Tính:

A = sin(a – 17°)cos(a + 13°) – sin(a + 13°)cos(a – 17°);

\(B = cos\left( {b + \frac{\pi }{3}} \right)\cos \left( {\frac{\pi }{6} - b} \right) - \sin \left( {b + \frac{\pi }{3}} \right)\sin \left( {\frac{\pi }{6} - b} \right)\).

Ta có:

A = sin(a – 17°)cos(a + 13°) – sin(a + 13°)cos(a – 17°)

    = sin(a – 17°)cos(a + 13°) – cos(a – 17°)sin(a + 13°)

    = sin[(a – 17°) – (a + 13°)]

    = sin(a – 17° – a – 13°)

    = sin(‒30°)

    = ‒ sin30°

    \( = - \frac{1}{2}\).

\(B = cos\left( {b + \frac{\pi }{3}} \right)\cos \left( {\frac{\pi }{6} - b} \right) - \sin \left( {b + \frac{\pi }{3}} \right)\sin \left( {\frac{\pi }{6} - b} \right)\)

   \[ = cos\left[ {\left( {b + \frac{\pi }{3}} \right) + \left( {\frac{\pi }{6} - b} \right)} \right]\]

   \[ = cos\left[ {b + \frac{\pi }{3} + \frac{\pi }{6} - b} \right]\]

   \[ = cos\frac{\pi }{2} = 0\].