Nhận dạng tam giác

N

nhocngo976

thử làm...

<=>4 cosA/2cosB/2cosC/2 - 4sinA/2sinB/2sinC/2 =2

(có 4 sinA/2sinB/2sinC/2 =cosA+cosB+cosC-1 cái này bạn có thể dùng lun hoặc cần thì cm nha)

<=> 4cosA/2cosB/2cosC/2 + 1-(cosA+cosB+cosC)=2

<=>1+cosA+cosB+cosC -4 cosA/2cosB/2cosC/2 =0

(có 1+cosA+cosB+cosC) =-4cosA/2cosB/2cosC/2 )

<=> -4cosA/2cosB/2cosC/2-4cosA/2cosB/2cosC/2=0

<=> cosA/2cosB/2cosC/2 =0

đến đây bạn tự làm típ đi...có j k đúng thì pm cho mình y!h nhocngo_maruco@yahoo.com
 
T

tomcangxanh

[TEX]cos{\frac{A}{2}}. cos {\frac{B}{2}}. cos( \frac{C}{2})- sin { \frac{A}{2}}. sin {\frac{B}{2}} sin.{ \frac{C}{2}}= \frac{1}{2} [/TEX]

[TEX]\Leftrightarrow \frac{1}{2} [cos {\frac{A-B}{2}} + cos {\frac{A+B}{2}} ] cos {\frac{C}{2}} - \frac{1}{2} [ (cos {\frac{A-B}{2}} - cos{\frac{A+B}{2}} ]. sin {\frac{C}{2}}= \frac{1}{2}[/TEX]

[TEX]\Leftrightarrow sin {\frac{C}{2}}. cos {\frac{C}{2}} + cos {\frac{A-B}{2}} . cos {\frac{C}{2}} - cos {\frac{A-B}{2}} . sin {\frac{C}{2}} + sin^2 {\frac{C}{2}} =1[/TEX]

[TEX]\Leftrightarrow sin {\frac{C}{2}}. cos {\frac{C}{2}} + cos {\frac{A-B}{2}} sin {\frac{C}{2}} - cos^2 {\frac{C}{2}} =0[/TEX]

[TEX]\Leftrightarrow sin {\frac{C}{2}} [ cos {\frac{C}{2}} + cos {\frac{A-B}{2}}] - cos {\frac{C}{2}}[ cos {\frac{C}{2}} - cos {\frac{A-B}{2}}]=0[/TEX]

[TEX]\Leftrightarrow [sin{\frac{C}{2}} - cos {\frac{C}{2}} ] [cos {\frac{C}{2}} - cos {\frac{A-B}{2}} ]=0[/TEX]

[TEX]\Leftrightarrow \left[{sin {\frac{C}{2}} =cos {\frac{C}{2}}\\{cos {\frac{C}{2}} = cos {\frac{A-B}{2}}[/TEX]

[TEX]\Leftrightarrow \left[{ tan {\frac{C}{2}}=1\\{cos {\frac{C}{2}} = cos {\frac{A-B}{2}}[/TEX]

[TEX]\Leftrightarrow \left[{\frac{C}{2}=\frac{\pi}{4}\\\frac{A-B}{2}=\frac{C}{2}\\{ \frac{A-B}{2}=-\frac{C}{2}(0<A,B,C<\pi)[/TEX]

[TEX]\Leftrightarrow \left[{C=\frac{\pi}{2}\\{A=B+C\\{B=A+C[/TEX]

[TEX]\Rightarrow[/TEX] tam giác ABC vuông
 
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