$cmr:\Delta ABC$ $\text{đều:}$
$\to a;a+b+c=2(a\cos A+b\cos B+c\cos C)$
$\to b;\angle A=60^0,a=10,r=\frac{5\sqrt{3}}{3}$
$cmr:\Delta ABC$ $\text{đều:}$
$\to a;a+b+c=2(a\cos A+b\cos B+c\cos C)$
$\to b;\angle A=60^0,a=10,r=\frac{5\sqrt{3}}{3}$
Hẹn ngày tái ngộ VMF thân yêu !
$cmr:\Delta ABC$ $\text{đều:}$
$\to a;a+b+c=2(a\cos A+b\cos B+c\cos C)$
Hệ thức đã cho tương đương:
$2R(sinA+sinB+sinC)=2R.2sinAcosA+2R.2sinBcosB+2R.2sinCcosC$
$\Leftrightarrow sinA+sinB+sinC=sin2A+sin2B+sin2C$
Ta có:
$sin2A+sin2B+sin2C=\frac{1}{2}[(sin2A+sin2B)+(sin2B+sin2C)+(sin2C+sin2A)]$
$=\frac{1}{2}[2sin(A+B)cos(A-B)+2sin(B+C)cos(B-C)+2sin(C+A)(C-A)]$
$=sinCcos(A-B)+sinAcos(B-C)+sinBcos(C-A) \leq sinC+sinA+sinB$
Dấu bằng xảy ra khi và chỉ khi $A=B=C$ hay $\Delta ABC$ đều
Facebook: https://www.facebook.com/ntn3004
$cmr:\Delta ABC$ $\text{đều:}$
$\to a;a+b+c=2(a\cos A+b\cos B+c\cos C)$
Cách 2:
Hệ thức tương đương $4sinAsinBsinC=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}$
$\Leftrightarrow cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}=8sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}$
$\Leftrightarrow 1=8sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}$
$\Leftrightarrow 1=4\left [ cos\frac{A-B}{2}-cos\frac{A+B}{2} \right ]cos\frac{A+B}{2}$
$\Leftrightarrow \left ( 2cos\frac{A+B}{2}-cos\frac{A-B}{2} \right )^2+sin^2\frac{A-B}{2}=0$
$\Leftrightarrow A=B=C\Leftrightarrow \Delta ABC$ đều
Edited by Forgive Yourself, 05-01-2015 - 15:47.
Facebook: https://www.facebook.com/ntn3004
0 members, 1 guests, 0 anonymous users