Question

Помогите решить lg (tg30°*tg32°*tg34°*...*tg58°*tg60°)

Answer 1

$lg(tg(30)*tg(32)*...*tg(58)*tg(60))$

$tg(\alpha)*tg(\beta) = \frac{cos(\alpha - \beta) - cos(\alpha + \beta)}{cos(\alpha - \beta) + cos(\alpha + \beta)}$

$tg(60) * tg(30) = \frac{cos(30) - cos(90)}{cos(30) + cos(90)} = \frac{cos(30)}{cos(30)} = 1 \\</p><p>tg(58) * tg(32) = \frac{cos(26) - cos(90)}{cos(26) + cos(90)} = \frac{cos(26)}{cos(26)} = 1 \\</p><p>... \\</p><p>tg(46)*tg(44) = \frac{cos(2) - cos(90)}{cos(2) + cos(90)} = \frac{cos(2)}{cos(2)} = 1$

$lg(tg(30)*tg(32)*...*tg(58)*tg(60)) = lg(1) = 0$