## 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 cos^{2} (A/2) + ca cos^{2} (B/2) + ab cos^{2} (C/2) = s^{2}, where 2s = a + b + c.

**Solution:**

bc cos^{2} (A/2) + ca cos^{2} (B/2) + ab cos^{2} (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) = 3s ^{2} – s(a + b + c) = 3s^{2} – s . 2s = 3s^{2} – 2s^{2} = s^{2} |