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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}