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Aratati ca in orice triunghi \( ABC \) are loc relatia:

\( \sin ^2\ \frac{A-B}{2}+\sin \ A\sin \ B+\sin ^2\ \frac{C}{2}=1 \)

ON 1980

Egalitatea se scrie:

\( \left\(\sin\frac{A}{2}\cos\frac{B}{2}-\sin\frac{B}{2}\cos\frac{A}{2}\right\)^2+4\sin\frac{A}{2}\cos\frac{A}{2}\sin\frac{B}{2}\cos\frac{B}{2}+\left\(\cos\frac{A}{2}\cos\frac{B}{2}-\sin\frac{A}{2}\sin\frac{B}{2}\right\)^2=1 \), deoarece \( \sin\frac{C}{2}=\cos\frac{A+B}{2} \)

\( \Longleftrightarrow \sin^2\frac{A}{2}\cos^2\frac{B}{2}+\sin^2\frac{B}{2}\cos^2\frac{A}{2}+\cos^2\frac{A}{2}\cos^2\frac{B}{2}+\sin^2\frac{A}{2}\sin^2\frac{B}{2}=1 \)

\( \Longleftrightarrow \sin^2\frac{A}{2}+\co^2\frac{A}{2}=1 \), adevarat.

alex2008 wrote:Aratati ca in orice triunghi \( ABC \) are loc relatia \( \sin ^2\ \frac{A-B}{2}+\sin \ A\sin \ B+\sin ^2\ \frac{C}{2}=1 \) (ON 1980).

Comentariu. Obisnuiam sa spun elevilor mei ca oridecateori intalnesc un produs (in particular puteri !)

de tipul \( \sin x\sin y \), \( \sin x\cos y \) sau \( \cos x\cos y \) sa inmulteasca cu doi relatia si sa aplice pachetul de

formule care exprima asemenea produse printr-o suma de functii trigonometrice de acelasi tip, adica

\( \underline{\overline{\left\|\ \begin{array}{c}
2\sin x\cos y=\sin (x+y)+\sin (x-y)\\\\\\\\
2\sin x\sin y=\cos (x-y)-\cos (x+y)\\\\\\\\
2\cos x\cos y=\cos (x+y)+\cos (x-y)\\\\\\\\
================\\\\\\\\
2\sin^2 x=1-\cos 2x\\\\\\\\
2\cos^2x=1+\cos 2x\\\\\\\\
2\sin x\cos x=\sin 2x\end{array}\ \right\|}} \) . Se observa ca \( \underline{\overline{\left\|\begin{array}{c}
\tan x=\frac {\sin 2x}{1+\cos 2x}=\frac {1-\cos 2x}{\sin 2x}\\\\\\\\
\sin^2x-\sin^2y=\sin (x+y)\sin (x-y)\end{array}\right\|}} \) .

Sa aplicam aceasta recomandare exercitiului propus de Alex. Nu prea imi vine sa cred ca este

de la ON - 1980. Nici pe vremea mea 1956-1960 (liceu) nu s-au dat asemenea probleme ...

Metoda 1. \( \overline{\underline {\left\|\begin{array}{c}
\sin ^2\ \frac{A-B}{2}+\sin A\sin B+\sin ^2\ \frac{C}{2}=1\\\\\\\\
2\sin ^2\ \frac{A-B}{2}+2\sin A\sin B+2\sin ^2\ \frac{C}{2}=2\\\\\\\\
\left[\underline 1-\underline{\underline{\cos (A-B)}}\right]+\left[\underline{\underline{\cos (A-B)}}-\underline{\underline{\underline{\cos (A+B)}}}\right]+\left[\underline 1-\underline{\underline{\underline{\cos C}}}\right]=\underline 2\\\\\\\\
\mathrm {O\ .\ K\ .}\end{array}\right\|}} \)

deoarece \( (A+B)+C=\pi\ \Longrightarrow\ \cos (A+B)=-\cos C \) .

Metoda 2. \( \overline{\underline {\left\|\begin{array}{c}
\sin ^2\ \frac{A-B}{2}+\sin A\sin B+\sin ^2\ \frac{C}{2}=1\\\\\\\\
\sin A\sin B=\sin^2\ \frac {A+B}{2}-\sin^2\ \frac {A-B}{2}\\\\\\\\
\sin A\sin B=\sin\ \frac {(A+B)+(A-B)}{2}\ \sin\ \frac {(A+B)-(A-B)}{2}\\\\\\\\
\mathrm {O\ .\ K\ .}\end{array}\right\|}} \)

deoarece \( \sin^2x-\sin^2y=\sin (x+y)\sin (x-y) \) si \( \frac {A+B}{2}+\frac C2=\frac {\pi}{2}\ \Longrightarrow\ \cos\frac C2=\sin\frac {A+B}{2} \) .

Va propun sa retineti si sa dovediti :

\( 1^{\circ}\ \ \overline{\underline{\left\|\begin{array}{c}
p(p-a)+(p-b)(p-c)=bc\\\\\\\\
p(p-a)-(p-b)(p-c)=bc\cdot \cos A\end{array}\right\|}}\ \Longrightarrow\ \underline{\overline{\left\|\begin{array}{c}
\sin\frac A2=\sqrt {\frac {p-b)(p-c)}{bc}}\\\\\\\\
\cos\frac A2=\sqrt {\frac {p(p-a)}{bc}}\\\\\\\\
\tan\frac A2=\sqrt {\frac {(p-b)(p-c)}{p(p-a)}}\end{array}\right\|}} \)

\( 2^{\circ}\ \ \underline{\overline{\left\|\begin{array}{c}
a=b\cdot\cos C+c\cdot \cos B\\\\\\\\
b=c\cdot\cos A+a\cdot\cos C\\\\\\\\
c=a\cdot\cos B+b\cdot\cos A\end{array}\right\|}}\ \left|\begin{array}{cccc}
\odot & (-a) & a & a\\\\\\\\
\odot & b & (-b) & b\\\\\\\\
\odot & c & c & (-c)\end{array}\right|\ \bigoplus\ \Longrightarrow\ \underline{\overline{\left\|\begin{array}{c}
-a^2+b^2+c^2=2bc\cdot\cos A\\\\\\\\
a^2-b^2+c^2=2ca\cdot\cos B\\\\\\\\
a^2+b^2-c^2=2ab\cdot\cos C\end{array}\right\|}} \)