Question

Prove that $\cos A+\cos B+\cos C$ is always positive in triangle $ABC$.

Answer 1

Determine the proving of the $\cos A+\cos B+\cos C$ that's always the positive in triangle $ABC$.

Solve the L.H.S part:

$$\begin{gathered} cosA+cosB+cosC=(cosA+cosB)+cosC \\ \Rightarrow =2·cos\left[\frac{(A+B)}{2}\right]·cos\left[\frac{(A-B)}{2}\right]+cosC \\ \Rightarrow =2·cos\left[\right(\frac{\pi }{2})–(\frac{C}{2}\left)\right]·cos\left[\frac{(A-B)}{2}\right]+cosC \\ \Rightarrow =2·sin\left(\frac{C}{2}\right)·cos\left[\frac{(A-B)}{2}\right]+1–2·sin²\left(\frac{C}{2}\right) \\ \Rightarrow =1+2sin\left(\frac{C}{2}\right)·cos\left[\frac{(A-B)}{2}\right]–sin\left(\frac{C}{2}\right) \\ \Rightarrow =1+2sin\left(\frac{C}{2}\right)·cos\left[\frac{(A-B)}{2}\right]–sin\left[\right(\frac{\pi }{2})–(\frac{(A+B)}{2}\left)\right] \\ \Rightarrow =1+2sin\left(\frac{C}{2}\right)·cos\left[\frac{(A-B)}{2}\right]–cos\left[\frac{(A+B)}{2}\right] \\ \Rightarrow =1+2sin\left(\frac{C}{2}\right)·2sin\left(\frac{A}{2}\right)·sin\left(\frac{B}{2}\right)...............................\left(ii\right) \\ \Rightarrow =1+4sin\left(\frac{A}{2}\right)sin\left(\frac{B}{2}\right)sin\left(\frac{C}{2}\right) \end{gathered}$$

Hence, the given expression is positive.