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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}