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

Publicat în aprilie 11, 2018mai 8, 2018 de MATEPEDIA

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

$\cos a+\cos b=2\cos\frac{a+b}{2}\cdot\cos\frac{a-b}{2}$
$\cos a-\cos b=-2\sin\frac{a-b}{2}\cdot\sin\frac{a+b}{2}$
$\sin a+\sin b=2\sin\frac{a+b}{2}\cdot\cos\frac{a-b}{2}$
$\sin a-\sin b=2\sin\frac{a-b}{2}\cdot\cos\frac{a+b}{2}$

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 $a=2R\cdot \sin A$ , $B=2R\cdot \sin B$ , $c=2R\cdot \sin C$

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

Publicat în iunie 16, 2014iulie 17, 2014 de MATEPEDIA

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:

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

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

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

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

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

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

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.

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:

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:

astfel 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:

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

Functiile trigonometrice ale unei sume si ale unei diferente de unghiuri

Publicat în martie 20, 2014 de MATEPEDIA

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

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

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

Mate Pedia

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