Cho $ΔABC$ có ba góc với: $\cot(\widehat{\dfrac{A}{2}})$; $\cot(\widehat{\dfrac{B}{2}})$; $\cot(\widehat{\dfrac{C}{2}})$ theo thứ tự đó lập thành một

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1. Giải thích các bước giải:

   Ta có:

   $\cot\left(\dfrac{A}{2}\right), \cot\left(\dfrac{B}{2}\right), \cot\left(\dfrac{C}{2}\right)$ lập thành cấp số cộng

   $\to \cot\left(\dfrac{A}{2}\right)+\cot\left(\dfrac{C}{2}\right)=2\cot\left(\dfrac{B}{2}\right)$

   $\to \dfrac{\cos\left(\dfrac{A}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}+\dfrac{\cos\left(\dfrac{C}{2}\right)}{\sin\left(\dfrac{C}{2}\right)}=2\dfrac{\cos\left(\dfrac{B}{2}\right)}{\sin\left(\dfrac{B}{2}\right)}$

   $\to \dfrac{\cos\left(\dfrac{A}{2}\right)\cdot\sin\left(\dfrac{C}{2}\right)+\cos\left(\dfrac{C}{2}\right)\cdot\sin\left(\dfrac{A}{2}\right)}{\sin\left(\dfrac{A}{2}\right)\cdot\sin\left(\dfrac{C}{2}\right)}=\dfrac{2\cos\left(\dfrac{B}{2}\right)}{\sin\left(\dfrac{B}{2}\right)}$

   $\to \dfrac{\cos\left(\dfrac{A}{2}\right)\cdot\sin\left(\dfrac{C}{2}\right)+\cos\left(\dfrac{C}{2}\right)\cdot\sin\left(\dfrac{A}{2}\right)}{\sin\left(\dfrac{A}{2}\right)\cdot\sin\left(\dfrac{C}{2}\right)}=\dfrac{2\cos\left(\dfrac{B}{2}\right)}{\sin\left(\dfrac{B}{2}\right)}$

   $\to \dfrac{\sin\left(\dfrac{A}{2}+\dfrac{C}{2}\right)}{\sin\left(\dfrac{A}{2}\right)\cdot\sin\left(\dfrac{C}{2}\right)}=\dfrac{2\sin\left(90^o-\dfrac{B}{2}\right)}{\cos\left(90^o-\dfrac{B}{2}\right)}$

   $\to \dfrac{\sin\left(\dfrac{A+C}{2}\right)}{\sin\left(\dfrac{A}{2}\right)\cdot\sin\left(\dfrac{B}{2}\right)}=\dfrac{2\sin\left(\dfrac{A+C}{2}\right)}{\cos\left(\dfrac{A+C}{2}\right)}$

   $\to 2\sin\left(\dfrac{A}{2}\right)\cdot\sin\left(\dfrac{C}{2}\right)=\cos\left(\dfrac{A+C}{2}\right)$

   $\to -\left(\cos\left(\dfrac{A+C}{2}\right)-\cos\left(\dfrac{A-C}{2}\right)\right)=\cos\left(\dfrac{A+C}{2}\right)$

   $\to \cos\left(\dfrac{A-C}{2}\right)=2\cos\left(\dfrac{A+C}{2}\right)$

   $\to \cos\left(\dfrac{A-C}{2}\right)\cdot\sin\left(\dfrac{A+C}{2}\right)=2\cos\left(\dfrac{A+C}{2}\right)\cdot\sin\left(\dfrac{A+C}{2}\right)$

   $\to \dfrac12\left(\sin\left(\dfrac{A-C}{2}+\dfrac{A+C}{2}\right)-\sin\left(\dfrac{A-C}{2}-\dfrac{A+C}{2}\right)\right)=\sin\left(A+C\right)$

   $\to \dfrac12\left(\sin\left(A\right)-\sin\left(-C\right)\right)=\sin\left(180^o-B\right)$

   $\to \dfrac12\left(\sin\left(A\right)+\sin\left(C\right)\right)=\sin\left(B\right)$

   $\to \sin\left(A\right)+\sin\left(C\right)=2\sin\left(B\right)$

   $\to \sin\left(A\right)\cdot 2R+\sin\left(C\right)\cdot 2R=2\sin\left(B\right)\cdot 2R$

   $\to a+c=2b$

   $\to a,b,c$ là cấp số cộng

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