Proof of the trig identities for half-angles

[chwala]

Homework Statement

see attached

Relevant Equations

Trigonometry

I was just checking this out the sin$\frac {A}{2}$ property, in doing so i picked a Right-Angled triangle, say $ABC$, with $AB=5cm$, $BC=4cm$ and $CA= 3cm$. From this i have,
$s=6cm$ now substituting this into the formula,
$sin\frac {A}{2}$= $\frac {1×3}{5×3}$=$\frac {3}{15}$=$0.2$
giving us angle $A=23.06^0$ which does not look correct to me because,

angle $A=53.13^0$ ...using trigonometry directly...$sin{A}$= $\frac {4}{5} =0.8$... i suspect a mix up in the indicated property...or i may have made a mistake.
i will amend my latex later using phone...


aaaaaaaaargh these guys are missing the square root sign! I had to check that from google.

 

[lurflurf]

The left hand sides should be squared or the right hand sides square-rooted
ie
$$\sin^2\left(\frac{A}{2}\right)=\frac{(s-b)(s-c)}{b c}$$
or
$$\sin\left(\frac{A}{2}\right)=\sqrt{\frac{(s-b)(s-c)}{b c}}$$