Question

Найдите tg2α , если cosα=4/5 и 0 меньше α меньше π/2

Answer 1

$0\ \textless \ \alpha \ \textless \ \frac{\pi}{2}$ - I четверть. В первой четверти все тригонометрические функции положительные

$\sin \alpha = \sqrt{1-\cos^2 \alpha } = \sqrt{1- \frac{16}{25} } = \frac{3}{5}$

Окончательное получим

$tg2\alpha = \frac{\sin2\alpha }{\cos2\alpha }= \frac{2\sin\alpha \cos\alpha }{\cos^2\alpha -\sin^2\alpha }= \frac{2\cdot \frac{3}{5}\cdot\frac{4}{5} }{(\frac{4}{5} )^2-(\frac{3}{5} )^2} = \frac{2\cdot3\cdot 4}{16-9} = \frac{24}{7}$

Answer 2

To find tg(2α) using the given information, we can use the double-angle trigonometric identity for tangent:

tg(2α) = 2 * tg(α) / (1 - tg²(α))

First, let's find tg(α) using the given value of cos(α):

cos²(α) + tg²(α) = 1

Since cos(α) = 4/5, we can use the Pythagorean identity:

tg²(α) = 1 - cos²(α) tg²(α) = 1 - (4/5)² tg²(α) = 1 - 16/25 tg²(α) = 9/25

Now, taking the positive square root (because 0 < α < π/2):

tg(α) = √(9/25) = 3/5

Now, we can find tg(2α):

tg(2α) = 2 * tg(α) / (1 - tg²(α)) tg(2α) = 2 * (3/5) / (1 - (3/5)²) tg(2α) = 2 * (3/5) / (1 - 9/25) tg(2α) = 2 * (3/5) / (16/25) tg(2α) = 2 * (3/5) * (25/16) tg(2α) = (3/5) * (25/8) tg(2α) = 75/40 tg(2α) = 15/8

So, tg(2α) = 15/8.