ตรีโกณพิสูจน์

[Pakpoom]

กำหนดให้ A+B+C=180 จงพิสูจน์ว่า
1.$cos^2A+cos^2B+cos^2C+2cosAcosBcosC=1$
2.$sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}=1-2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}$
3.$tanA+tanB+tanC=tanAtanBtanC$
4.$cotBcotC+cotCcotA+cotAcotB=1$
5.$tan\frac{B}{2}\frac{C}{2}+tan\frac{C}{2}tan\frac{A}{2}+tan\frac{A}{2}tan\frac{B}{2}=1$

[~ArT_Ty~]

ข้อ 3 ใช้สูตร Tan ผลบวกมุมครับ

[sahaete]

แนวคิด ข้อ 1
$\begin{array}{l}
{\cos ^2}A + {\cos ^2}B + {\cos ^2}C\\
= \dfrac{1}{2}\left( {1 + \cos 2A} \right) + \dfrac{1}{2}\left( {1 + \cos 2B} \right) + \cos C\cos C\\
= 1 + \dfrac{1}{2}\left( {\cos 2A + \cos 2B} \right) + \cos C\cos \left[ {\pi - \left( {A + B} \right)} \right]\\
= 1 + \dfrac{1}{2}\left( {2\cos \dfrac{{2A + 2B}}{2}\cos \dfrac{{2A - 2B}}{2}} \right) - \cos C\cos \left( {A + B} \right)\\
= 1 + \cos \left( {A + B} \right)\cos \left( {A - B} \right) - \cos C\cos \left( {A + B} \right)\\
= 1 + \cos \left( {\pi - C} \right)\cos \left( {A - B} \right) - \cos C\cos \left( {A + B} \right)\\
= 1 - \cos C\cos \left( {A - B} \right) - \cos C\cos \left( {A + B} \right)\\
= 1 - \cos C\left[ {\cos \left( {A - B} \right) + \cos \left( {A + B} \right)} \right]\\
= 1 - \cos C\left[ {2\cos \dfrac{{\left( {A - B} \right) + \left( {A + B} \right)}}{2}\cos \dfrac{{\left( {A - B} \right) - \left( {A + B} \right)}}{2}} \right]\\
= 1 - 2\cos C\cos A\cos B
\end{array}$

[sahaete]

แนวคิด ข้อ 2
$\begin{array}{l}
{\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}\\
= \frac{1}{2}\left( {1 - \cos A} \right) + \frac{1}{2}\left( {1 - \cos B} \right) + \sin \frac{C}{2}\sin \frac{C}{2}\\
= 1 - \frac{1}{2}\left( {\cos A + \cos B} \right) + \sin \frac{C}{2}\sin \left( {\frac{\pi }{2} - \frac{{A + B}}{2}} \right)\\
= 1 - \frac{1}{2}\left( {2\cos \frac{{A + B}}{2}\cos \frac{{A - B}}{2}} \right) + \sin \frac{C}{2}\cos \frac{{A + B}}{2}\\
= 1 - \cos \left( {\frac{\pi }{2} - \frac{C}{2}} \right)\cos \frac{{A - B}}{2} + \sin \frac{C}{2}\cos \frac{{A + B}}{2}\\
= 1 - \sin \frac{C}{2}\cos \frac{{A - B}}{2} + \sin \frac{C}{2}\cos \frac{{A + B}}{2}\\
= 1 - \sin \frac{C}{2}\left( {\cos \frac{{A - B}}{2} -\cos \frac{{A + B}}{2}} \right)\\
= 1 - \sin \frac{C}{2}\left( {2\sin \frac{A}{2}\sin \frac{B}{2}} \right)\\
= 1 - 2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}
\end{array}$

[sahaete]

แนวคิด ข้อ 3
$\begin{array}{l}
\tan \left( {A + B} \right) = \tan \left( {\pi - C} \right)\\
\dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}} = - \tan C\\
\tan A + \tan B = - \tan C + \tan A\tan B\tan C\\
\tan A + \tan B + \tan C = \tan A\tan B\tan C
\end{array}$

[sahaete]

แนวคิด ข้อ 4
$\begin{array}{l}
\cot \left( {A + B} \right) = \cot \left( {\pi - C} \right)\\
\dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}} = - \cot C\\
\cot A\cot B - 1 = - \cot C\cot A - \cot C\cot B\\
\cot A\cot B + \cot B\cot C + \cot C\cot A = 1
\end{array}$

[sahaete]

แนวคิด ข้อ 5
$\begin{array}{l}
A + B = \pi - C\\
\dfrac{A}{2} + \dfrac{B}{2} = \dfrac{\pi }{2} - \dfrac{C}{2}\\
\tan \left( {\dfrac{A}{2} + \dfrac{B}{2}} \right) = \tan \left( {\dfrac{\pi }{2} - \dfrac{C}{2}} \right)\\
\dfrac{{\tan \dfrac{A}{2} + \tan \dfrac{B}{2}}}{{1 - \tan \dfrac{A}{2}\tan \dfrac{B}{2}}} = - \tan \dfrac{C}{2}\\
\tan \dfrac{A}{2} + \tan \dfrac{B}{2} = - \tan \dfrac{C}{2} + \tan \dfrac{A}{2}\tan \dfrac{B}{2}\tan \dfrac{C}{2}\\
\tan \dfrac{A}{2} + \tan \dfrac{B}{2} + \tan \dfrac{C}{2} = \tan \dfrac{A}{2}\tan \dfrac{B}{2}\tan \dfrac{C}{2}
\end{array}$