Answer:
In ∆ABC, prove that
[tex]sin \frac{A}{2} + sin \frac{B}{2} - sin \frac{C}{2} =
- 1 + 4 cos \frac{\pi - A}{4} cos \frac{\pi - B}{4} sin \frac{\pi - C}{2}
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Answers & Comments
Answer:
In ∆ABC, prove that
[tex]sin \frac{A}{2} + sin \frac{B}{2} - sin \frac{C}{2} =
- 1 + 4 cos \frac{\pi - A}{4} cos \frac{\pi - B}{4} sin \frac{\pi - C}{2}