[Toán 10] Lượng giác

L

lp_qt

$cosA+cosB+cosC+cos\dfrac{\pi }{3}$

$=2.cos\dfrac{A+B}{2}.cos\dfrac{A-B}{2}+2.cos(\dfrac{C}{2}+\dfrac{\pi }{6}).cos(\dfrac{C}{2}-\dfrac{\pi }{6})$

$\le 2.cos\dfrac{A+B}{2}+2.cos(\dfrac{C}{2}+\dfrac{\pi }{6})=4.cos(\dfrac{A+B+C}{12}+\dfrac{\pi }{12}).cos(\dfrac{A+B+C}{12}-\dfrac{\pi }{12})$

$\le 4.cos(\dfrac{A+B+C}{12}+\dfrac{\pi }{12})=4.cos\dfrac{\pi }{3}$

\Rightarrow $cosA+cosB+cosC+cos \le 3.cos\dfrac{\pi }{3}=\dfrac{3}{2}$

dấu = xảy ra khi $A=B=C$
 
L

lp_qt

cách khác

$cosA+cosB+cosC \le \dfrac{3}{2}$

\Leftrightarrow $2.cos\dfrac{A+B}{2}.cos\dfrac{A-B}{2}+1-2.sin^2\dfrac{C}{2}+\dfrac{3}{2} \le 0$

\Leftrightarrow $-2.sin^2\dfrac{C}{2}+2.sin\dfrac{C}{2}.cos\dfrac{A-B}{2}-\dfrac{1}{2} \le 0$

$f(t)=-2t^2+2t.cos\dfrac{A-B}{2}-\dfrac{1}{2}$

$\Delta '=cos^2\dfrac{A-B}{2}-1\le 0$

\Rightarrow $f(t) \le 0$
\Rightarrow đpcm
 
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