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Trigonometric Equations

Prove that ...

Question

Prove that ${(a - b)}^{2} {cos}^{2} \frac{C}{2} + {(a + b)}^{2} {sin}^{2} \frac{C}{2} = c^{2}$

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Solution

L.H.S $= {(a - b)}^{2} {cos}^{2} \frac{C}{2} + {(a + b)}^{2} {sin}^{2} \frac{C}{2}$

expanding

$= (a^{2} + b^{2} - 2 a b) {cos}^{2} \frac{C}{2} + (a^{2} + b^{2} + 2 a b) {sin}^{2} \frac{C}{2}$

rearranging

$= (a^{2} + b^{2}) + 2 a b ({sin}^{2} \frac{C}{2} - {cos}^{2} \frac{C}{2})$
$= a^{2} + b^{2} - 2 a b cos C = c^{2} =$ R.H.S
Hence proved.

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