Question

Дано: cos t= -5/13 pi/2 < t < pi Найти: sin t,tg t и ctg t

Answer 1

$sin \alpha = \sqrt{1-cos^2 \alpha } \\ \\ sint= \sqrt{1-\frac{25}{169} } = \sqrt{ \frac{169-25}{169} } = \sqrt{ \frac{144}{169} }=\frac{12}{13} \\ \\ tg \alpha = \frac{sin \alpha }{cos \alpha } \\ \\ tgt= \frac{\frac{12}{13}}{\frac{-5}{13} } =\frac{12}{13}*\frac{13}{-5}=\frac{12}{-5}=-2,4 \\ \\ ctg \alpha = \frac{cos \alpha }{sin \alpha } \\ \\ ctgt=\frac{\frac{-5}{13}}{\frac{12}{13}}=\frac{-5}{13}*\frac{13}{12}=\frac{-5}{12}=-\frac{5}{12}$

Answer 2

Given Information:

We are given that:

cos(t) = -5/13 π/2 < t < π

Finding sin(t), tan(t), and cot(t):

To find sin(t), we can use the identity sin^2(t) + cos^2(t) = 1.

Since we know cos(t) = -5/13, we can substitute it into the identity:

sin^2(t) + (-5/13)^2 = 1

Simplifying the equation:

sin^2(t) + 25/169 = 1

Subtracting 25/169 from both sides:

sin^2(t) = 144/169

Taking the square root of both sides:

sin(t) = ±12/13

Since we know that π/2 < t < π, t must be in the second or third quadrant. In these quadrants, sin(t) is negative. Therefore, sin(t) = -12/13.

To find tan(t), we can use the identity tan(t) = sin(t)/cos(t). Substituting the values we found:

tan(t) = (-12/13) / (-5/13)

Dividing fractions:

tan(t) = 12/5

To find cot(t), we can use the identity cot(t) = 1 / tan(t). Substituting the value we found:

cot(t) = 1 / (12/5)

Dividing fractions:

cot(t) = 5/12

Solution:

For the given values of t (π/2 < t < π), we have:

sin(t) = -12/13 tan(t) = 12/5 cot(t) = 5/12