Cho cos(a + 2b) = 2cos a. Chứng minh rằng: tan(a + b) tan b = \(\frac{{ - 1}}{3}\).

Ta có cos(a + 2b) = 2cos a

⇔ cos[(a + b) + b] = 2cos[(a + b) – b]

⇔ cos(a + b) . cos b – sin(a + b) . sin b = 2[cos(a + b) . cos b + sin(a + b) . sin b]

⇔ cos(a + b) . cos b – 2 cos(a + b) . cos b = 2 sin(a + b) . sin b + sin(a + b) . sin b

⇔ – cos(a + b) . cos b = 3 sin(a + b) . sin b

⇔ sin(a + b) . sin b = \( - \frac{1}{3}\) cos(a + b) . cos b

\( \Leftrightarrow \frac{{\sin \left( {a + b} \right)\sin b}}{{\cos \left( {a + b} \right)\cos b}} = - \frac{1}{3}\)