

5

♥ 1 11 1 .

11

b

a a an n nb aa qnn

a an

b

b

n an

+ += = = =++

itd. ■

6

Dokaz 545


, , , ...1 2 3n n n

a a a

Teorija


1 .a a qnn = ⋅+


.1

an qa

n

=

−

Dijeljenje potencija jednakih eksponenata:

( ) ( ), , , .: : : :n nn n

a a a an nn n n na b a b a b a bn n

b bb b= = = =

��


3 12 4 ... ... .1 2 3

a aa an q

a a a an

+= = = = = =


♥ 2 2 .11

nna a n

qn aa

= =

♥ 3 3 .22

nna a n

qn aa

= =

♥ 4 4 .33

nna a n

qn aa

= =

…

♥ 1 1 .nn

a a nn n qn aa nn

+ += =

itd. ■

7

Dokaz 546


a1 + a2, a2 + a3, a3 + a4, …

Teorija


1 .a a qnn = ⋅+


.1

an qa

n

=

−

Zakon distribucije množenja prema zbrajanju.

( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ + Skratiti razlomak znači brojnik i nazivnik tog razlomka podijeliti istim brojem različitim od nule i jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

��


3 12 4 ... ... .1 2 3

a aa an q

a a a an

+= = = = = =


♥ ( )

( )

( )

( )

12 3 2 2 2 2 2 .11 2 1 1 1

1

1 11

a a a a q a q a aq

a a a a q a

q

qq a a

+ + ⋅ ⋅ + ⋅= = = = =

+ + ⋅ ⋅

+

+ +⋅

♥ ( )

( )

( )

( )

13 4 3 3 3 3 3 .12 3 2 2 2

1

2 12

a a a a q a q a aq

a a a a q a

q

qq a a

+ + ⋅ ⋅ + ⋅= = = = =

+ + ⋅ ⋅

+

+ +⋅

♥ ( )

( )

( )

( )

154 4 4 4 4 4 .13 4 3 3 3

1

3 13

a a a a q a q a aq

a a a a q a

q

qq a a

+ + ⋅ ⋅ + ⋅= = = = =

+ + ⋅ ⋅

+

+ +⋅

…

♥ ( )

( )

( )

( )

11 2 1 1 1 1 1 .11

1

1

a a a a q a q a an n n n n n n qa a a a q a q a an n n n q n

q

nn

+ + ⋅ ⋅ + ⋅+ + + + + + += = = = =

+ + ⋅ ⋅ ++

+

+⋅ itd. ■

8

Dokaz 547


a1 ∙ a2, a2 ∙ a3, a3 ∙ a4, …

Teorija


1 .a a qnn = ⋅+


.1

an qa

n

=

−


( ) ( ), , , .: : : :n nn n


b bb b= = = =


, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

Svojstvo potencije: 1 1

, .a a a a= =

Množenje potencija jednakih baza:

, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅

��


3 12 4 ... ... .1 2 3

a aa an q

a a a an

+= = = = = =


♥

22 2 222 3 2 2 2 2 2 2 .

2 2 21 2 1 1 11 1 1

a a a a q a q a a aq

a a a a q aa q a a

q

q

⋅ ⋅ ⋅ ⋅ ⋅ = = = = = = ⋅ ⋅ ⋅ ⋅ ⋅

♥

22 2 223 4 3 3 3 3 3 3 .

2 2 22 3 2 2 22 2 2


a a a a q aa q a a

q

q

⋅ ⋅ ⋅ ⋅ ⋅ = = = = = = ⋅ ⋅ ⋅ ⋅ ⋅

♥

22 2 2254 4 4 4 4 4 4 .

2 2 23 4 3 3 33 3 3


a a a a q aa q a a

q

q

⋅ ⋅ ⋅ ⋅ ⋅ = = = = = = ⋅ ⋅ ⋅ ⋅ ⋅

…

♥ 22 2 2

21 2 1 1 1 1 1 1 .2 2 2

1

a a a a q a q a a an n n n n n n n qa a a a q an n n na q a an n n n

q

q

⋅ ⋅ ⋅ ⋅ ⋅ + + + + + + + += = = = = = ⋅ ⋅ ⋅ ⋅ ⋅ +

itd. ■

9

Dokaz 548

Dokažite ako brojevi a, b, c čine aritmetički niz jednadžba a ∙ x2 – 2 ∙ b ∙ x + c = 0 ima korijen 1.

Teorija

Niz (slijed) je aritmetički ako je razlika svakog člana niza (osim prvog) i člana ispred njega stalna i iznosi d.

5 52 1 3 2 4 3 4 1, , , , , ... , . .6 .a a d a a d a a d a a d a a da d an n−− = =− = − = − = = −−

, .21a a d nn n− = ≥−

Broj d naziva se razlika (diferencija) aritmetičkog niza. Aritmetički niz je jednoznačno određen ako znamo prvi član a1 i razliku d. Zakon distribucije množenja prema zbrajanju.

( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ + Svojstvo potencije:

1 1, .a a a a= =

Dijeljenje potencija jednakih baza:

, .: :n m n m n m n m

a a a a a a− −

= =

Da bi umnožak bio jednak nuli, dovoljno je da jedan faktor bude jednak nuli. 0 0 ili 0 il .i 0a b a b a b⋅ = ⇔ = = = =

��

1.inačica Ako brojevi a, b, c čine aritmetički niz, onda je razlika između člana i člana pred njim stalna. Vrijedi:

2 2 2 0.b a c b b b a c b a c a c b a b c− = − + = + ⋅ = + + = ⋅ − ⋅ + =

Usporedbom 2 2 0

2 0

a x b x c

a b c

⋅ − ⋅ ⋅ + =

− ⋅ + =

vidi se da je x = 1 korijen zadane jednadžbe. ■ 2.inačica Ako brojevi a, b, c čine aritmetički niz, onda je razlika između člana i člana pred njim stalna. Vrijedi:

2 .b a c b b b a c b a c− = − + = + ⋅ = +

Dokažimo da je x = 1 korijen zadane jednadžbe.

[ ] ( )2 20 022 b aa x b x c a x a c x cc⋅ − ⋅ ⋅ + = ⋅ − + ⋅+ += =⋅

( ) ( )2 20 0a x a x c x c a x a x c x c ⋅ − ⋅ − ⋅ + = ⋅ − ⋅ + − ⋅ + =

( ) ( ) ( ) ( )1 0

1 1 0 1 00

xa x x c x x a x c

a x c

− = ⋅ ⋅ − − ⋅ − = − ⋅ ⋅ − =

⋅ − =

/:

111 1.

2

xx x

ca x c a x c x

aa

= = =

⋅ = ⋅ = =

■

10

Dokaz 549

Dokažite ako brojevi a, b, c čine geometrijski niz, onda jednadžba a ∙ x2 + 2 ∙ b ∙ x + c = 0 ima

dvostruki korijen.

Teorija


1 .a a qnn = ⋅+


.1

an qa

n

=

−


, .a a a a= =


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅


( ) ( ), .n nn n n n

a b a b a b a b⋅ = ⋅ ⋅ = ⋅

Diskriminanta kvadratne jednadžbe 2

0a x b x c⋅ + ⋅ + = je broj

2 4 .D b a c= − ⋅ ⋅ ♥ Ako je D > 0, jednadžba ima dva realna rješenja. ♥ Ako je D = 0, jednadžba ima jedno dvostruko realno rješenje. ♥ Ako je D < 0, jednadžba ima kompleksno – konjugirana rješenja. Kvadrat zbroja:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b+ = + ⋅ ⋅ + + ⋅ ⋅ + = +

Drugi korijen pozitivnog broja a je pozitivni broj koji pomnožen sam sa sobom daje broj a. Vrijedi:

( ) ( )2 2

, .a a a a= =

Množenje drugih korijena:

, , 0, .a b a b a b a b a b⋅ = ⋅ ⋅ = ⋅ ≥

Kvadrat realnog broja je nenegativan broj. 2 2

0 , , 0 0.a a R a a≥ ∈ = =

Dijeljenje drugih korijena:

, ., , 0a aa a

a bb bb b

= = ≥

��

1.inačica Ako brojevi a, b, c čine geometrijski niz, onda je kvocijent između člana i člana pred njim stalan. Vrijedi:

/ 2 .b c b c

b a cb a b

aa

b⋅ ⋅= = = ⋅

11

Pokažimo da je diskriminanta zadane jednadžbe jednaka nuli. 22 2 02 0

, 2

2 4,

a x b x ca x b x c

a a b b c cD b a c

⋅ + ⋅ ⋅ + = ⋅ + ⋅ ⋅ + = = = ⋅= − ⋅ ⋅

=

( ) 24 22

2 4 4D b a c D a b a cb c = ⋅ − ⋅ ⋅ = ⋅ − ⋅ ⋅

=

⋅

4 4 0.4 4D a c a a Dac cD c = ⋅ ⋅ − ⋅ ⋅ = ⋅ ⋅ − =⋅ ⋅ ■ 2.inačica Ako brojevi a, b, c čine geometrijski niz, onda je kvocijent između člana i člana pred njim stalan. Vrijedi:

/ ./2 2b c b c

b a c b a c b a c b a ca b a

ab

b= = = ⋅ = ⋅ = ⋅ = ⋅⋅ ⋅

Dalje slijedi:

( ) ( )2 22 22 0 2 0a x b x c a xb ca ac x c ⋅ + ⋅ ⋅ + = ⋅ + ⋅ ⋅ ⋅= +⋅ =

( ) /:2

0 0a x c a x c a x c a x c a ⋅ + = ⋅ + = ⋅ = − ⋅ = −

.1,2c c

x xaa

= − = − ■

12

Dokaz 550

Dokažite da je funkcija inverzna linearnoj funkciji i sama linearna.

Teorija

Kompozicija funkcija:

( )( ) ( )( ) ( ) ( ) ( )( ), .f g x f g x g f x g f x= =� � Funkcija :g K D→ inverzna je funkciji :f D K→ ako vrijedi:

( )( ) za sveg f x x x D= ∈

( )( ) za sve .f g y y y K= ∈ Inverzna funkcija g piše se f -1.

( )( ) ( )( ), .1 1f f x x f f x x− −= =

f -1f

KD K D

1

2

3

3

5

77

5

3

3

2

1

Linearna funkcija je realna funkcija zadana jednadžbom f(x) = a ∙ x + b, a ≠ 0.

��

1.inačica 1 1

/l

y k x l k x l y k x y l k x y l x yk kk

= ⋅ + ⋅ + = ⋅ = − ⋅ = − = ⋅ −⋅

[ ]

zamje

11.

na

ly x yx y a

k

l

b

b

ak

k

xk

↔ =

= −

= ⋅ − = ⋅ +

■

2.inačica

[ ]1

/y k x l x k y l k y l x k y xx y lk

l k y x= ⋅ + = ⋅ + ⋅ + = ⋅ = − ⋅ = − ⋅↔

zamje

.11

na

ak

lb

k

ly x y a x b

k k

= ⋅ − = ⋅ +

=

= −

■

3.inačica

( )( ) ( )( ) ( )1 1f f x x f f x x f x k x l= ⋅ +− −= = � ( ) ( ) ( ) ( )

1 11 1 1/1l

k f x l x k f x x l k f xk

x l f x xk k

− − − − ⋅ + = ⋅ = − ⋅ = − = ⋅ −⋅

13

( )

zamjena

1 1 .ak

l

a b

bk

f x x

− = ⋅ +

=

= −

■

14

Dokaz 551

Why sin x can't be greater than 2 cos x+ ? (Mala Sirena, studies mathematics)

Teorija

Množenje nejednakosti pozitivnim brojem: 0 .,a b c a c b c≤ > ⋅ ≤ ⋅

Temeljni trigonometrijski identitet 2 2 2 2cos sin 1 1 co i, .s s nx x x x+ = = +

Dijeljenje drugih korijena:

, .a aa a

b bb b= =

Za realni broj x njegova je apsolutna vrijednost (modul) broj │x│ koji određujemo na ovaj način:

, .

, 0

0

x xx

x x

≥=

− 0 vrijedi: 2 2

.2 2

a b a b+ +≤

Broj 2

a b+ zove se aritmetička sredina, a broj

2 2

2

a b+ kvadratna sredina brojeva a i b.

��

You can use the AM – QM inequality:

2 2sin cos sin cos sin cossin cos 1 1

2 2 2 2 2 2

x x x x x xx x+ + ++≤ ≤ ≤

sin cos sin cos/

2 2sin co

2 22 s 2

2 2

x x x xx x⋅

+ + ≤ ≤ + ≤

sin 2 cos sin sin 2 cos 2 cosx x x x x x ≤ − ≤ ≤ − ≤ +

because, for

1 32 , 2

2 2x k kπ π

∈ + ⋅ ⋅ + ⋅ ⋅

2 cos 2 cosx x− ≤ +

and, for

1 12 , 2

2 2x k kπ π

∈ − + ⋅ ⋅ + ⋅ ⋅

2 cos 2 cos 2 cos .x x x− = − ≤ + ■

15

Dokaz 552

Dokažite identitet ( )2

sin 2 .xtg x ctg x

⋅ =+

Teorija

Formula za sinus dvostrukog kuta

( ) ( )sin 2 2 sin cos 2 sin cos sin, .2α α α α α α⋅ = ⋅ ⋅ ⋅ ⋅ = ⋅

Definicija tangensa pomoću sinusa i kosinusa: sin sin

cos co, .

s

x xtg x tg x

x x= =

Definicija kotangensa pomoću sinusa i kosinusa:

,cos cos

sins si.

n

x xctg x ctg x

x x= =


Dvojni razlomak:

.

a

a db

c b c

d

⋅=

⋅

Zbrajanje razlomaka jednakih nazivnika:

, .a b a b a b a b

n n n n n n

+ += + + =


, .a a a a= =


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅


, .n n

a an m n ma am m

a a

− −= =


, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

Svaki cijeli broj je racionalan broj:

1, .

1

n nn n= =

Zbrajanje razlomaka:

.a c a d b c

b d b d

⋅ + ⋅+ =

⋅

��

1.inačica

16

( )2 2 2

sin 2 2 sin cos2 2 2 21 sin cos sin cos

sin cos sin cos sin cos sin cos

x x xx x x x

x x x x x x x x

⋅ = ⋅ ⋅ = = = =+

+⋅ ⋅ ⋅ ⋅

2 2 2.

sin cossin coscos sinc

2 2

sin cosos sinx x

x x tg x ctg xx xx xx x

= = =+++

⋅ ⋅

■

2.inačica 2

2 2 2 2 12 2sin cos 1 1sin cos

cos sin sin cos sin cossin cos

x xtg x ctg x x xx x x x x xx x

= = = = =+ ++

⋅ ⋅⋅

( )2 sin cos sin 2 .x x x= ⋅ ⋅ = ⋅ ■

17

Dokaz 553

U trokutu ABC vrijedi 2 .ABC ACB∠ = ⋅∠ Dokažite da je 2 2

.BC AB AC AB⋅ = −

Teorija

Trokut je dio ravnine omeđen s tri dužine. Te dužine zovemo stranice trokuta. Nasuprot jednakih stranica leže jednaki kutovi. Kut koji je susjedni kut nekog unutarnjeg kuta trokuta nazivamo vanjskim kutom tog trokuta. Vanjski kut trokuta jednak je zbroju njemu nasuprotnih unutarnjih kutova. Jednakokračni trokut je onaj kome su dvije stranice jednakih duǉina i te stranice se nazivaju krakovi, dok je treća stranica (osnovica) različite duǉine od duǉine kraka. Sličnost trokuta Kažemo da su dva trokuta slična ako postoji pridruživanje vrhova jednog vrhovima drugog tako da su odgovarajući kutovi jednaki, a odgovarajuće stranice proporcionalne.

,, ,1 1 11 1 1

.a b c

ka b c

α α β β γ γ= = = = = =

Omjer stranica sličnih trokuta k zovemo koeficijent sličnosti.

b1

c1

a1

c

b a

C1

A B

C

A1 B1

Prvi poučak sličnosti (K – K) Dva su trokuta slična ako se podudaraju u dva kuta. Drugi poučak sličnosti (S – K – S) Dva su trokuta slična ako se podudaraju u jednom kutu, a stranice koje određuju taj kut su proporcionalne. Treći poučak sličnosti (S – S – S) Dva su trokuta slična ako su im sve odgovarajuće stranice proporcionalne. Četvrti poučak sličnosti (S – S – K) Dva su trokuta slična ako su im dvije stranice proporcionalne, a podudaraju se u kutu nasuprot većoj stranici. Svojstvo potencije:

1 1, .a a a a= =


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

��

18

2 ⋅⋅⋅⋅ γγγγ

γγγγ

B

C

A

Produžimo stranicu AB preko točke B do točke D tako da je

.BC BD=

γγγγ

γγγγ

2 ⋅⋅⋅⋅ γγγγ

γγγγ

D

B

C

A

Trokut BDC je jednakokračan pa vrijedi

.BDC DCB γ∠ = ∠ =

Sa slike vidi se:

.AD AB BD AD AB BC= + = +

Iz sličnosti trokuta ΔABC i ΔADC (podudaraju se u dva kuta, K – K) slijedi razmjer: 2

/AD AC AD AC

AD AB ACAC AB AC AB

AC AB⋅ ⋅⋅= = =

( )2

AD AB B AB BC AB ACC + ⋅ = = +

2 2 2 2.AB BC AB AC BC AB AC AB + ⋅ = ⋅ = − ■

Zahvaljujem Lauri Župčić, studentici PMF – a, na dokazu!

19

Dokaz 554

Dokažite ako je a b a b→ → → →

+ = − da su vektori okomiti.

Teorija


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ + Kvadrat razlike:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b− = − ⋅ ⋅ + − ⋅ ⋅ + = −

Kvadrat zbroja:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b+ = + ⋅ ⋅ + + ⋅ ⋅ + = +

Drugi korijen pozitivnog broja a je pozitivni broj koji pomnožen sam sa sobom daje broj a. Vrijedi:

( ) ( )2 2

, .a a a a= =

Formula za skalarni produkt vektora = i =a a i a j b b i b jx y x y→ → → → → →

⋅ + ⋅ ⋅ + ⋅ pomoću njihovih

komponenata u pravokutnom koordinatnom sustavu (Kartezijevu koordinatnom sustavu) glasi:

.a b a b a bx x y y→ →

= ⋅ + ⋅�

Dva su vektora ia b→ →

okomita ako im je skalarni produkt jednak nuli:

0.a b a b a bx x y y→ →

⊥ ⋅ + ⋅ =

Duljina vektora definira 2 2se= .a a i a j a a ax y x y→ → → →

⋅ + ⋅ = +

��

Neka su zadani vektori

= i =a a i a j b b i b jx y x y→ → → → → →

⋅ + ⋅ ⋅ + ⋅

a b a b a i a j b i b j a i a j b i b jx y x y x y x y

→ → → → → → → → → → → → + = − ⋅ + ⋅ + ⋅ + ⋅ = ⋅ + ⋅ − ⋅ + ⋅

a i a j b i b j a i a j b i b jx y x y x y x y

→ → → → → → → → ⋅ + ⋅ + ⋅ + ⋅ = ⋅ + ⋅ − ⋅ − ⋅

( ) ( ) ( ) ( )a b i a b j a b i a b jx x y y x x y y→ → → →

+ ⋅ + + ⋅ = − ⋅ + − ⋅

( ) ( ) ( ) ( )2 22 2

a b a b a b a bx x y y x x y y + + + = − + −

( ) ( ) ( ) ( )2 2 2/

2 2a b a b a b a bx x y y x x y y + + + = − + −

( ) ( ) ( ) ( )2 2

2 22 2a b a b a b a bx x y y x x y y

+ + + = − + −

( ) ( ) ( ) ( )2 22 2

a b a b a b a bx x y y x x y y + + + = − + −

20

2 2 2 2 2 2 2 22 2 2 2a a b b a a b b a a b b a a b bx x x x y y y y x x x x y y y y + ⋅ ⋅ + + + ⋅ ⋅ + = − ⋅ ⋅ + + − ⋅ ⋅ +

2 2 2 2 2 2 22 2 2 22a b a b a b a bx x y y xa b a b a b a bx x y y x x y yx y y + ⋅ ⋅ + ⋅ ⋅ = − ⋅+ + + + + ⋅ +⋅ − ⋅

2 2 2 2a b a b a b a bx x y y x x y y ⋅ ⋅ + ⋅ ⋅ = − ⋅ ⋅ − ⋅ ⋅

2 2 2 2 0 4 4 0a b a b a b a b a b a bx x y y x x y y x x y y ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ = ⋅ ⋅ + ⋅ ⋅ =

/4 4 0 0.: 4a b a b a b a bx x y y x x y y ⋅ ⋅ + ⋅ ⋅ = ⋅ + ⋅ = ■

21

Dokaz 555

Dokažite ako je P period od funkcija f i g, onda je P period i od funkcije ,f gα β⋅ + ⋅ gdje su α i β

bilo koji realni brojevi.

Teorija

Ako za funkciju :f D R→ postoji P > 0 takav da je

( ) ( ) za sva ,i, kf x P f x x D+ = ∈ tada funkciju f nazivamo periodična funkcija. Pozitivni brojevi P za koje vrijedi gornja formula nazivaju se periodi funkcije f. Ako postoji najmanji takav pozitivni broj P, tada se on naziva temeljni period.

��

( )( ) ( ) ( )( ) ( )

( ) ( )( ) ( )

f x P f x

g x P g xf g x P f x P g x P f x g xα β α β α β

+ = ⋅ + ⋅ + = ⋅ + + ⋅ + = = ⋅ + ⋅ =

+ =

( )( ),f g xα β= ⋅ + ⋅ tj. P je period od .f gα β⋅ + ⋅ ■

22

Dokaz 556

Dokažite ako je P period od funkcija f i g, onda je P period i od funkcije .f g⋅

Teorija



��

( )( ) ( ) ( )( ) ( )

( ) ( )( ) ( ) ( )( ),

f x P f x

g x P g xf g x P f x P g x P f x g x f g x

+ =

+

⋅ + = + ⋅ + = = ⋅ = ⋅

=

tj. P je period od .f g⋅ ■

23

Dokaz 557

Dokažite ako je P period od funkcija f i g, onda je P period i od funkcije ( ), 0, .f

g x x Rg

≠ ∀ ∈

Teorija



��

( )( )( )

( ) ( )

( ) ( )( )( )

( ),f x P f x

g x P g x

f x P f xf fx P x

g g x P g x g

+ = + + = = = =

+ + =

tj. P je period od .f

g ■

24

Dokaz 558

Dokažite u svakom pravokutnom trokutu vrijedi relacija ( )2

2 .2 2

a btg

a b

β⋅ ⋅

⋅ =−

Teorija

Trokut je dio ravnine omeđen s tri dužine. Te dužine zovemo stranice trokuta. Pravokutni trokuti imaju jedan pravi kut (kut od 90º). Stranice koje zatvaraju pravi kut zovu se katete, a najdulja stranica je hipotenuza pravokutnog trokuta. Sinus šiljastog kuta pravokutnog trokuta jednak je omjeru duljine katete nasuprot tog kuta i duljine hipotenuze. Kosinus šiljastog kuta pravokutnog trokuta jednak je omjeru duljine katete uz taj kut i duljine hipotenuze. Formule za sinus i kosinus dvostrukog kuta

( ) ( )sin 2 2 sin cos 2 sin cos sin, .2α α α α α α⋅ = ⋅ ⋅ ⋅ ⋅ = ⋅

( ) ( )2 2 2 2cos 2 cos sin cos sin, c s 2 .oα α α α α α⋅ = − − = ⋅ Svojstvo potencije:

1 1, .a a a a= =


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅


cos co, .

s

x xtg x tg x

x x= =



, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


( ) ( ), .n nn n n n

a b a b a b a b⋅ = ⋅ ⋅ = ⋅

��

ββββ

αααα

cb

a

Sa slike vidi se:

sin sin sinsin

.cos

c

/

/os cos cos

b b b

b cc c c

a a a a c

c cc

c

cβ β ββ

ββ β β

= = = = ⋅

= ⋅ = =

⋅

⋅=

25

Sada je

( ) ( )

( )

( )

22 2 cos sin2 2 cos sin 2 cos sin2 2 2 2 2 2 22 2 2 2cos sin cos sincos sin

ca b c c c

a b c c cc c

β ββ β β β

β β β ββ β

⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅= = = =

− ⋅ − ⋅ ⋅ −⋅ − ⋅

( )

( )( )( )

( )2 cos sin sin 2

2 .2 2 cos 2cos s

2

2 in

c

ctg

β β ββ

ββ β

⋅ ⋅ ⋅ ⋅= = = ⋅

⋅⋅ − ■

26

Dokaz 559

Dokažite ako u trokutu vrijedi relacija ( )2 2

cos 2 ,2

a b

cβ

−⋅ = a > b, onda je trokut pravokutan.

Teorija

Trokut je dio ravnine omeđen s tri dužine. Te dužine zovemo stranice trokuta. Pravokutni trokuti imaju jedan pravi kut (kut od 90º). Stranice koje zatvaraju pravi kut zovu se katete, a najdulja stranica je hipotenuza pravokutnog trokuta. Pitagorin poučak Trokut ABC je pravokutan ako i samo ako je kvadrat nad hipotenuzom jednak zbroju kvadrata nad katetama. Kosinus šiljastog kuta pravokutnog trokuta jednak je omjeru duljine katete uz taj kut i duljine hipotenuze. Poučak o kosinusu (kosinusov poučak) U trokutu ABC vrijede ove jednakosti:

2 2 22 cos ,a b c b c α= + − ⋅ ⋅ ⋅

2 2 22 cos ,b a c a c β= + − ⋅ ⋅ ⋅

2 2 22 cos .c a b a b γ= + − ⋅ ⋅ ⋅

Formula za kosinus dvostrukog kuta

( ) ( )2 2cos 2 2 cos 1 2 cos 1 cos, .2α α α α⋅ = ⋅ − ⋅ − = ⋅ Svojstvo potencije:

1 1, .a a a a= =


, .: :n m n m n m n m

a a a a a a− −

= =


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ + Da bi umnožak bio jednak nuli, dovoljno je da jedan faktor bude jednak nuli.

0 0 ili 0 il .i 0a b a b a b⋅ = ⇔ = = = =

��

ββββ

αααα

cb

a

( )2 2 2 2 2 2

2 22cos 2 2 cos 1 2 c 1 /os2 2 2

a b a b a b

c c ccβ β β

− − −⋅ = ⋅ − = = ⋅⋅ −

2 2 2 2 2 2 2 22 22 cos 2 cos .c c a b b a c cβ β ⋅ ⋅ − = − = + − ⋅ ⋅

Iz kosinusova poučka dobijemo 2 2 2 2 cos .b a c a c β= + − ⋅ ⋅ ⋅

27

Sada je 2 2 2 2 22 cos 2 2 2 2 222 cos 2 cos2 2 2 2 cos

b a c ca c c a c a c

b a c a c

ββ β

β

= + − ⋅ ⋅ + − ⋅ ⋅ = + − ⋅ ⋅ ⋅

= + − ⋅ ⋅ ⋅ 2 22 22 cos 2 cos2 2 2 2 cos cos2 2a c a cc a c c a cβ β β β − ⋅ ⋅ = − ⋅ ⋅ ⋅ − ⋅ ⋅ = − ⋅ ⋅ ⋅+ +

( )2 22 22 cos 2 cos cos co/: s2c a c c a cβ β β β − ⋅ ⋅ = − ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅−

( )0

2 2cos cos 0 cos cos 0

nema smisla

cos 0

cos 0

c

c a c c c a

c a

β β β β β

β

=

⋅ − ⋅ ⋅ = ⋅ ⋅ ⋅ − = = ⋅ − =

cos 0.

cos 0c a

β

β

=

⋅ − =

Iz cos β = 0 slijedi 2

πβ = pa bi stranica b trebala biti hipotenuza. Zbog uvjeta a > b to nije moguće.

Iz cos 0c aβ⋅ − = slijedi

cos 0 cos cos c1

/ os .a

c a c a cc

ac

β β β β⋅ − = ⋅ = ⋅ = =⋅

Sada je stranica c hipotenuza pa Pitagorin poučak glasi: 2 2 2.c a b= + ■

28

Dokaz 560

Dokažite log log 1b aa b⋅ = , gdje su a i b pozitivni brojevi različiti od 1.

Teorija

Logaritam broja a po bazi b je broj c kojim treba potencirati bazu b da se dobije broj a. Mnemotehničko pravilo za pamćenje osnovne veze eksponencijalne i logaritamske funkcije:

llog ogb

c ca c a b a b

b=

→= =

Svojstva:

log log log lo, .gn na n a n a ab b b b

= ⋅ ⋅ =

log 1 1 lo, .gb bb b

= =

��

Iz jednakosti slijedi: log

.log

mx ma x a m na b

nx n x bb

= = = = =

Jednakost am = bn logaritmiramo najprije po bazi a, onda po bazi b.

log log log log

log loglog lo

/ lo

g

g

/ log

m n m na b a b m a n ba a a a

m n m n m a n ba b a b b bb b

a

b

= = ⋅ = ⋅ ⋅ = ⋅= =

1 log loglo

metoda

zamjeng log

log e1 log

m n b m n ba an b a na bm a n m a n

b b

⋅ = ⋅ = ⋅ ⋅ ⋅ = ⋅ = ⋅ ⋅ =

/ :log log log log 1.nn b a n b aa ab b ⋅ ⋅ = ⋅ = ■

29

Dokaz 561

Dokažite u svakom pravokutnom trokutu zbroj polumjera opisane i upisane kružnice je aritmetička

sredina kateta.

.2

a bR r

++ =

Teorija

Trokut je dio ravnine omeđen s tri dužine. Te dužine zovemo stranice trokuta. Pravokutni trokuti imaju jedan pravi kut (kut od 90º). Stranice koje zatvaraju pravi kut zovu se katete, a najdulja stranica je hipotenuza pravokutnog trokuta. Ploština pravokutnog trokuta izračunava se po formuli

2,

a bP

⋅=

gdje su a i b duljine kateta. Ploština trokuta:

• ,P r s= ⋅ gdje je r polumjer upisane kružnice, s poluopseg 2

.a b c

s+ +

=

• 4

,a b c

PR

⋅ ⋅=

⋅ gdje su a, b, c duljine stranica, R polumjer opisane kružnice.



, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


.a c a d b c

b d b d

⋅ + ⋅+ =

⋅


, .a a a a= =


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅


.:n m n m

a a a−

=

Kvadrat zbroja:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b+ = + ⋅ ⋅ + + ⋅ ⋅ + = +

Aritmetička sredina (prosjek, srednja vrijednost) Aritmetička sredina brojeva a i b je njihov poluzbroj.

.2

Aa b

=+

Aritmetičku sredinu računamo tako da zbrojimo sve brojeve i dobiveni zbroj podijelimo s brojem koliko ima zadanih brojeva.

��

30

4/

14

/

4 .4

a b c a b ca b c R

P

a b c R RP P P P

R RP

P r s r s P r s rs

Ps

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = = = = ⋅ ⋅

⋅ ⋅ = ⋅ ⋅ = ⋅ = =

⋅

⋅

Računamo zbroj R + r.

2

2

24 4

2 2

a b

a b c P a

a bP

a b

b cR

cs

r R ra b a b cP s

⋅ ⋅ ⋅ ⋅ ⋅

+ = + + = + ⋅ + +⋅ ⋅

⋅=

+ +=

( )( )

2

42 2

2

2 2

a bc a b c a bc c a b

R r Ra b

ar R r

a b c a b cb a b c

⋅⋅ + + + ⋅ ⋅⋅ ⋅ ⋅

+ = + + = + + = ⋅ + + + + ⋅ + +⋅

( )2 2 2

22

2c a c b c

c a ba b

R ra b c

⋅ + ⋅ + + ⋅ ⋅ + = ⋅

= ++ +

( ) ( )

2 2 2 22 2

2 2

c a c b a b a b c a c b a a b bR r R r

a b c a b c

⋅ + ⋅ + + + ⋅ ⋅ ⋅ + ⋅ + + ⋅ ⋅ + + = + =

⋅ + + ⋅ + +

( ) ( )( )

( ) ( )( )

( ) ( )( )

2

2 2 2

a b cc a b a b a b c a b a bR r R

ar R r

a b c b ca b c

⋅ + + + + ⋅ + ++ + ⋅ + = + = + =

⋅ + + ⋅ + + + +⋅

+

.2

a bR r

+ + = ■

31

Dokaz 562

Dokažite da iz ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2

2 2 2a b c a b c a b c a b c b c a− + − + − = + − ⋅ + − ⋅ + + + − ⋅ slijedi

.a b c= =

Teorija

Kvadrat razlike:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b− = − ⋅ ⋅ + − ⋅ ⋅ + = −

Kvadrat trinoma:

( ) ,2 2 2 2

2 2 2a b c a b c a b a c b c+ + = + + + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅

( )2

2 .2 2 2

2 2a b c a b a c b c a b c+ + + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ = + +


( ) ( ), .n nn n n n

a b a b a b a b⋅ = ⋅ ⋅ = ⋅

Zbroj kvadrata realnih brojeva jednak je nuli, ako je svaki član jednak nuli. 2 2

0 0.a b a b+ = ⇔ = =

��

Preoblikujemo lijevu stranu jednakosti.

( ) ( ) ( )2 2 2 2 2 2 2 2 22 2 2L a b c a b c L a a b b c c a a b b c c= − + − + − = − ⋅ ⋅ + + − ⋅ ⋅ + + − ⋅ ⋅ +

2 2 22 2 2 2 2 2 .L a b c a b a c b c = ⋅ + ⋅ + ⋅ − ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅

Preoblikujemo desnu stranu jednakosti.

( ) ( ) ( )2 2 2

2 2 2D a b c a b c b c a= + − ⋅ + − ⋅ + + + − ⋅

2 2 2 2 2 24 2 4 4 4 4 2 4D a b c a b a c b c a b c a b a c b c = + + ⋅ + ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅ + + ⋅ + − ⋅ ⋅ + ⋅ ⋅ − ⋅ ⋅ +

2 2 24 2 4 4b c a b c a b a c+ + + ⋅ + ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅

2 2 26 6 6 6 6 6 .D a b c a b a c b c = ⋅ + ⋅ + ⋅ − ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅

Zbog uvjeta L = D slijedi: 2 2 2 2 2 22 2 2 2 2 2 6 6 6 6 6 6a b c a b a c b c a b c a b a c b c⋅ + ⋅ + ⋅ − ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅ = ⋅ + ⋅ + ⋅ − ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅

2 2 22 2 2 2 2 2a b c a b a c b c ⋅ + ⋅ + ⋅ − ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅ −

2 26 6 6 6 6 0b c a b a c b c− ⋅ − ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ =

2 2 24 4 4 4 4 4 0a b c a b a c b c − ⋅ − ⋅ − ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ =

( )2 2 24 4 4 4 /4 4 0 : 2a b c a b a c b c − ⋅ − ⋅ − ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ = −⋅ 2 2 2

2 2 2 2 2 2 0a b c a b a c b c ⋅ + ⋅ + ⋅ − ⋅ ⋅ − ⋅ ⋅ − ⋅ ⋅ =

2 2 2 2 2 22 2 2 0a a b b a a c c b b c c − ⋅ ⋅ + + − ⋅ ⋅ + + − ⋅ ⋅ + =

( ) ( ) ( )2 2 2 2 2 22 2 2 0a a b b a a c c b b c c − ⋅ ⋅ + + − ⋅ ⋅ + + − ⋅ ⋅ + = ( ) ( ) ( )

2 2 20a b a c b c − + − + − =

32

0

0 .

0

a b a b

a c a c a b c

b c b c

− = =

− = = = = − = =

■

33

Dokaz 563

Dokažite da je n2 + 3 ∙ n – 4 paran broj za svaki n prirodan broj.

Teorija

Kvadrat zbroja:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b+ = + ⋅ ⋅ + + ⋅ ⋅ + = +


( ) ( ), .n nn n n n

a b a b a b a b⋅ = ⋅ ⋅ = ⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ + Skup prirodnih brojeva označavamo slovom N, a zapisujemo

{ }1, 2, 3, 4, 5, ... , 1, , 1, . ...N n n n= − + Prirodni brojevi dijele se na parne i neparne brojeve. Da je proizvoljni prirodni broj m neparan znači da se može napisati u obliku

( )2 neki prirodni broj 1 2 1 .,,m m k k N= ⋅ + = ⋅ + ∈ Da je proizvoljni prirodni broj n paran znači da se može napisati u obliku

( )2 neki prirodni , 2 .broj ,n n k k N= ⋅ = ⋅ ∈

��

Neka je n paran broj. 2 , .n k k N= ⋅ ∈

Slijedi:

[ ] ( )22 2

3 4 2 3 2 4 6 42 4n kn n k k k k+ ⋅ − = = ⋅ + ⋅ ⋅ − = ⋅ + ⋅⋅ −= =

( ) 22 3 222 2 3 2 2 , .k k m mm Nk k= ⋅ += ⋅ ⋅ + ⋅ − = = ∈⋅ − ⋅ Neka je n neparan broj.

2 1 , .n k k N= ⋅ + ∈

Slijedi:

[ ] ( ) ( )22 2

3 4 2 1 3 22 1 1 4 4 4 1 6 3 4n n k k k k kn k+ ⋅ − = = ⋅ + + ⋅ ⋅ + − = ⋅ + ⋅ + + ⋅+ +⋅ −= =

( )2 2 2 24 4 6 4 4 6 4 101 3 4 2 2 5k k k k k k k k k k= ⋅ + ⋅ + ⋅ = ⋅ + ⋅ + ⋅ = ⋅ + ⋅ = ⋅ ⋅+ ++ ⋅− = 2

5 22 , .m k k m m N= ⋅ ⋅= = ∈+ ⋅ ■

34

Dokaz 564

Dokažite identitet 3 3cos sin

sin cos 1 0.cos sin

x xx x

x x

++ ⋅ − =

+

Teorija

Kub zbroja:

( ) ( )3 33 2 2 3 3 2 2 3

3 3 3 3, .a b a a b a b b a a b a b b a b+ = + ⋅ ⋅ + ⋅ ⋅ + + ⋅ ⋅ + ⋅ ⋅ + = +


, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅



1, .

1

n nn n= =


.a c a d b c

b d b d

⋅ + ⋅+ =

⋅

Oduzimanje razlomaka:

.a c a d b c

b d b d

⋅ − ⋅− =

⋅


, .a a a a= =


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ + Razlomak je jednak nuli ako je brojnik nula.

0 , 0 0.a

a nn

= ≠ =

Množenje zagrada

( ) ( ) .a b c d a c a d b c b d+ ⋅ + = ⋅ + ⋅ + ⋅ + ⋅

��

1.inačica

( ) ( )2 2cos sin cos cos sin sin3 3cos sinsin cos 1 sin cos 1

cos sin cos sin

x x x x x xx x

x x x xx x x x

+ ⋅ − ⋅ +++ ⋅ − = + ⋅ − =

+ +

( ) ( )2 2cos cos sin scos sincos si

i

n

nsin cos 1

x xx

x x

x x xx x

⋅ − ⋅ += + ⋅

+−

+=

2 2 2 2cos cos sin sin sin cos co1 cos sins sin sin co 1sx x x x xx x xx x x x= − ⋅ + + ⋅ − = + ⋅+⋅ −− =

35

2 2cos sin 1 1 1 0.x x= + − = − = ■ 2.inačica

3 3 3 3cos sin cos sin sin cos 1sin cos 1

cos sin cos sin 1 1

x x x x x xx x

x x x x

+ + ⋅+ ⋅ − = + − =

+ +

( ) ( )3 3cos sin sin cos cos sin cos sincos sin

x x x x x x x x

x x

+ + ⋅ ⋅ + − += =

+

3 3 2 2cos sin sin cos sin cos cos sin

cos sin

x x x x x x x x

x x

+ + ⋅ + ⋅ − −= =

+

( ) ( )3 3 2 2cos sin sin 1 sin 1 cos cos cos sincos sin

x x x x x x x x

x x

+ + ⋅ − + − ⋅ − −= =

+

3 3 3 3cos sin sin sin cos cos cos sin

cos sin

x x x x x x x x

x x

+ + − + − − −= =

+

3 3 3 3cos sin sin sin cos co 00.

cos sin cos sin

s cos sinx x x x x x

x x x

x

x

x= = =

+ +

+ + − + − − − ■

3.inačica

( )3 3 3 3cos sin cos sin 2 2sin cos 1 sin cos cos sin

cos sin cos sin

x x x xx x x x x x

x x x x

+ ++ ⋅ − = + ⋅ − + =

+ +

3 3 2 2cos sin sin cos cos sin

cos sin 1 1

x x x x x x

x x

+ ⋅ += + − =

+

( ) ( ) ( )3 3 2 2cos sin sin cos cos sin cos sin cos sincos sin

x x x x x x x x x x

x x

+ + ⋅ ⋅ + − + ⋅ += =

+

( )3 3 2 2 3 2 2 3cos sin sin cos sin cos cos cos sin sin cos sincos sin

x x x x x x x x x x x x

x x

+ + ⋅ + ⋅ − + ⋅ + ⋅ += =

+

3 3 2 2 3 2 2 3cos sin sin cos sin cos cos cos sin sin cos sin

cos sin


x x

+ + ⋅ + ⋅ − − ⋅ − ⋅ −= =

+

3 3 2 2

co

3 2 2 3cos sin sin cos sin cos cos cos sin sin cos sin

s sin


x x

+ + ⋅ +=

⋅ − ⋅

+

− ⋅ − −=

00.

cos sinx x= =

+ ■

4.inačica 3 3cos sin

sin cos 1cos sin

x xx x

x x

++ ⋅ − =

+

( ) ( )( )

cos sin

c

2 2cos cos sin sin2 2sin cos cos si

osn

sin

x x

x

x x x xx x x

xx

⋅ − ⋅ += + − +

+

+⋅ =

2 2 2 2cos cos sin sin sin cos cos sinx x x x x x x x= − ⋅ + + ⋅ − − =

2 2 2 2cos cos sin sin sin cos cos s 0.inx x x x x x x x− ⋅ + + ⋅ − =−= ■

36

Dokaz 565

Dokažite da su svaka dva jednakokračna trokuta sukladna ako se podudaraju u jednom kraku i

kutu nasuprot osnovici.

Teorija

Trokut je dio ravnine omeđen s tri dužine. Te dužine zovemo stranice trokuta. Trokut koji ima dvije sukladne stranice zove se jednakokračan trokut. Sukladne stranice su kraci, a treća stranica zove se osnovica trokuta. Sukladnost trokuta Kažemo da su dva trokuta sukladna ako postoji pridruživanje vrhova jednog vrhovima drugog tako da su odgovarajući kutovi jednaki, a odgovarajuće stranice jednakih duljina.

, , , ,1 1 1 1 1 1, .a a b b c cα α β β γ γ= = = = = =

Prvi poučak sukladnosti (S – S – S) Dva su trokuta sukladna ako se podudaraju u sve tri stranice. Drugi poučak sukladnosti (S – K – S) Dva su trokuta sukladna ako se podudaraju u dvije stranice i kutu između njih. Treći poučak sukladnosti (K – S – K) Dva su trokuta sukladna ako se podudaraju u jednoj stranici i oba kuta na toj stranici. Četvrti poučak sukladnosti (S – S – K) Dva su trokuta sukladna ako se podudaraju u dvije stranice i kutu nasuprot većoj stranici.

��

αααα

bb

Dva jednakokračna trokuta podudaraju se u jednom kraku. To ima za posljedicu da se podudaraju u oba kraka. Ti su trokuti sukladni po drugom poučku sukladnosti (S – K – S). ■

37

Dokaz 566

Dokažite jednakost ( ) ( ) ( ) ( ) ( ) ( )

2 2 2 2 2 2.

b c c a a ba b b c c a

a b a c b c b a c a c b

⋅ ⋅ ⋅+ + = ⋅ + ⋅ + ⋅

− ⋅ − − ⋅ − − ⋅ −

Teorija



, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


.a c a d b c

b d b d

⋅ + ⋅+ =

⋅


.a c a d b c

b d b d

⋅ − ⋅− =

⋅


, .a a a a= =


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅


.:n m n m

a a a−

=

Razlika kubova:

( ) ( ) ( ) ( )3 3 2 2 2 2, .3 3a b a b a a b b a b a a b b a b− = − ⋅ + ⋅ + − ⋅ + ⋅ + = − Razlika kvadrata:

( ) ( ) ( ) ( ),2 .2 2 2a b a b a b a b a b a b− = − ⋅ + − ⋅ + = −

��

( ) ( ) ( ) ( ) ( ) ( )

2 2 2 2 2 2b c c a a b

a b a c b c b a c a c b

⋅ ⋅ ⋅+ + =

− ⋅ − − ⋅ − − ⋅ −

( ) ( ) ( ) ( )( ) ( )( ) ( )( )

2 2 2 2 2 2b c c a a b

a b a c b c a b a c b c

⋅ ⋅ ⋅= + + =

− ⋅ − − ⋅ − − − − ⋅ − −

( ) ( ) ( ) ( ) ( ) ( )

2 2 2 2 2 2b c c a a b

a b a c b c a b a c b c

⋅ ⋅ ⋅= − + =

− ⋅ − − ⋅ − − ⋅ −

( ) ( ) ( )( ) ( ) ( )

2 2 2 2 2 2b c b c c a a c a b a b

a b a c b c

⋅ ⋅ − − ⋅ ⋅ − + ⋅ ⋅ −= =

− ⋅ − ⋅ −

( ) ( ) ( )

3 2 2 3 2 3 3 2 3 2 2 3b c b c c a c a a b a b

a b a c b c

⋅ − ⋅ − ⋅ + ⋅ + ⋅ − ⋅= =

− ⋅ − ⋅ −

38

( ) ( ) ( )( ) ( ) ( )

3 2 2 3 3 2 3 2 2 3 2 3b c b c a b a c a b a c

a b a c b c

⋅ − ⋅ + ⋅ − ⋅ + − ⋅ + ⋅= =

− ⋅ − ⋅ −

( ) ( ) ( )( ) ( ) ( )

2 2 3 2 2 2 3 3b c b c a b c a b c

a b a c b c

⋅ ⋅ − + ⋅ − − ⋅ −= =

− ⋅ − ⋅ −

( ) ( ) ( ) ( ) ( )( ) ( ) ( )

2 2 3 2 2 2b c b c a b c b c a b c b b c c

a b a c b c

⋅ ⋅ − + ⋅ − ⋅ + − ⋅ − ⋅ + ⋅ += =

− ⋅ − ⋅ −

( ) ( ) ( )( ) ( ) ( )

2 2 3 2 2 2b c b c a b c a b b c c

a b a c b c

− ⋅ ⋅ + ⋅ + − ⋅ + ⋅ + = =

− ⋅ − ⋅ −

( ) ( ) ( )( ) ( ) ( )

2 2 3 2 2 2b c a b c a b b c c

a b

b c

ba c c

⋅ ⋅ + ⋅ + − ⋅ + ⋅ + = =

− ⋅ − ⋅

−

−

( ) ( )( ) ( )

2 2 3 2 2 2b c a b c a b b c c

a b a c

⋅ + ⋅ + − ⋅ + ⋅ += =

− ⋅ −

( ) ( )

2 2 3 3 2 2 2 2 2b c a b a c a b a b c a c

a b a c

⋅ + ⋅ + ⋅ − ⋅ − ⋅ ⋅ − ⋅= =

− ⋅ −

( ) ( ) ( )( ) ( )

2 2 2 2 3 2 2 3 2a c b c a b a b a c a b c

a b a c

− ⋅ + ⋅ + ⋅ − ⋅ + ⋅ − ⋅ ⋅= =

− ⋅ −

( ) ( ) ( )( ) ( )

2 2 2 2 2c a b a b a b a c a b

a b a c

− ⋅ − + ⋅ ⋅ − + ⋅ ⋅ −= =

− ⋅ −

( ) ( ) ( ) ( )( ) ( )

2 2 2c a b a b a b a b a c a b

a b a c

− ⋅ − ⋅ + + ⋅ ⋅ − + ⋅ ⋅ −= =

− ⋅ −

( ) ( )( )( ) ( )

( ) ( )( )( ) ( )

2 2 2 2 2 2a b c a b a b a c c a b a b a c

a b a c a

a b

b ca

− ⋅ − ⋅ + + ⋅ + ⋅ ⋅ − ⋅ + + ⋅ + ⋅= =

−

−=

− ⋅ − ⋅ −

( )2 2 2 2 2 2 2c a b a b a c c a c b a b a ca c a c

− ⋅ + + ⋅ + ⋅ − ⋅ − ⋅ + ⋅ + ⋅= = =

− −

( ) ( ) ( ) ( )2 2 2 2 2 2a c a c a b b c a c a c b a ca c a c

⋅ − ⋅ + ⋅ − ⋅ ⋅ ⋅ − + ⋅ −= = =

− −

( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )a c a c b a c a c b a ca c a c b a c a cc c

c

a a

a

a c

− ⋅ ⋅ + ⋅ + ⋅ ⋅ + ⋅ +⋅ ⋅ − + ⋅ − ⋅ +=

−=

− − −= =

( ) .a c b a c a c b a b c a b b c c a= ⋅ + ⋅ + = ⋅ + ⋅ + ⋅ = ⋅ + ⋅ + ⋅ ■

39

Dokaz 567

Dokažite jednakost 1 2cos .

21 tgα

α=

+

Teorija


cos co, .

s

x xtg x tg x

x x= =


( ) ( ), , , .: : : :n nn n


b bb b= = = =


1, .

1

n nn n= =


.a c a d b c

b d b d

⋅ + ⋅+ =

⋅


Dvojni razlomak:

.

a

a db

c b c

d

⋅=

⋅

Trokut je dio ravnine omeđen s tri dužine. Te dužine zovemo stranice trokuta. Pravokutni trokuti imaju jedan pravi kut (kut od 90º). Stranice koje zatvaraju pravi kut zovu se katete, a najdulja stranica je hipotenuza pravokutnog trokuta. Pitagorin poučak Trokut ABC je pravokutan ako i samo ako je kvadrat nad hipotenuzom jednak zbroju kvadrata nad katetama. Tangens šiljastog kuta pravokutnog trokuta jednak je omjeru duljine katete nasuprot tog kuta i duljine katete uz taj kut. Kosinus šiljastog kuta pravokutnog trokuta jednak je omjeru duljine katete uz taj kut i duljine hipotenuze.

��

1.inačica 1

1 1 1 1 1 21 cos .2 2 2 2 2 1 11 sin 1 sin cos sin

1 2 22 2 2 cos cos1cos cos cos

tgα

α α α α αα αα α α

= = = = = =+ +

+ +

■

2.inačica

40

ββββ

αααα

c

b

a

1 1 1 1 1 1

2 2 2 2 2 2 21 111 2 2

2 2

2 21

2

tg a a b a ca

b b

at

b

a

b

g c bb

b

αα

= = = = = = = = + + + ++

= = +

1 2221 cos .

2

o2

c s2

b

c

b b

cc c

b

αα

= = = = = =

■

41

Dokaz 568

Dokažite jednakost 1 2sin .

21 ctgα

α=

+

Teorija

Definicija kotangensa pomoću sinusa i kosinusa: cos cos

si i.

n s n,

x xctg x ctg x

x x= =


( ) ( ), , , .: : : :n nn n


b bb b= = = =


1, .

1

n nn n= =


.a c a d b c

b d b d

⋅ + ⋅+ =

⋅


Dvojni razlomak:

.

a

a db

c b c

d

⋅=

⋅

Trokut je dio ravnine omeđen s tri dužine. Te dužine zovemo stranice trokuta. Pravokutni trokuti imaju jedan pravi kut (kut od 90º). Stranice koje zatvaraju pravi kut zovu se katete, a najdulja stranica je hipotenuza pravokutnog trokuta. Pitagorin poučak Trokut ABC je pravokutan ako i samo ako je kvadrat nad hipotenuzom jednak zbroju kvadrata nad katetama. Kotangens šiljastog kuta pravokutnog trokuta jednak je omjeru duljine katete uz taj kut i duljine katete nasuprot tog kuta. Sinus šiljastog kuta pravokutnog trokuta jednak je omjeru duljine katete nasuprot tog kuta i duljine hipotenuze.

��

1.inačica 1

1 1 1 1 1 21 sin .2 2 2 2 2 1 11 cos 1 cos sin cos

1 2 22 2 2 sin sin1sin sin sin

ctgα

α α α α αα αα α α

= = = = = =+ +

+ +

■

2.inačica

42

ββββ

αααα

c

b

a

1 1 1 1 1 1

2 2 2 2 2 2 21 111 2 2 2 21

2 2 2

ctg b b a b cb

a a a a

bctg c a b

a

a

αα

= = = = = = = = = + +

=

+ ++

+

1 2221 sin .

2

i2

s n2

a

c

a a

cc c

a

αα

= = = = = =

■

43

Dokaz 569

Dokažite jednakost 21 cos

.sin cos

tgα

αα α

−=

⋅

Teorija



, .a a a a= =


.n

a n mam

a

−=


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅


, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


cos co, .

s

x xtg x tg x

x x= =

��

1.inačica 2 21 cos sin sin sin

.sin cos sin cos cos con s

2

sitg

α α α αα

α α α α α αα

−= = = =

⋅ ⋅ ⋅ ■

2.inačica 2 2 21 cos 1 cos sin 1 cos sin

sin cos sin cos cos sin c/ sin c

os cosostg

α α α α αα

α α α α α α α αα α

− − −= = =

⋅⋅ ⋅

⋅ ⋅

2 2 2 21 cos sin sin sin .α α α α − = = ■

44

Dokaz 570

Dokažite jednakost 21 2 sin

1.22 cos 1

α

α

− ⋅=

⋅ −

Teorija




, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

��

1.inačica

( )2 2 2 2 2 2 2 21 2 sin cos sin 2 sin cos sin cos sin

2 2 2 2 2 22 2 22 cos 1 2 cos cos sin cos sin2 cos cos sin

α α α α α α α α

α α α α α αα α α

− ⋅ + − ⋅ − −= = = =

⋅ − ⋅ − − −⋅ − +

2 2cos sin2 2cos sin

1.α α

α α

−= =

− ■

2.inačica

( )21 2 1 cos2 2 21 2 sin 1 2 2 cos 2 cos 11.

2 2 2 22 cos 1 2 cos 1 2 cos

22 cos 122 c1 2 ocos 1 s 1

αα α α

α α α

α

αα

− ⋅ −− ⋅ − + ⋅ ⋅ −= = = = =

⋅ − ⋅

⋅ −

⋅ −− ⋅ − ⋅ − ■

3.inačica

( )2 2 2 21 2 sin 1 2 sin 1 2 sin 1 2 sin

1.2 2 222 cos 1 2 2 sin 1 1 2 sin2 1

21 2 sin2s 1in 1 2 sin

α α α α

α αα

α

α α

− ⋅ − ⋅ − ⋅ − ⋅= = = = =

⋅ − − ⋅ − −

−

⋅

⋅

− ⋅⋅ − − ■

4.inačica

( )2 21 2 sin 1 2 sin 2 21 1 1 2 sin 2 cos 1

2 22 cos 1 2

2/ 2 cos 1cos 1

α αα α

α αα

− ⋅ − ⋅= = − ⋅ = ⋅ −

⋅ −⋅ −

−⋅

⋅

( )2 2 2 21 2 sin 2 cos 1 0 2 2 sin cos 0 2 2 1 0α α α α − ⋅ − ⋅ + = − ⋅ + = − ⋅ = 2 2 0 0 0. − = = ■

45

Dokaz 571

Dokažite ako su x1 i x2 rješenja kvadratne jednadžbe 2 0,a x b x c⋅ + ⋅ + = tada vrijedi:

( ) ( )2 .1 2a x b x c a x x x x⋅ + ⋅ + = ⋅ − ⋅ −

Teorija

Množenje zagrada:

( ) ( ) .a b c d a c a d b c b d+ ⋅ + = ⋅ + ⋅ + ⋅ + ⋅ Svojstvo potencije:

1 1, .a a a a= =


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅

Rješenja x1 i x2 kvadratne jednadžbe 2

0a x b x c⋅ + ⋅ + = zadovoljavaju Vièteove formule:

, 2 .1 2 1b c

x x x xa a

+ = − ⋅ =



, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

��

( ) ( ) ( ) ( )( )2 21 2 2 1 1 2 1 2 1 2a x x x x a x x x x x x x a x x x x x x⋅ − ⋅ − = ⋅ − ⋅ − ⋅ + ⋅ = ⋅ − + ⋅ + ⋅ =

1 2

1

2

2

2 2.

b c ca x x a x x a x b x c

a a

bx x

a

cx x

a

a

b

a= = ⋅ − − ⋅ + = ⋅ ⋅ + = ⋅ + +

+ = −

⋅ =

⋅

+

■

46

Dokaz 572

Dokažite jednakost ( ) ( ) ( )1 1 2 2f x f x f x− + + = ⋅ + za funkciju f(x) = x2.

Teorija

Kvadrat razlike:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b− = − ⋅ ⋅ + − ⋅ ⋅ + = −

Kvadrat zbroja:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b+ = + ⋅ ⋅ + + ⋅ ⋅ + = +

��

( ) ( ) ( ) ( )2 2 2 2 2 21 1 1 1 2 1 2 21 121f x f x x x x x x x x xx x− + + = − + + = − ⋅ + + + ⋅ + − ⋅ + ⋅= + + + =

( )2 2 21 1 2 2 2 2.x x x f x= + + + = ⋅ + = ⋅ + ■

47

Dokaz 573

Dokažite jednakost ( ) ( ) ( )2

1 sin cos 2 1 sin 1 cos .α α α α+ + = ⋅ + ⋅ +

Teorija

Kvadrat trinoma:

( ) .2 2 2 2

2 2 2a b c a b c a b a c b c+ + = + + + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅

( )2

2 .2 2 2

2 2a b c a b a c b c a b c+ + + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ = + +



( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

��

( )2 2 21 sin cos 1 sin cos 2 sin 2 cos 2 sin cosα α α α α α α α+ + = + + + ⋅ + ⋅ + ⋅ ⋅ =

1 1 2 sin 2 cos 2 sin cos 2 2 sin 2 cos 2 sin cosα α α α α α α α= + + ⋅ + ⋅ + ⋅ ⋅ = + ⋅ + ⋅ + ⋅ ⋅ =

( ) ( ) ( )( )2 1 sin cos sin cos 2 1 sin cos sin cosα α α α α α α α= ⋅ + + + ⋅ = ⋅ + + + ⋅ =

( ) ( )( ) ( ) ( )2 1 sin cos 1 sin 2 1 sin 1 cos .α α α α α= ⋅ + + ⋅ + = ⋅ + ⋅ + ■

48

Dokaz 574

Dokažite jednakost ( ) ( ) ( )2

sin cos 1 2 1 sin 1 cos .α α α α+ − = ⋅ − ⋅ −

Teorija

Kvadrat trinoma:

( ) .2 2 2 2

2 2 2a b c a b c a b a c b c+ + = + + + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅

( )2

2 .2 2 2

2 2a b c a b a c b c a b c+ + + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ = + +



( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

��

( )2 2 2sin cos 1 sin cos 1 2 sin cos 2 sin 2 cosα α α α α α α α+ − = + + + ⋅ ⋅ − ⋅ − ⋅ =

1 1 2 sin cos 2 sin 2 cos 2 2 sin cos 2 sin 2 cosα α α α α α α α= + + ⋅ ⋅ − ⋅ − ⋅ = + ⋅ ⋅ − ⋅ − ⋅ =

( ) ( ) ( )( )2 1 sin cos sin cos 2 1 sin sin cos cosα α α α α α α α= ⋅ + ⋅ − − = ⋅ − + ⋅ − =

( ) ( )( ) ( ) ( )2 1 sin cos 1 sin 2 1 sin 1 cos .α α α α α= ⋅ − − ⋅ − = ⋅ − ⋅ − ■

49

Dokaz 575

Dokažite ako je ax = ay, tada je x = y.

Teorija

Ako je a > 1, tada f raste, tj. za x < y vrijedi ax < ay. Ako je 0 < a < 1, tada f pada, tj. za x < y vrijedi ax > ay.

�� Pretpostavimo suprotno, tj. da je x ≠ y. Tada je x < y ili x > y. ♥ Ako je x < y, tada vrijedi

1.

0 1

x ya a za a

x ya a za a

< >

> < <

U svakom slučaju ne vrijedi jednakost ax = ay. ♥ Ako je x > y, tada vrijedi

1.

0 1

x ya a za a

x ya a za a

> >

< <

<

U svakom slučaju ne vrijedi jednakost ax = ay. Došli smo u suprotnost s početnom pretpostavkom, tj. da je ax = ay. ■

50

Dokaz 576

Dokažite ako su zbroj i umnožak kompleksnih brojeva ,z a b i w c d i= + ⋅ = + ⋅ realni brojevi, tada

je .w z=

Teorija

Kompleksan broj je broj oblika ,z x y i= + ⋅

gdje su x i y realni brojevi, a i imaginarna jedinica. Broj x zove se realni dio kompleksnog broja z, a broj y imaginarni dio kompleksnog broja z. Pišemo:

I .Re m,x z y z= =

Standardni ili algebarski oblik kompleksnog broja je oblika ,z x y i= + ⋅

gdje su x i y realni brojevi. Za kompleksne brojeve x + y ∙ i i x – y ∙ i kažemo da su kompleksno konjugirani jedan drugome. Simbol konjugiranja jest povlaka iznad broja koji se konjugira:

.z x y i z x y i= + ⋅ = − ⋅

Ako je imaginarni dio kompleksnog broja jednak 0, onda je broj realan.

[ ]Im 0 .z x y i C z y z R= + ⋅ ∈ = = ∈ Množenje zagrada:


1 1, .a a a a= =


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ + Kvadrat imaginarne jedinice:

2 21 , 1 .i i= − − =

�� ♥ Zbroj kompleksnih brojeva z i w je realan broj.

( ) ( ) ( ) ( )( )Im 0 Im 0 Im 0z w a b i c d i a c b d i+ = + ⋅ + + ⋅ = + + + ⋅ = 0 .b d d b + = = −

♥ Umnožak kompleksnih brojeva z i w je realan broj.

( ) ( ) ( )( ) ( )2Im 0 Im 0 Im 0z w a b i c d i a c a d i b c i b d i⋅ = + ⋅ ⋅ + ⋅ = ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ = ( )( ) ( )Im 1 0 Im 0a c a d i b c i b d a c a d i b c i b d ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ − = ⋅ + ⋅ ⋅ + ⋅ ⋅ − ⋅ =

( ) ( )( )Im 0 0.a c b d a d b c i a d b c ⋅ − ⋅ + ⋅ + ⋅ ⋅ = ⋅ + ⋅ = Promatramo sustav.

( ) ( )/0 0 :00

d ba b b c a b b c a b b c

d b cb

a

= − ⋅ − + ⋅ = − ⋅ + ⋅ = − ⋅ + ⋅ = −

⋅ + ⋅ =

0 .a c a c − = =

Sada slijedi:

51

.c a

d bw c d i w a b i w z= + ⋅ = =

−

=−

=⋅

■

52

Dokaz 577

Dokažite da je zbroj dvaju međusobno kompleksno konjugiranih brojeva jednak dvostrukom

realnom dijelu tih brojeva.

Teorija



I .Re m,x z y z= =



.z x y i z x y i= + ⋅ = − ⋅

��

bz a b i

z z a b i a b i z z az b

i iaa i

b= + ⋅

+ = + ⋅ + ⋅+ − ⋅ + = + =

⋅⋅

−−

2 2 Re .z z a a z z a z z z + = + + = ⋅ + = ⋅ ■

53

Dokaz 578

Dokažite da je umnožak dvaju međusobno kompleksno konjugiranih brojeva uvijek nenegativan

broj.

Teorija



I .Re m,x z y z= =



.z x y i z x y i= + ⋅ = − ⋅


, .a a a a= =


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅

Kvadrat imaginarne jedinice: 2 21 , 1 .i i= − − =


( ) ( ), .n nn n n n

a b a b a b a b⋅ = ⋅ ⋅ = ⋅

Razlika kvadrata:

( ) ( ) ( ) ( ),2 .2 2 2a b a b a b a b a b a b− = − ⋅ + − ⋅ + = − Kvadrat realnog broja je nenegativan broj.

20 , .a a R≥ ∈

��

( ) ( ) ( )22 2 2 2z a b i

z z a b i a b i z z a b i z z a b iz a b i

= + ⋅ ⋅ = + ⋅ ⋅ − ⋅ ⋅ = − ⋅ ⋅ = − ⋅

= − ⋅

( )2 2 2 21 0.z z a b z z a b ⋅ = − ⋅ − ⋅ = + ≥ ■

54

Dokaz 579

Dokažite da vrijedi z w z w⋅ = ⋅ , gdje su ,z a b i w c d i= + ⋅ = + ⋅ kompleksni brojevi.

Teorija



I .Re m,x z y z= =


gdje su x i y realni brojevi. Množenje zagrada:


1 1, .a a a a= =


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅

Kvadrat imaginarne jedinice: 2 21 , 1 .i i= − − =


( ) ( ), .n nn n n n

a b a b a b a b⋅ = ⋅ ⋅ = ⋅

Kvadrat razlike:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b− = − ⋅ ⋅ + − ⋅ ⋅ + = −

Kvadrat zbroja:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b+ = + ⋅ ⋅ + + ⋅ ⋅ + = +

Množenje drugih korijena:

, , 0, .a b a b a b a b a b⋅ = ⋅ ⋅ = ⋅ ≥

Modul ili apsolutna vrijednost kompleksnog broja z = x + y · i definira se formulom

2 2.z x y= +

�� Neka su , .z a b i w c d i= + ⋅ = + ⋅ Tada je:

♥

( ) ( ) 2z w a b i c d i z w a c a d i b c i b d i⋅ = + ⋅ ⋅ + ⋅ ⋅ = ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅

( )1z w a c a d i b c i b d z w a c a d i b c i b d ⋅ = ⋅ + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ − ⋅ = ⋅ + ⋅ ⋅ + ⋅ ⋅ − ⋅

( ) ( ) ( ) ( )2 2

z w a c b d a d b c i z w a c b d a d b c ⋅ = ⋅ − ⋅ + ⋅ + ⋅ ⋅ ⋅ = ⋅ − ⋅ + ⋅ + ⋅

2 2 2 2 2 2 2 22 2z w a c a c b d b d a d a d b c b c ⋅ = ⋅ − ⋅ ⋅ ⋅ ⋅ + ⋅ + ⋅ + ⋅ ⋅ ⋅ ⋅ + ⋅

2 22 2 2 2 2 2 2 2

z w a c b da c b d a d b ca d b c− ⋅ ⋅ ⋅ ⋅ + ⋅ = ⋅ + ⋅ + ⋅ +⋅ ⋅ ⋅⋅ ⋅

55

2 2 2 2 2 2 2 2z w a c b d a d b c ⋅ = ⋅ + ⋅ + ⋅ + ⋅

2 2 2 2 2 2 2 2.z w a c a d b c b d ⋅ = ⋅ + ⋅ + ⋅ + ⋅

♥ 2 2 2 2

z w a b i c d i z w a b c d⋅ = + ⋅ ⋅ + ⋅ ⋅ = + ⋅ +

( ) ( )2 2 2 2 2 2 2 2 2 2 2 2 .z w a b c d z w a c a d b c b d ⋅ = + ⋅ + ⋅ = ⋅ + ⋅ + ⋅ + ⋅ Konačno dobijemo:

2 2 2 2 2 2 2 2

.2 2 2 2 2 2 2 2

z w a c a d b c b dz w z w

z w a c a d b c b d

⋅ = ⋅ + ⋅ + ⋅ + ⋅ ⋅ = ⋅

⋅ = ⋅ + ⋅ + ⋅ + ⋅

■

56

Dokaz 580

Dokažite formulu ( ) 21 cos 2 cos .2

xx+ = ⋅

Teorija

Temeljni trigonometrijski identitet

( ) ( ) ( ) ( )2 2 2 2cos sin 1 1 co i, .s s nx x x x+ = = +

2 2 2 2cos sin 1 1 cos sin2 2 2

, .2

x x x x + = = +

Funkcija polovičnog kuta

( ) ( )2 2 2 2cos cos sin cos sin cos2 2 2

, .2

x x x xx x

= − − =

�� Budući da je

( ) 2 2 2 2cos cos sin i cos sin 1,2 2 2 2

x x x xx

= − + =

dobijemo

( ) 2 2 2 21 cos cos sin cos sin2 2 2 2

x x x xx

+ = + + −

( ) ( )2 2sin sin2 2 21 cos cos cos 1 cos 2 cos .2 2 22 2

x x xx xx x

+ = + + = ⋅

+ −

■

57

Dokaz 581

Dokažite formulu ( ) 21 cos 2 sin .2

xx− = ⋅

Teorija

Temeljni trigonometrijski identitet

( ) ( ) ( ) ( )2 2 2 2cos sin 1 1 co i, .s s nx x x x+ = = +

2 2 2 2cos sin 1 1 cos sin2 2 2

, .2

x x x x + = = +

Funkcija polovičnog kuta

( ) ( )2 2 2 2cos cos sin cos sin cos2 2 2

, .2

x x x xx x

= − − =


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

�� Budući da je

( ) 2 2 2 2cos cos sin i cos sin 1,2 2 2 2

x x x xx

= − + =

dobijemo


x x x xx

− = + − −


x x x xx

− = + − +

( ) ( )2 2 21 cos sin sin 1 cos 2 sin .2

2 2cos cos2 2 2 2

x xx x xx x

−

− = + + − = ⋅

■

58

Dokaz 582

Dokažite identitet ( )( )

( )( )

sin 6 cos 61.

2 sin 2 2 cos 2

x x

x x

⋅ ⋅− =

⋅ ⋅ ⋅ ⋅

Teorija


.a c a d b c

b d b d

⋅ − ⋅− =

⋅

Adicijska formula za sinus razlike Za svaka dva realna broja x i y vrijedi

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )sin sin cos cos sin sin cos cos sin s, in .x y x y x y x y x y x y− = ⋅ − ⋅ ⋅ − ⋅ = − Formule za sinus dvostrukog kuta

( ) ( ) ( ) ( ) ( ) ( )sin 2 2 sin cos 2 sin cos sin, .2α α α α α α⋅ = ⋅ ⋅ ⋅ ⋅ = ⋅ Skratiti razlomak znači brojnik i nazivnik tog razlomka podijeliti istim brojem različitim od nule i jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

��

( )( )

( )( )

( ) ( ) ( ) ( )( ) ( )

sin 6 cos 6 sin 6 cos 2 cos 6 sin 2

2 sin 2 2 cos 2 2 sin 2 cos 2

x x x x x x

x x x x

⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅− = =

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

( )( )

( )( )

( )( )

sin 6 2 sin 41.

sin 2

sin 4

sin 42 sin 4

x

x

x x x

x x

⋅ − ⋅ ⋅ ⋅

⋅= = = =

⋅ ⋅ ⋅ ■

59

Dokaz 583

Dokažite identitet sin cos .4 4

x xπ π

+ = −

Teorija

Trigonometrijske funkcije šiljastog kuta

sin cos4

.4

π π =

Adicijska formula za sinus zbroja Za svaka dva realna broja x i y vrijedi

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )sin sin cos cos sin sin cos cos sin s, in .x y x y x y x y x y x y+ = ⋅ + ⋅ ⋅ + ⋅ = + Adicijska formula za kosinus razlike Za svaka dva realna broja x i y vrijedi

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )cos cos cos sin sin cos cos sin sin c, os .x y x y x y x y x y x y− = ⋅ + ⋅ ⋅ + ⋅ = −

Formula redukcije za argument 2

xπ

−

( ) ( )sin cos cos si, n2

.2

x x x xπ π

= − − =


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ + Oduzimanje razlomaka:

.a c a d b c

b d b d

⋅ − ⋅− =

⋅

�� 1.inačica

( ) ( ) sin csin sin cos c osos s4

in4 44 4

x x xπ π ππ π

+ = ⋅ + ⋅ == =

( ) ( )cos cos sin sin cos .4 4 4

x x xπ π π

= ⋅ + ⋅ = −

■

2.inačica

2sin cos cos cos cos .

4 2 4 2 4 4 4x x x x x

π π π π π π π π⋅ −+ = − + = − − = − = −

■

60

Dokaz 584

Dokažite jednakost ( )sin cos

4 4.

sin cos .4 4

x x

tg x

x x

π π

π π

+ − +

=

+ + +

Teorija

Adicijska formula za sinus zbroja Za svaka dva realna broja x i y vrijedi

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )sin sin cos cos sin sin cos cos sin s, in .x y x y x y x y x y x y+ = ⋅ + ⋅ ⋅ + ⋅ = + Adicijska formula za kosinus zbroja Za svaka dva realna broja x i y vrijedi

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )cos cos cos sin sin cos cos sin sin c, os .x y x y x y x y x y x y+ = ⋅ − ⋅ ⋅ − ⋅ = + Zakon distribucije množenja prema zbrajanju.

( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ + Trigonometrijske funkcije šiljastog kuta

2sin cos

4 4 2.

π π = =


, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

��

sin cos4 4

sin cos .4 4

x x

x x

π π

π π

+ − +

=

+ + +

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

sin cos cos sin cos cos sin sin4 4 4 4


x x x x

x x x x

π π π π

π π π π

⋅ + ⋅ − ⋅ − ⋅

= =

⋅ + ⋅ + ⋅ − ⋅

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )



x x x x

x x x x

π π π π

π π π π

⋅ + ⋅ − ⋅ + ⋅

= =

⋅ + ⋅ + ⋅ − ⋅

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 2 2 2cos sin cos sin

2 2 2 22 2 2 2

cos sin cos sin2 2 2 2

x x x x

x x x x

⋅ + ⋅ − ⋅ + ⋅

= =

⋅ + ⋅ + ⋅ − ⋅

61

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )

( )

2 2cos cos

2 22 2

sin sin2

2 2 2sin sin sin

2 2 22 2 2

cos c2

os cos2 2 2

2

2

x x

x x

x x x

x x x

+ ⋅ + ⋅ ⋅

= = =

⋅ + ⋅ ⋅

⋅ − ⋅

+ ⋅ − ⋅

⋅

⋅

( )

( )

( )( )

( )sin sin

.cos

co

22

22

2s2

xx

tg xx

x

⋅ ⋅

= = =

⋅ ⋅

■

62

Dokaz 585

Dokažite jednakost ( )

( )( )

2sin 2 .21

tg xx

tg x

⋅= ⋅

+

Teorija

Definicija tangensa pomoću sinusa i kosinusa:

( )( )( )

( )( )

( )sin sin

cos co, .

s

x xtg x tg x

x x= =


1, .

1

n nn n= =


.a c a d b c

b d b d

⋅ + ⋅+ =

⋅

Množenje cijelog broja i razlomka:

, .b a b a a c

a cc c b b

⋅ ⋅⋅ = ⋅ =

Dvojni razlomak:

.

a

a db

c b c

d

⋅=

⋅


.n

a n mam

a

−=

Formula za sinus dvostrukog kuta

( ) ( ) ( ) ( ) ( ) ( )sin 2 2 sin cos 2 sin cos sin, .2α α α α α α⋅ = ⋅ ⋅ ⋅ ⋅ = ⋅ Skratiti razlomak znači brojnik i nazivnik tog razlomka podijeliti istim brojem različitim od nule i jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

��

( )

( )

( )( )

( )

( )

( )( )

( )

( )

( )( )

( ) ( )

( )

( )( )

( )

sin 2 sin 2 sin 2 sin2

2 cos cos cos cos2 2 2 2 2 11 sin sin cos sin1

1 22 2 2 cos1cos cos cos

x x x x

tg x x x x x

tg x x x x x

xx x x

⋅ ⋅ ⋅⋅

⋅= = = = =

+ ++ +

( ) ( )( )

( ) ( )( )

( ) ( ) ( )22 sin cos 2 sin cos2

cos2 sin cos sin 2 .

cos

x x x xx x x

xx

⋅ ⋅ ⋅ ⋅= = = ⋅ ⋅ = ⋅ ■

63

Dokaz 586

Dokažite jednakost ( )( )

( )( )

1 1 sin 2.

1 cos 2

tg x x

tg x x

− − ⋅=

+ ⋅

Teorija


( )( )( )

( )( )

( )sin sin

cos co, .

s

x xtg x tg x

x x= =


1, .

1

n nn n= =


.a c a d b c

b d b d

⋅ − ⋅− =

⋅


.a c a d b c

b d b d

⋅ + ⋅+ =

⋅

Dvojni razlomak:

.

a

a db

c b c

d

⋅=

⋅


( ) ( ) ( ) ( ) ( ) ( )2 2 2 2cos 2 cos sin cos sin, c s 2 .oα α α α α α⋅ = − − = ⋅ Skratiti razlomak znači brojnik i nazivnik tog razlomka podijeliti istim brojem različitim od nule i jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

Proširiti razlomak znači brojnik i nazivnik tog razlomka pomnožiti istim brojem različitim od nule i jedinice

, 0 ., 1a a n

n nb b n

⋅= ≠ ≠

⋅


, .a a a a= =


, .n m n m n m n m

a a a a a a+ +

⋅ = = ⋅

Množenje razlomaka:

.a c a c

b d b d

⋅⋅ =

⋅

Kvadrat razlike:

( ) ( )2 22 2 2 2

2 , .2a b a a b b a a b b a b− = − ⋅ ⋅ + − ⋅ ⋅ + = −

Razlika kvadrata:

( ) ( ) ( ) ( ),2 .2 2 2a b a b a b a b a b a b− = − ⋅ + − ⋅ + = −

64

Osnovna trigonometrijska relacija

( ) ( ) ( ) ( )2 2 2 2cos sin 1 1 co i, .s s nα α α α+ = = + Zbrajanje razlomaka jednakih nazivnika:

, .a b a b a b a b

n n n n n n

+ += + + =

Oduzimanje razlomaka jednakih nazivnika:

, .a b a b a b a b

n n n n n n

− −= − − =

�� 1.inačica

( )( )

( )( )( )( )

( ) ( )( )

( ) ( )( )

( ) ( )( )

( ) ( )( )

( ) ( )( ) ( )

sin cos sin cos sin1

1 1 cos cos cos sin

sin cos si

cos

n cos sin1 1 cos sin

1 cos cos sco

x x x x x

tg x x x x x

x x x x xtg x x x

x x

x

x

− −−

− −= = = = =

+ ++ ++

( ) ( )( ) ( )

( ) ( )( ) ( )

( ) ( )( )( ) ( )

( ) ( ) ( ) ( )( )

2 2 2cos sincos sin cos sin cos 2 cos sin sin2 2cos sin cos sin cos 2cos sin

x xx x x x x x x x

x x x x xx x

−− − − ⋅ ⋅ += ⋅ = = =

+ − ⋅−

( ) ( )( )

( )( )

1 2 cos sin 1 sin 2.

cos 2 cos 2

x x x

x x

− ⋅ ⋅ − ⋅= =

⋅ ⋅ ■

2.inačica

( )( )

( ) ( ) ( ) ( )

( ) ( )

2 21 sin 2 cos sin 2 sin cos2 2cos 2 cos sin

x x x x x

x x x

− ⋅ + − ⋅ ⋅= =

⋅ −

( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )( )

( ) ( )( )( ) ( )( ) ( ) ( )( )

22 2 cos sincos 2 sin cos sin

cos sin cos sin cos sin cos sin

x xx x x x

x x x x x x x x

−− ⋅ ⋅ += = =

− ⋅ + − ⋅ +

( ) ( )( )( ) ( )( ) ( ) ( )( )

( ) ( )( ) ( )

( ) ( )( )

( ) ( )( )

( )( )

( )( )

( )( )

( )( )

cos sin cos sin

cos sin cos sin cos cos cos

cos sin cos sincos sin co

2

cos si s sin

cos cos cos

n

x x x x

x x x x x x x

x x x xx x x x

x

x

x x

x

−−

− −= =

−= = =

+⋅ + ++

( )( )

( )( )

( )( )

( )( )

( )( )

cos

cos

cos

co

sin

cos 1.

sin

oss

1

c

x

x tg x

x tg x

x

x

x

x

x

−−

= =+

+

■

65

Dokaz 587


( )( )

21cos 2 .21

tg xx

tg x

−= ⋅

+

Teorija


( )( )( )

( )( )

( )sin sin

cos co, .

s

x xtg x tg x

x x= =


1, .

1

n nn n= =


.a c a d b c

b d b d

⋅ − ⋅− =

⋅


.a c a d b c

b d b d

⋅ + ⋅+ =

⋅

Dvojni razlomak:

.

a

a db

c b c

d

⋅=

⋅


( ) ( ) ( ) ( ) ( ) ( )2 2 2 2cos 2 cos sin cos sin, c s 2 .oα α α α α α⋅ = − − = ⋅ Skratiti razlomak znači brojnik i nazivnik tog razlomka podijeliti istim brojem različitim od nule i jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

Proširiti razlomak znači brojnik i nazivnik tog razlomka pomnožiti istim brojem različitim od nule i jedinice

, 0 ., 1a a n

n nb b n

⋅= ≠ ≠

⋅

Osnovna trigonometrijska relacija

( ) ( ) ( ) ( )2 2 2 2cos sin 1 1 co i, .s s nα α α α+ = = + Zbrajanje razlomaka jednakih nazivnika:

, .a b a b a b a b

n n n n n n

+ += + + =

Oduzimanje razlomaka jednakih nazivnika:

, .a b a b a b a b

n n n n n n

− −= − − =

��

1.inačica

66

( )

( )

( )

( )

( )

( )

( ) ( )

( )

( ) ( )

( )

( )

( )

( )

( )

( )

( )

( )

2 2 2sin cos sin cos 2 cos 212 2 2 211 cos cos cos

cos 2 .2 2 2 2 1 11 sin cos sin122 2 c

2cos

2cosos1 cos cos

x x x x x

tg x x x xx

tg x x x x

x

x

xx x

− ⋅ ⋅−

−= = = = = ⋅

+ ++

■

2.inačica

( )( ) ( ) ( )

( ) ( )

( ) ( )

( )

( ) ( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

2 2 2 2cos sin cos sin2 2 2 2 2cos 2 cos sin cos cos cos

cos 2 2 2 2 2 2 21 cos sin cos sin cos sin2 2 2cos cos cos

x x x x

x x x x x xx

x x x x x x

x x x

−−

⋅ −⋅ = = = = =

+ ++

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

2cos2cos2cos2co

2sin2 2cos 1

.2 2sin

cos

12s

x

x tg x

x tg x

x

x

x

x

x

−−

= =+

+

■

67

Dokaz 588


( )( )( ) ( ) ( )

sin cos 1.

1 1 sin cos

541 … dokaz 541 dokažite ako je p prost broj razli čit od ... · 3 dokaz 543 dokažite ako...

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