Question

Cho tam giác ABC có ba góc với : \(cot\left(\widehat{\dfrac{A}{2}}\right);cot\dfrac{\widehat{B}}{2};cot\left(\widehat{\dfrac{C}{2}}\right)\) theo thứ tự đó lập thành một cấp số cộng. Chứng minh ba cạnh tương ứng theo thứ tự đó cũng tạo thành một cấp số cộng

Answer 1

gọi a,b,c là 3 cạnh của tam giác.

Ta có :\(cot\left(\dfrac{A}{2}\right)+cot\left(\dfrac{C}{2}\right)=2cot\left(\dfrac{B}{2}\right)\) <=> \(\dfrac{cot\left(\dfrac{A}{2}\right)}{sin\left(\dfrac{A}{2}\right)}+\dfrac{cos\left(\dfrac{C}{2}\right)}{sin\left(\dfrac{C}{2}\right)}=\dfrac{2.cos\left(\dfrac{B}{2}\right)}{sin\left(\dfrac{B}{2}\right)}\)

<=> \(\dfrac{sin\left(\dfrac{C}{2}\right)cos\left(\dfrac{A}{2}\right)+cos\left(\dfrac{C}{2}\right)sin\left(\dfrac{A}{2}\right)}{sin\left(\dfrac{A}{2}\right).sin\left(\dfrac{C}{2}\right)}=2.\dfrac{cos\left(\dfrac{B}{2}\right)}{sin\left(\dfrac{C}{2}\right)}\)

<=> \(\dfrac{sin\left(\dfrac{A}{2}+\dfrac{C}{2}\right)}{sin\left(\dfrac{A}{2}\right)sin\left(\dfrac{C}{2}\right)}=2.\dfrac{cos\left(\dfrac{B}{2}\right)}{sin\left(\dfrac{B}{2}\right)}\) <=> \(\dfrac{cos\left(\dfrac{B}{2}\right)}{sin\left(\dfrac{A}{2}\right)sin\left(\dfrac{C}{2}\right)}=2.\dfrac{cos\left(\dfrac{B}{2}\right)}{sin\left(\dfrac{B}{2}\right)}\)

<=> \(sin\left(\dfrac{B}{2}\right).cos\left(\dfrac{B}{2}\right)=2sin\left(\dfrac{A}{2}\right)sin\left(\dfrac{C}{2}\right)cos\left(\dfrac{B}{2}\right)\)

<=> \(\dfrac{1}{2}sinB=\left[cos\left(\dfrac{A}{2}-\dfrac{C}{2}\right)-cos\left(\dfrac{A}{2}+\dfrac{C}{2}\right)\right]cos\left(\dfrac{B}{2}\right)\)

<=>\(\dfrac{1}{2}sinB=cos\left(\dfrac{A}{2}-\dfrac{C}{2}\right).cos\left(\dfrac{B}{2}\right)-sin\left(\dfrac{B}{2}\right)cos\left(\dfrac{B}{2}\right)\)

<=> \(\dfrac{1}{2}sinB=cos\left(\dfrac{A}{2}-\dfrac{C}{2}\right)sin\left(\dfrac{A}{2}+\dfrac{C}{2}\right)-\dfrac{1}{2}sinB\)

<=> sinB = \(\dfrac{1}{2}\left(sinA+sinC\right)\) <=> \(2sinB=sinA+sinC\)

<=> \(2.\dfrac{b}{2R}=\dfrac{a}{2R}+\dfrac{c}{2R}\)

<=> a+c =2b

=> 3 cạnh của tam giác tạo thành cấp số cộng.