[กิตติ]

ลองวาดรูปตามโจทย์



จะได้ว่า $\theta =\hat A +\hat C =\pi-\hat B$
$\frac{\theta}{2}=\frac{\pi}{2}-\frac{\hat B}{2} $
$\sin (\frac{\theta}{2})=\cos (\frac{\hat B}{2})$
$\cos (\frac{\theta}{2})=\sin (\frac{\hat B}{2})$

$\cos \hat B=\frac{21^2+10^2-17^2}{2(21)(10)} $
$=\frac{3}{5} $
$\sin \hat B=\frac{4}{5}$

$\cos \hat B=2\cos^2 (\frac{\hat B}{2})-1$
$\cos (\frac{\hat B}{2})=\frac{2}{\sqrt{5} } $
$\sin (\frac{\hat B}{2})=\frac{1}{\sqrt{5} }$

$\sin (\frac{\theta}{2})-\cos (\frac{\theta}{2})=\cos (\frac{\hat B}{2})-\sin (\frac{\hat B}{2})$
$=\frac{1}{\sqrt{5} }$