[Ne[S]zA]

ข้อ3) $\sin\frac{25\pi}{6}+\cos\frac{25\pi}{3}-\tan\frac{25\pi}{4}$
$=\sin(4\pi+\frac{\pi}{6})+\cos(8\pi+\frac{\pi}{3})-\tan(6\pi+\frac{\pi}{4})$
$=\sin\frac{\pi}{6}+\cos\frac{\pi}{3}-\tan\frac{\pi}{4}$
$=\cos\frac{\pi}{2}$