Saved Bookmarks
| 1. |
. What is the value of\[\cos \left(\frac{A+B}{2}\right) \cos \left(\frac{B+C}{2}\right) \cos \left(\frac{C+A}{2}\right) ?\] |
|
Answer» If ABC is a triangle Then \(A + B + C = \pi\) \(\therefore cos \left(\frac{A + B}2\right) cos \left(\frac{B +C}2\right) cos\left(\frac{C +A}2\right)\) \(= cos \left(\frac{\pi - C}2\right) cos \left(\frac{\pi - A}2\right) cos\left(\frac{\pi - B}2\right)\) \(= cos \left(\frac\pi2 - \frac C2\right) cos \left(\frac\pi2 - \frac A2\right) cos\left(\frac\pi2 - \frac B2\right)\) \(= sin\left(\frac C2\right) sin\left(\frac A2\right) sin\left(\frac B2\right)\) |
|