Problem 3 - Precalculus Essentials

Question

In Exercises \(1-8,\) a point on the terminal side of angle \(\theta\) is given. Find the exact value of each of the six trigonometric functions of \(\theta\). $$(2,3)$$

Step-by-Step Solution

Answer

The exact values of the six trigonometric functions for the given point (2,3) are: \(sin = \frac{3}{\sqrt{13}}\), \(cos = \frac{2}{\sqrt{13}}\), \(tan = \frac{3}{2}\), \(csc = \frac{\sqrt{13}}{3}\), \(sec = \frac{\sqrt{13}}{2}\), \(cot = \frac{2}{3}\).

Step 1: Find the hypotenuse

The hypotenuse of the right triangle is given by the square root of the sum of the squares of the two other sides. So, apply the Pythagorean theorem, \(r = \sqrt{x^2 + y^2}\), to get \(r = \sqrt{2^2 + 3^2} = \sqrt{13}\).

Step 2: Find sine, cosine and tangent

Once you have the length of the sides ready, you can then calculate the sine, cosine, and tangent of the angle using their respective formulae. The sine (sin) is \(\frac{y}{r} = \frac{3}{\sqrt{13}}\), the cosine (cos) is \(\frac{x}{r} = \frac{2}{\sqrt{13}}\), and the tangent (tan) is \(\frac{y}{x} = \frac{3}{2}\).

Step 3: Find the reciprocals

The reciprocals of sine, cosine, and tangent yield the other three trigonometric functions. The cosecant (csc) is the reciprocal of sin, so \(csc = \frac{r}{y} = \frac{\sqrt{13}}{3}\). The secant (sec) is the reciprocal of cos, so \(sec = \frac{r}{x} = \frac{\sqrt{13}}{2}\). The cotangent (cot) is the reciprocal of tan, so \(cot = \frac{x}{y} = \frac{2}{3}\).