Functii trigonometrice. Transformarea sumelor in produs si a produselor in sume

Pentru a transforma sumele in produs trebuie sa stim urmatoarele formule:

  1. \cos a+\cos b=2\cos\frac{a+b}{2}\cdot\cos\frac{a-b}{2}
  2.    \cos a-\cos b=-2\sin\frac{a-b}{2}\cdot\sin\frac{a+b}{2}
  3.  \sin a+\sin b=2\sin\frac{a+b}{2}\cdot\cos\frac{a-b}{2}
  4.  \sin a-\sin b=2\sin\frac{a-b}{2}\cdot\cos\frac{a+b}{2}

Aplicatii!

Sa se descompuna in produs:

a) \sin 3x+2\sin 2x+\sin x

Pentru a transforma sumele in produs observam ca trebuie sa avem la suma termeni de numar par, asfel rescriind expresia obtinem \sin 3x+2\sin 2x+\sin x=(\sin 3x+\sin 2x)+(\sin 2x+\sin x)

Am scris termenul al doilea ca suma de doi termeni.

Astfel folosind formula a treia obtinem (\sin 3x+\sin 2x)+(\sin 2x+\sin x)=2\sin\frac{3x+2x}{2}\cdot\cos\frac{3x-2x}{2}+2\cdot\sin\frac{2x+x}{2}\cdot\cos\frac{2x-x}{2}=2\sin\frac{5x}{2}\cdot\cos\frac{x}{2}+2\cdot \sin\frac{3x}{2}\cdot\cos\frac{x}{2}

Dand factor comun 2\cos\frac{x}{2} obtinem 2\cos\frac{x}{2}\left(\sin\frac{5x}{2}+\sin\frac{3x}{2}\right)

Aplicand din nou formula obtinem 2\cos\frac{x}{2}\left(2\sin\frac{\frac{5x}{2}+\frac{3x}{2}}{2}\cdot\cos\frac{\frac{5x}{2}-\frac{3x}{2}}{2}\right)

AsTfel obtinem 2\cos\frac{x}{2}\left(2\sin\frac{\frac{5x+3x}{2}}{2}\cdot\cos\frac{\frac{5x-3x}{2}}{2}\right)

Efectuand calculele obtinem:

2\cos\frac{x}{2}\left(2\sin\frac{\frac{8x}{2}}{2}\cdot\cos\frac{\frac{2x}{2}}{2}\right)

2\cos\frac{x}{2}\left(2\sin\frac{4x}{2}\cdot\cos\frac{x}{2}\right)

2\cos\frac{x}{2}\cdot 2\sin 2x\cdot cos\frac{x}{2}

4\cos^{2}\frac{x}{2}\cdot sin {2x}

b) \sin x+\sin y+\sin\left(x-y\right)

Asa cum am spus si mai sus, in cazul termenilor din suma trebuie sa avem un numar par, dupa cum bine vedeti noi avem 3 termeni.

Sa nu incercam sa aplicam formula de la diferenta de unghiuri, pentru ca ne complicam, fiind mult prea mult de calculat, astfel stim ca \sin 0=0 si astfel rescriind expresia obtinem un numar par de termeni, astfel obtinem:

\sin x+\sin y+\sin\left(x-y\right)= (\sin x+\sin y)+(\sin\left(x-y\right)+\sin 0)

Aplicand din nou formulele obtinem: 2\sin\frac{x+y}{2}\cdot\cos\frac{x-y}{2}+2\sin\frac{x-y+0}{2}\cdot\cos\frac{x-y-0}{2}

Astfel efectuand calculele obtinem: 2\sin\frac{x+y}{2}\cdot\cos\frac{x-y}{2}+2\sin\frac{x-y}{2}\cdot\cos\frac{x-y}{2}

La fel ca si mai sus dand factor comun pe 2\cos\frac{x-y}{2} obtinem: 2\cos\frac{x-y}{2}\left(\sin\frac{x+y}{2}+\sin\frac{x-y}{2}\right)

Aplicand din nou formula pentru transformarea sumelor in produs obtinem: 2\sin\frac{x-y}{2}\left(2\cdot\sin\frac{\frac{x+y}{2}+\frac{x-y}{2}}{2}\cdot\cos\frac{\frac{x+y}{2}-\frac{x-y}{2}}{2}\right)

Efectuand calcule in paranteza rotunda obtinem: 2\sin\frac{x-y}{2}\left(2\cdot\sin\frac{\frac{x+y+x-y}{2}}{2}\cdot\cos\frac{\frac{x+y-x+y}{2}}{2}\right)

Astfel, obtinem: 2\sin\frac{x-y}{2}\left(2\cdot\sin\frac{\frac{2x}{2}}{2}\cdot\cos\frac{\frac{2y}{2}}{2}\right)

2\sin\frac{x-y}{2}\cdot 2\cdot\sin\frac{x}{2}\cdot\cos\frac{y}{2}

4\cdot\sin\frac{x-y}{2}\cdot\sin\frac{x}{2}\cdot\cos\frac{y}{2}.

2. Aratati  ca daca x+y=z atunci sunt adevarate relatiile:

\cos^{2}x+\cos^{2}y+\cos^{z}=1+2\cos x\cdot cos y\cdot\cos z

Rezolvare!

x+y=z obtinem \cos(x+y)=cos z. Calculand obtinem:

\cos(x+y)=\cos x\cdot \cos y-\sin x\cdot\sin y

adica \cos x\cdot \cos y-\sin x\cdot\sin y=\cos z\Rightarrow \cos x\cdot \cos y-\cos z=\sin x\cdot sin y

Ridicand la patrat egalitatea obtinem:

\cos x\cdot \cos y-\cos z=\sin x\cdot sin y|^{2}

\Rightarrow \left(\cos x\cdot \cos y-\cos z\right)^{2}=\left(\sin x\cdot\sin y\right)^{2}\Rightarrow

\cos^{2}x\cdot\cos^{2}y-2\cdot\cos x\cdot\cos y\cdot\cos z+\cos^{2}z=\sin^{2}x+\cdot\sin^{2}y

\cos^{2}x\cdot\cos^{2}y-2\cdot\cos x\cdot\cos y\cdot\cos z+\cos^{2}z=\left(1-\cos^{2}x\right)\cdot\left(1-\cos^{2}y\right).

Efectuand produsul in membrul drept obtinem: \cos^{2}x\cdot\cos^{2}y-2\cdot\cos x\cdot\cos y\cdot\cos z+\cos^{2}z=1-\cos^{2}y-\cos^{2}x+\cos^{2}x\cdot\cos^{2}y

\cos^{2}x\cdot\cos^{2}y-2\cdot\cos x\cdot\cos y\cdot\cos z+\cos^{2}z-1+\cos^{2}y+\cos^{2}x-\cos^{2}x\cdot\cos^{2}y=0

Dupa ce am redus termenii asemenea obtinem:

\cos^{2} x+\cos^{2} y+\cos^{2} z=1+2\cos x\cdot\cos y\cdot cos z.

Si astfel am demonstrat egalitatea.

3. Se consideră triunghiul ABC. Să se demonstreze că \sin A+\sin B+\sin C=4\cos\frac{A}{2}\cdot\cos\frac{B}{2}\cdot\cos\frac{C}{2}, unde, A,B,C sunt masurile unghiurilor unui triunghi.

Demonstratie:

Stim ca suma masurilor unghiurilor intr-un triunghi este de 180^{0} sau \pi, astfel avem ca A+B+C=\pi, de unde obtinem C=\pi-(A+B)

Astfel in relatia de mai sus obtinem:

\sin A+\sin B+\sin C=\sin A+\sin B+\sin(\pi-(A+B))

Dar stim \sin(\pi-(A+B))=\sin (A+B)

\sin A+\sin B+\sin (A+B)+\sin 0=2\cdot\sin\frac{A+B}{2}\cdot\cos\frac{A-B}{2}+2\sin\frac{A+B+0}{2}\cdot\frac{A+B-0}{2}

\sin A+\sin B+\sin (A+B)+\sin 0=2\cdot\sin\frac{A+B}{2}\cdot\cos\frac{A-B}{2}+2\sin\frac{A+B}{2}\cdot\cos\frac{A+B}{2}

Astfel, dand factor comun in membrul drept al egalitatii, obtinem:

2\cdot\sin\frac{A+B}{2}\cdot\cos\frac{A-B}{2}+2\sin\frac{A+B}{2}\cdot\cos\frac{A+B}{2}=

2\cdot\sin\frac{A+B}{2}\left(\cos\frac{A-B}{2}+\cos\frac{A+B}{2}\right)

Aplicand din nou formula pentru transformarea sumei in produs obtinem:

2\cdot\sin\frac{A+B}{2}\left(2\cdot\cos\frac{\frac{A-B}{2}+\frac{A+B}{2}}{2}\cdot\cos\frac{\frac{A-B}{2}-\frac{A+B}{2}}{2} \right)

Astfel obtinem 4\cdot\sin\frac{A+B}{2}\cos\frac{\frac{2A}{2}}{2}\cdot\cos\frac{\frac{-2B}{2}}{2}

Functia cos,  fiind functie para, obtinem 4\cdot\sin\frac{A+B}{2}\cos\frac{A}{2}\cdot\cos\frac{B}{2}

Stim ca A+B=\pi-C

Si astfel avem ca \sin\frac{\pi-C}{2}=\sin\left(\frac{\pi}{2}-\frac{C}{2}\right)=\cos\frac{C}{2}

Asadar obtinem 4\cdot\cos\frac{C}{2}\cdot\cos\frac{A}{2}\cdot\cos\frac{B}{2}

4. Demonstrati ca daca intre elementele unui triunghi exista relatia \frac{a+b}{c}=\cot\frac{C}{2}, atunci triunghiul este dreptunghic.

Solutie:

Din teorema sinusului stim ca \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R, unde a, b ,c sunt laturile triunghiului, astfel avem ca a=2R\cdot \sin AB=2R\cdot \sin Bc=2R\cdot \sin C

Rescriind relatia obtinem:

\frac{2R\sin A+2R\sin B}{2R\sin C}=\frac{\cos\frac{C}{2}}{\sin\frac{C}{2}}

\frac{2R\left(\sin A+\sin B\right)}{2R\sin C}=\frac{\cos\frac{C}{2}}{\sin\frac{C}{2}}

\frac{\sin A+\sin B}{\sin C}=\frac{\cos\frac{C}{2}}{\sin\frac{C}{2}}

Folosind transformarea sumei in produs obtinem:

\frac{2\cdot\sin\frac{A+B}{2}\cos\frac{A-B}{2}}{2\cdot\sin\frac{C}{2}\cdot\cos\frac{C}{2}}\frac{\cos\frac{C}{2}}{\sin\frac{C}{2}}

Dar stim ca A+B+C=\pi\Rightarrow A+B=\pi-C

\frac{2\cdot\cos\frac{C}{2}\cdot\cos\frac{A-B}{2}}{2\cdot\sin\frac{C}{2}\cdot\cos\frac{C}{2}}\frac{\cos\frac{C}{2}}{\sin\frac{C}{2}}

Si astfel obtinem \frac{\cos\frac{A-B}{2}}{\sin \frac{C}{2}}=\frac{\cos\frac{C}{2}}{\sin\frac{C}{2}}

\cos\frac{A-B}{2}=\cos\frac{C}{2}\Rightarrow A-B=C\Rightarrow C=A-B\Rightarrow m\left(\widehat{A}\right)=\frac{\pi}{2}.

5. Demonstrati ca daca intr-un triunghi ABC, are loc relația (b^{2}+c^{2})\cdot\sin(C-B)=(c^{2}-b^{2})\cdot\sin(C+B) atunci triunghiul este isoscel sau dreptunghic.

Demonstrație:

(b^{2}+c^{2})\cdot\sin(C-B)=(c^{2}-b^{2})\cdot\sin(C+B)

\Rightarrow (c^{2}-b^{2})\left(\sin C\cdot\cos B+\sin B\cdot\cos C\right)=\left(c^{2}-b^{2}\right)\left(\sin C\cdot\cos B+\sin B\cdot\cos C\right)

\Rightarrow \left(2R\sin B\right)^{2}\cdot\sin C\cdot\sin B=\left(2R\sin C\right)^{2}\sin B\cdot \cos C\Rightarrow \sin B\cdot\cos B=\sin C\cdot\cos C\Rightarrow\sin 2B=\sin 2C\Rightarrow 2sin (B-C)\cdot\cos(B+C)=0\Rightarrow \sin(B-C)\cdot\cos A=0

Egalitatea are loc daca B=C sau m\left(\widehat{A}\right)=\frac{\pi}{2}

6. Demonstrati ca daca x+y+z=2\pi atunci este adevarata relatia \sin x+\sin y+\sin z=4\sin\frac{x}{2}\cdot\sin\frac{y}{2}\cdot\sin\frac{z}{2}

Demonstratie:

\sin x+\sin y+\sin z=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}-\sin\left(x+y\right)=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}-\sin\left(x+y\right)+\sin 0=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}-2\sin\frac{x+y}{2}\cos\frac{x+y}{2}=2\sin\frac{x+y}{2}\left(\cos\frac{x-y}{2}-\cos\frac{x+y}{2}\right)=2\sin\frac{2\pi-z}{2}\cdot\left[-2\sin\left(\frac{-y}{2}\right)\cdot\sin\frac{x}{2}\right]=2\sin\frac{z}{2}\cdot 2\sin\frac{y}{2}\cdot\sin\frac{x}{2}=4\cdot\sin\frac{x}{2}\cdot\sin\frac{y}{2}\cdot\frac{z}{2}.

7. Aratati ca \cot2^{k-1} x-\cot2^{k}x=\frac{1}{\sin 2^{k}x}

Demonstratie:

Stim ca \cot x=\frac{\cos x}{\sin x}, astfel obtinem ca \cot2^{k-1} x-\cot2^{k}x=\frac{\cos 2^{k-1}x}{\sin 2^{k-1}x}-\frac{\cos 2^{k}x}{\sin 2^{k}x}

Aducand la acelasi numitor  obtinem:

\frac{\cos 2^{k-1}x}{\sin 2^{k-1}x}-\frac{\cos 2^{k}x}{\sin 2^{k}x}=\frac{\sin 2^{k}x\cdot\cos 2^{k-1}x-\sin 2^{k} x\cdot\cos 2^{k-1}x}{\sin 2^{k-1}x\cdot\sin 2^{k}x}

\frac{\sin\left(2^{k}x-2^{k-1}x\right)}{\sin 2^{k-1}x\cdot\sin 2^{k}x}=

\frac{\sin 2^{k-1}x}{\sin 2^{k-1}x\cdot\sin 2^{k}x}=\frac{1}{\sin 2^{k}x}

Aplicatii trigonometrice in geometria plana

O  aplicatie a trigonometriei in geometria plana o reprezinta rezolvarea triunghiurilor.

Astfel fie ABC un triunghi. Numerele a=BC, b=AC, c=AB  si A=m\left(\widehat{BAC}\right), B=m\left(\widehat{ABC}\right), C=m\left(\widehat{ACD}\right), care sunt elementele triunghiului.

Triunghiul ABC este bine determinat daca se cunosc elementele sale.

A rezolva un triunghi inseamna a determina elementele triunghiului cunoscand trei dintre acestea.

Astfel avem mai multe cazuri de congruente:

a) Rezolvarea triunghiului dreptunghic cand se cunosc laturile (cazul de congruenta L.L.L)

In acest caz elementele cunoscute sunt a,b, c si elementele necunoscute sunt A, B, C.

Astfel din teorema cosinusului avem ca:

cum aplicam teorema cosinusuluiBC^{2}=AB^{2}+AC^{2}-2\cdot AB\cdot AC\cdot\cos A\Rightarrow a^{2}=

c^{2}+b^{2}-2\cdot c\cdot b\cdot\cos A\Rightarrow

a^{2}-c^{2}-b^{2}=-2\cdot c\cdot b\cdot \cos A\Rightarrow

\cos A=\frac{a^{2}-c^{2}-b^{2}}{-2\cdot c\cdot b}\Rightarrow

\cos A=\frac{\left(-a^{2}+c^{2}+b^{2}\right)}{-2\cdot c\cdot b}

\Rightarrow \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}

La fel obtinem pentru

\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}

Dar si

cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}, relatii care conduc la aflarea unghiurilor triunghiului cand stim laturile.

b) Rezolvarea triunghiului cand se cunosc doua unghiuri si o latura comuna (cazul de congruenta U.L.U)

In acest caz elementele cunoscute sunt, de exemplu: a, B, C si elementele necunoscute sunt b, c, A.

Teorema sinusului

In acest caz ca sa aflam masura unghiului , stim ca

A+B+C=180^{0}

In cazul de mai sus

A=180^{0}-B-C sau A=\pi-B-C, iar din teorema sinusului obtinem ca:

\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}

Astfel obtinem ca

b=\frac{a\cdot \sin B}{\sin A}=\frac{a\sin B}{\sin\left(B+C\right)}

c) Rezolvarea triunghiului cand se cunosc doua laturi si unghiul cuprins intre ele (cazul de congruenta L.U.L)

In acest ca putem sa aplicam Teorema cosinusului pentru a afla cea de-a treia latura si Teorema sinusului pentru a afla unghiurile pe care le cunoastem.

Aplicatii:

1) Fie triunghiul  ABC, calculati lungimea laturii [BC], stiind ca m\left(\widehat{A}\right)=60^{0}, AB=4\;\; cm\;\; AC=6\;\; cm

Demonstratie:

aplicatii cu teorema cosinusului

Observati ca suntem in cazul de congruenta  L.U.L. Astfel daca in triunghiul ABC aplicam Teorema cosinusului obtinem :

BC^{2}=AB^{2}+AC^{2}-2\cdot AB\cdot AC\cdot \cos\widehat{A}\Rightarrow

BC^{2}=4^{2}+6^{2}-2\cdot 4\cdot 6\cdot \cos 60^{0}

\Rightarrow BC^{2}=16+36-48\cdot\frac{1}{2}

\Rightarrow BC^{2}=52-24=28\Rightarrow BC=\sqrt{28}\Rightarrow BC=2\sqrt{7}

Fie ABC un triunghi dreptunghic in A si CB=26 cm, \sin B=\frac{12}{13}. Aflati Perimetrul triunghiului ABC

Demonstratie

Stim ca triunghiul ABC este dreptunghic in A, deci putem aplica notiunile trigonometrice invatate in clasele la mici, astfel avem ca:

 

cum aplicam functiile trigonometriceastfel avem ca:

\sin B=\frac{12}{13}\Rightarrow \frac{AC}{BC}=\frac{12}{13}\Rightarrow \frac{AC}{26}=\frac{12}{13}\Rightarrow 13\cdot AC=26\cdot 12\Rightarrow AC=\frac{26\cdot 12}{13}=\frac{2\cdot 12}{1}=24\;\; cm

Acum daca aplicam Teorema lui Pitagora in triunghiul dreptunghic ABC, gasim ca:

AB^{2}=BC^{2}-AC^{2}\Rightarrow AB^{2}=26^{2}-24^{2}\Rightarrow AB^{2}=676-576\Rightarrow AB^{2}=100\Rightarrow AB=\sqrt{100}\Rightarrow AB=10\;\; cm

 

Astfel

P_{\Delta ABC}=AB+AC+BC=10+24+26=34+26=60

3) Rezolvati triunghiul ABC in cazul:

R=4\;\; cm; A=\frac{2\pi}{3}, C=\frac{\pi}{12}

Observati ca in cazul de sus stim doua unghiuri, iar intr-un triunghi suma masurii unghiurilor este de \pi

Astfel avem ca

A+B+C=\pi\Rightarrow \frac{2\pi}{3}+B+\frac{\pi}{12}=\pi\Rightarrow B=\pi-\frac{2\pi}{3}-\frac{\pi}{12}\Rightarrow B=\frac{12\cdot \pi-4\cdot 2\pi-1\cdot \pi}{12}\Rightarrow B=\frac{12\pi-8\pi-\pi}{12}=\frac{3\pi}{12}^{(3}=\frac{\pi}{4}

Deci B=\frac{\pi}{4}

Acum in triunghiul ABC putem aplica Teorema sinusului:

\frac{BC}{\sin A}=\frac{AB}{\sin C}=\frac{AC}{\sin B}=2\cdot R\Rightarrow    \frac{BC}{\sin \frac{2\pi}{3}}=\frac{AB}{\sin \frac{\pi}{12}}=\frac{AC}{\sin\frac{\pi}{4}}=2\cdot 4

Astfel stim ca

\frac{AC}{\sin\frac{\pi}{4}}=2\cdot 4\Rightarrow \frac{AC}{\frac{\sqrt{2}}{2}}=8\Rightarrow AC=\frac{\sqrt{2}}{2}\cdot 8\Rightarrow AC=4\sqrt{2}

Dar acum stim si ca

\frac{AB}{\sin\frac{\pi}{12}}=8\Rightarrow AB=\sin \frac{\pi}{12}\cdot 8(*)

Dar mai intai sa aflam \sin \frac{\pi}{12}=\sin\left(\frac{\pi}{3}-\frac{\pi}{4}\right)=\sin\frac{\pi}{3}\cdot \cos\frac{\pi}{4}-\sin\frac{\pi}{4}\cdot \cos\frac{\pi}{3}=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\cdot \frac{1}{2}=\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\frac{\sqrt{6}-\sqrt{2}}{4}

Acum daca inlocuim in (*), obtinem ca:

AB=\frac{\sqrt{6}-\sqrt{2}}{4}\cdot 8=\frac{\sqrt{6}-\sqrt{2}}{1}\cdot 2=2\left(\sqrt{6}-\sqrt{2}\right)

Acum ca sa aflam BC, stim ca

\frac{BC}{sin\frac{2\pi}{3}}=8\Rightarrow BC=\sin\frac{2\pi}{3}\cdot 8=(**)

Dar mai intai calculam

\sin\frac{2\pi}{3}

Observati ca suntem in cadranul II, deci face reducerea la primul cadran si obtinem:

\sin\frac{2\pi}{3}=\sin\left(\pi-\frac{2\pi}{3}\right)=\sin\frac{3\pi-2\pi}{3}=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}

Deci in (**) obtinem ca:

BC=\sin\frac{2\pi}{3}\cdot 8=\sin\frac{\pi}{3}\cdot 8=\frac{\sqrt{3}}{2}\cdot 8=4\sqrt{3}

 

Teorema sinusului

Functiile trigonometrice ale unei sume si ale unei diferente de unghiuri

Astazi o sa invatam sa calculam Functiile trigonometrice ale unei sume si ale unei diferente de unghiuri, astfel

Teorema. Pentru oricare numere reale x si y au loc egalitatatile:

\cos{\left(x+y\right)}=\cos x\cdot \cos y-\sin x\cdot\sin y    \\ \cos{\left(x-y\right)}=\cos x\cdot\cos y+\sin x\cdot\sin y    \\ \ sin{\left(x+y\right)}=\sin x\cdot\cos y+\sin y\cdot \cos x    \\ \sin{\left(x-y\right)}=\sin x\cdot\cos y-\sin y\cdot\cos x

Consecinta:

Au loc relatiile:

\cos{\left(x+x\right)}=\cos x\cdot cos x-\sin x\cdot \sin x=\cos^{2} x-\sin^{2} x

Deci gasim ca \cos{ 2x}=\cos^{2} x-\sin^{2} x

Dar mai stim si ca: \sin{2x}=\sin{\left(x+x\right)}=\sin x\cdot \cos x+\sin x\cdot\cos x=2\sin x\cos x

Astfel avem ca: \sin{2x}=2\sin x\cos x

Pentru x\in R, unde R este multimea numerelor reale.

Teorema fundamentala a trigonometriei

cos^{2} x+\sin^{2} x=1

Observati ca cu ajutorul Teoremei fundamentale a trigonometriei, daca scoatem \sin^{2} x=1-\cos^{2} x obtinem

\cos{2x}=\cos^{2} x-\sin^{2} x=\cos^{2} x-\left(1-cos^{2} x\right)=cos^{2} x-1+\cos^{2} x=2\cos^{2} x-1

Sau

Daca scoatem din Teorema fundamentala a trigonometriei $latex \cos^{2} x=1-\sin^{2} x$ obtinem:

\cos{2x}=\cos^{2} x-\sin^{2} x=1-\sin^{2} x-\sin^{2} x=1-2\sin^{2} x

Exemplu:

Sa se calculeze  \cos{\left(a+b\right)}, \cos{\left(a-b\right)} daca:

a) \sin a=\frac{3}{5}, \sin b=\frac{5}{13}, a,b\in\left(0,\frac{\pi}{2}\right)

Observati ca stim sin a si sin b, deci trebuie sa aflam cos a si cos de b, pentru ca

cos{\left(a+b\right)}=\cos a\cdot \cos b-\sin a\cdot sin b

Astfel cu ajutorul Teoremei fundamentale a trigonometriei avem ca:

cos^{2} a+\sin^{2} a=1\Rightarrow \cos^{2} a+\left(\frac{3}{5}\right)^{2}=1\Rightarrow \cos^{2} a+\frac{9}{25}=1\Rightarrow \cos^{2} a=1-\frac{9}{25}\Rightarrow\cos^{2} a=\frac{25-9}{25}\Rightarrow cos^{2}=\frac{16}{25}\Rightarrow \cos a=\pm\sqrt{\frac{16}{25}}\Rightarrow \cos a=\pm\frac{4}{5}

In cazul nostru a\in \left(0,\frac{\pi}{2}\right), deci cosinusul este pozitiv, astfel gasim ca \cos a=\frac{4}{5}

Acum sa aflam cos b

Stim ca \cos^{2} b+\sin^{2} b=1\Rightarrow \cos^{2} b+\left(\frac{5}{13}\right)^{2}=1\Rightarrow \cos^{2} b=1-\frac{25}{169}\Rightarrow \cos^{2} b=\frac{169-25}{169}\Rightarrow \cos^{2} b=\frac{144}{169}\Rightarrow cos b=\pm\sqrt{\frac{144}{169}}\Rightarrow \cos b=\pm\frac{12}{13}

Stim ca b\in\left(0,\frac{\pi}{2}\right), deci cos b=\frac{12}{13}

Acum sa calculam \cos{\left(a+b\right)}=\cos a\cdot cos b-\sin a\cdot \sin b=\frac{4}{5}\cdot\frac{12}{13}-\frac{3}{5}\cdot\frac{5}{13}=\frac{48}{65}-\frac{15}{65}=\frac{33}{65}

Acum calculam \cos{\left(a-b\right)}=\cos a\cdot cos b+\sin a\cdot \sin b=\frac{4}{5}\cdot\frac{12}{13}+\frac{3}{5}\cdot\frac{5}{13}=\frac{48}{65}+\frac{15}{65}=\frac{63}{65}

Putem sa calculam si

\sin\left(a+b\right)=\sin a\cdot \cos b+\sin b\cdot\cos a=\frac{3}{5}\cdot \frac{12}{13}+\frac{5}{13}\cdot \frac{4}{5}=\frac{36}{65}+\frac{20}{65}=\frac{36+20}{65}=\frac{56}{65}.

b) \tan a=\frac{3}{4}, \cos b=\frac{5}{13}, a, b\in\left(0,\frac{\pi}{2}\right)

Cu teorema fundamentala a trigonometriei stim ca :

\cos^{2} b+\sin^{2} b=1\Rightarrow \left(\frac{5}{13}\right)^{2}+\sin^{2} b=1\Rightarrow \sin^{2} b=1-\frac{25}{169}\Rightarrow \sin^{2} b=\frac{169-25}{169}\Rightarrow \sin b=\pm\sqrt{\frac{144}{169}}\Rightarrow \sin b=\pm\frac{12}{13}

Cum b\in \left(0,\frac{\pi}{2}\right)\Rightarrow \sin b=\frac{12}{13}

Stim sin b, cos b, deci trebuie sa aflam sin a, cos a.

Stim ca \tan a=\frac{3}{4}

Mai stim si ca \tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\tan a\cdot\cos a

Iar cu teorema fundamentala a trigonometriei stim ca:

\sin^{2} a+\cos^{2} a=1\Rightarrow \tan^{2} a\cdot\cos^{2} a +cos^{2} a=1\Rightarrow

\cos^{2}\left(\tan^{2} a+1\right)=1\Rightarrow \cos^{2} a=\frac{1}{\tan^{2} a+1}\Rightarrow

\cos a=\pm\sqrt{\frac{1}{\tan^{2} a+1}}\Rightarrow

\cos a=\pm\frac{1}{\sqrt{\tan^{2} a+1}}

Noi stim ca

a\in\left(0,\frac{\pi}{2}\right), deci \cos a=\frac{1}{\sqrt{\tan^{2} a+1}}

Deci gasim

\cos a=\frac{1}{\sqrt{\left(\frac{3}{4}\right)^{2}+1}}=\frac{1}{\sqrt{\frac{9}{16}+1}}=\frac{1}{\sqrt{\frac{9+16}{16}}}=\frac{1}{\frac{\sqrt{25}}{\sqrt{16}}}=\frac{1}{\frac{5}{4}}=\frac{4}{5}

Acum sa aflam sin a

Stim ca \sin^{2} a+\cos^{2} a=1\Rightarrow \sin^{2} a+\left(\frac{4}{5}\right)^{2}=1\Rightarrow \sin^{2} a=1-\frac{16}{25}\Rightarrow \sin^{2} a=\frac{25-16}{25}\Rightarrow \sin a=\pm\sqrt{\frac{4}{25}}=\pm\frac{2}{5}

Cum a\in\left(0,\frac{\pi}{2}\right) deci \sin a=\frac{2}{5}

Acum calculam \cos{\left(a+b\right)}=\cos a\cdot\cos b-\sin a\cdot \sin b=\frac{4}{5}\cdot\frac{5}{13}-\frac{2}{5}\cdot\frac{12}{13}=\frac{20}{65}-\frac{24}{65}=\frac{20-24}{65}=-\frac{4}{65}

Iar \cos{\left(a-b\right)}=\cos a\cdot\cos b+\sin a\cdot \sin b=\frac{4}{5}\cdot\frac{5}{13}+\frac{2}{5}\cdot\frac{12}{13}=\frac{20}{65}+\frac{24}{65}=\frac{20+24}{65}=\frac{44}{65}