Problem 9 - Precalculus

Question

The terminal side of an angle \(\theta\) in standard position passes through values of the six trigonometric functions for angle \(\theta\) $$(-\sqrt{2}, \sqrt{3})$$

Step-by-Step Solution

Answer

The trigonometric functions are: \(\sin \theta = \frac{\sqrt{3}}{\sqrt{5}}, \cos \theta = \frac{-\sqrt{2}}{\sqrt{5}}, \tan \theta = -\frac{\sqrt{3}}{\sqrt{2}}; \csc \theta = \frac{\sqrt{5}}{\sqrt{3}}, \sec \theta = \frac{-\sqrt{5}}{\sqrt{2}}, \cot \theta = -\frac{\sqrt{2}}{\sqrt{3}}.\)

Step 1: Understand the Coordinates

The terminal side of the angle \(\theta\) passes through the point \((-\sqrt{2}, \sqrt{3})\). This means that the \(x\)-coordinate is \(-\sqrt{2}\) and the \(y\)-coordinate is \(\sqrt{3}\).

Step 2: Compute the Radius (r)

To find the radius \(r\), which is the distance from the origin to the point \((-\sqrt{2}, \sqrt{3})\), use the formula for the distance of a point from the origin: \[ r = \sqrt{(-\sqrt{2})^2 + (\sqrt{3})^2} = \sqrt{2 + 3} = \sqrt{5}. \]

Step 3: Determine Sine (sin)

The sine function is defined as the ratio of the opposite side to the hypotenuse in a right triangle. Here it is \(\sin \theta = \frac{y}{r} = \frac{\sqrt{3}}{\sqrt{5}} = \frac{\sqrt{15}}{5}.\)

Step 4: Determine Cosine (cos)

The cosine function is the ratio of the adjacent side to the hypotenuse. It is given by \(\cos \theta = \frac{x}{r} = \frac{-\sqrt{2}}{\sqrt{5}} = -\frac{\sqrt{10}}{5}.\)

Step 5: Determine Tangent (tan)

Tangent is the ratio of the opposite side to the adjacent side. It is computed by \(\tan \theta = \frac{y}{x} = \frac{\sqrt{3}}{-\sqrt{2}} = -\frac{\sqrt{6}}{2}.\)

Step 6: Find Cosecant (csc)

Cosecant is the reciprocal of sine: \(\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{\sqrt{3}} = \frac{\sqrt{15}}{3}.\)

Step 7: Find Secant (sec)

Secant is the reciprocal of cosine: \(\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{-\sqrt{2}} = -\frac{\sqrt{10}}{2}.\)

Step 8: Find Cotangent (cot)

Cotangent is the reciprocal of tangent: \(\cot \theta = \frac{x}{y} = \frac{-\sqrt{2}}{\sqrt{3}} = -\frac{\sqrt{6}}{3}.\)

Key Concepts

Sine

The sine function is one of the primary trigonometric functions, which helps us understand angles in relation to a right triangle. In the context of right triangles, the sine of an angle \( \theta \) is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. If we have an angle with point coordinates \((x, y)\) and a radius or hypotenuse \(r\), then the sine is given as:

• \( \sin \theta = \frac{y}{r} \)

In our problem, given the point \((-\sqrt{2}, \sqrt{3})\), we first calculated the hypotenuse \(r\). With \(r = \sqrt{5}\), the sine function becomes:

• \( \sin \theta = \frac{\sqrt{3}}{\sqrt{5}} = \frac{\sqrt{15}}{5} \)

This tells us how the y-coordinate (opposite side) relates to the hypotenuse, allowing us to understand the angle's position on the unit circle. Sine is helpful in modeling wave-like phenomena, such as sound and light waves.

Cosine

The cosine function is closely related to the sine function and also plays a crucial role in trigonometry. It pertains to the ratio of the length of the adjacent side of an angle to its hypotenuse. Using the same point coordinates, \((x, y)\), and radius \(r\), the formula for cosine is:

• \( \cos \theta = \frac{x}{r} \)

Based on the coordinates \((-\sqrt{2}, \sqrt{3})\), with the computed hypotenuse \(r = \sqrt{5}\), the cosine function becomes:

• \( \cos \theta = \frac{-\sqrt{2}}{\sqrt{5}} = -\frac{\sqrt{10}}{5} \)

The cosine function describes the angle's relationship against the x-axis, essentially telling us how far along the x-axis the point is projected. It's particularly useful in calculating the horizontal component of forces in physics and engineering.

Tangent

The tangent function offers a different perspective in trigonometry by examining the ratio of the opposite side to the adjacent side within a right triangle. This function is represented by:

• \( \tan \theta = \frac{y}{x} \)

For the specified point \((-\sqrt{2}, \sqrt{3})\), the tangent of angle \(\theta\) is:

• \( \tan \theta = \frac{\sqrt{3}}{-\sqrt{2}} = -\frac{\sqrt{6}}{2} \)

In the broader sense, the tangent function can tell us about the angle's slope, providing insights into its steepness. Often used in calculus, tangents assist in interpreting the rate of change or gradient within certain mathematical contexts.