Half-Angle Formulae in Trigonometry:
(i) In △ABC with BC = a, CA = b and AB = c,
The other two formulae can also be deduced similarly.
(ii) In △ABC with BC = a, CA = b and AB = c,
The other two formulae can also be deduced in the same manner.
(iii) In △ABC with BC = a, CA = b and AB = c,
Similarly, the other two of the set can also be deduced.
Example- In any △ABC, prove that bc cos2 (A/2) + ca cos2 (B/2) + ab cos2 (C/2) = s2, where 2s = a + b + c.
Solution:
| bc cos2 (A/2) + ca cos2 (B/2) + ab cos2 (C/2) = bc. s(s – a)/bc + ca. s(s – b)/ca + ab. s(s – c)/ab = s(s – a) + s(s – b) + s(s – c) = 3s2 – s(a + b + c) = 3s2 – s . 2s = 3s2 – 2s2 = s2 |
