A. \(\cos a + \cos b = 2\cos \frac{{a + b}}{2}\cos \frac{{a - b}}{2}\)
B. \(\sin a - \sin b = 2\cos \frac{{a + b}}{2}\sin \frac{{a - b}}{2}\)
C. \(\sin a + \sin b = 2\sin \frac{{a + b}}{2}\cos \frac{{a - b}}{2}\)
D. \(\cos a - \cos b = 2\sin \frac{{a + b}}{2}\sin \frac{{a - b}}{2}\)
D
Ta có: \(\left\{ \begin{array}{l}\cos a - \cos b = 2\cos \frac{{a + b}}{2}\cos \frac{{a - b}}{2}\\\sin a - \sin b = 2\cos \frac{{a + b}}{2}\sin \frac{{a - b}}{2}\\\sin a + \sin b = 2\sin \frac{{a + b}}{2}\cos \frac{{a - b}}{2}\\\cos a - \cos b = - 2\sin \frac{{a + b}}{2}\sin \frac{{a - b}}{2}\end{array} \right.\)
Vậy D sai.
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