For a standard-position angle determined by the point (x,y), what are the values of the trigonometric functions. For the point (5,12) find csc theta and sec theta

Answer 1

To find the trigonometric functions for the point (5,12), we first need to determine the values of the hypotenuse and the adjacent side of the right triangle formed by the point (5,12) and the origin.

Using the Pythagorean theorem, we have:

\(r = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\).

Now, we can determine the values of the trigonometric functions:

\(\sin(\theta) = \frac{y}{r} = \frac{12}{13} = \frac{12}{13}\),
\(\cos(\theta) = \frac{x}{r} = \frac{5}{13} = \frac{5}{13}\),
\(\tan(\theta) = \frac{y}{x} = \frac{12}{5}\),
\(\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{\frac{12}{13}} = \frac{13}{12}\),
\(\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{\frac{5}{13}} = \frac{13}{5}\).

Therefore, the values of the trigonometric functions for the point (5,12) are:
\(\sin(\theta) = \frac{12}{13}\),
\(\cos(\theta) = \frac{5}{13}\),
\(\tan(\theta) = \frac{12}{5}\),
\(\csc(\theta) = \frac{13}{12}\),
\(\sec(\theta) = \frac{13}{5}\).