Question

A point on the terminal side of an angle \(\theta\) in standard position is given. Find the exact value of each of the six trigonometric functions of \(\theta .\) $$ \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) $$

Short answer

sin(θ) = \( \frac{\sqrt{3}}{2} \), cos(θ) = \( -\frac{1}{2} \), tan(θ) = \( -\sqrt{3} \), csc(θ) = \( \frac{2\sqrt{3}}{3} \), sec(θ) = -2, cot(θ) = \( -\frac{\sqrt{3}}{3} \).

Step-by-step solution

Determine the radius (r)

Use the formula for the radius in the coordinate plane: \[ r = \sqrt{x^2 + y^2} \] Substitute \( x = -\frac{1}{2} \) and \( y = \frac{\sqrt{3}}{2} \): \[ r = \sqrt{\left( -\frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

Evaluate sin(θ) and cos(θ)

Trigonometric functions are defined as: \[ \sin(\theta) = \frac{y}{r} \] \[ \cos(\theta) = \frac{x}{r} \] Given that \( y = \frac{\sqrt{3}}{2} \) and \( x = -\frac{1}{2} \): \[ \sin(\theta) = \frac{\frac{\sqrt{3}}{2}}{1} = \frac{\sqrt{3}}{2} \] \[ \cos(\theta) = \frac{\frac{-1}{2}}{1} = -\frac{1}{2} \]

Evaluate tan(θ)

The tangent function is defined as: \[ \tan(\theta) = \frac{y}{x} \] Using \( y = \frac{\sqrt{3}}{2} \) and \( x = -\frac{1}{2} \): \[ \tan(\theta) = \frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}} = -\sqrt{3} \]

Evaluate csc(θ)

The cosecant function is the reciprocal of the sine function: \[ \csc(\theta) = \frac{1}{\sin(\theta)} \] From Step 2, \( \sin(\theta) = \frac{\sqrt{3}}{2} \): \[ \csc(\theta) = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \]

Evaluate sec(θ)

The secant function is the reciprocal of the cosine function: \[ \sec(\theta) = \frac{1}{\cos(\theta)} \] From Step 2, \( \cos(\theta) = -\frac{1}{2} \): \[ \sec(\theta) = \frac{1}{-\frac{1}{2}} = -2 \]

Evaluate cot(θ)

The cotangent function is the reciprocal of the tangent function: \[ \cot(\theta) = \frac{1}{\tan(\theta)} \] From Step 3, \( \tan(\theta) = -\sqrt{3} \): \[ \cot(\theta) = \frac{1}{-\sqrt{3}} = -\frac{\sqrt{3}}{3} \]

Key concepts

sine function

The sine function, written as \(\text{sin}(\theta)\), is one of the most fundamental trigonometric functions. It calculates the ratio of the length of the opposite side of a right triangle to the hypotenuse. In other words, for any given angle \(\theta\):

• The opposite side is the side that is opposite to the angle \(\theta\).
• The hypotenuse is the longest side of the triangle, opposite the right angle.

The formula is: \[ \text{sin}(\theta) = \frac{y}{r} \] where \(y\) is the y-coordinate of the point, and \(r\) is the radius or the hypotenuse. In our exercise, \(y = \frac{\text{√}3}{2}\) and \(r = 1\), hence: \[ \text{sin}(\theta) = \frac{\frac{\text{√}3}{2}}{1} = \frac{\text{√}3}{2} \] Understanding sine is crucial because it's used to find other functions and in various applications.

cosine function

The cosine function, noted as \(\text{cos}(\theta)\), is another key trigonometric function. It represents the ratio of the length of the adjacent side of a right triangle to the hypotenuse. For any angle \(\theta\):

• The adjacent side lies next to the angle \(\theta\).
• The hypotenuse is, again, the triangle's longest side.

The formula is: \[ \text{cos}(\theta) = \frac{x}{r} \] where \(x\) is the x-coordinate of the point, and \(r\) is the radius or the hypotenuse. Given \(x = -\frac{1}{2}\) and \(r = 1\) from our problem: \[ \text{cos}(\theta) = \frac{-\frac{1}{2}}{1} = -\frac{1}{2} \] Cosine is widely used in geometry and physics, especially for finding distances and in wave equations.

tangent function

The tangent function, indicated as \(\text{tan}(\theta)\), is the ratio of the sine function to the cosine function. It effectively relates the opposite side to the adjacent side of a right-angle triangle:
\[ \text{tan}(\theta) = \frac{\text{sin}(\theta)}{\text{cos}(\theta)} \] Alternatively, it can be directly calculated using: \[ \text{tan}(\theta) = \frac{y}{x} \] In this case, with \(y = \frac{\text{√}3}{2}\) and \(x = -\frac{1}{2}\), we get: \[ \text{tan}(\theta) = \frac{\frac{\text{√}3}{2}}{\frac{-1}{2}} = -\text{√}3 \] Tangent is particularly useful in various fields including engineering and navigation, where angle calculations are paramount.

radius in coordinate plane

The radius in the coordinate plane, often represented as \(r\), is the distance from the origin (0,0) to a given point (x,y). This distance forms the hypotenuse of a right-angle triangle with sides x and y. To calculate the radius, we use the Pythagorean theorem: \[ r = \text{√}(x^2 + y^2) \] For the point \((- \frac{1}{2}, \frac{\text{√}3}{2})\):

• Square both x and y coordinates: \(\text{√}(-\frac{1}{2})^2 = \frac{1}{4}\) and \( \text{√}(\frac{\text{√}3}{2})^2 = \frac{3}{4}\)


• Add the results: \( \frac{1}{4} + \frac{3}{4} = 1 \)


• Take the square root: \( \text{√1} = 1 \)

The radius is 1, which simplifies calculations in trigonometric functions significantly.

reciprocal trigonometric functions

Reciprocal trigonometric functions are inverses of the primary trigonometric functions. These include:

• **Cosecant (csc):** Reciprocal of sine, \( \text{csc}(\theta) = \frac{1}{\text{sin}(\theta)} \)
• **Secant (sec):** Reciprocal of cosine, \( \text{sec}(\theta) = \frac{1}{\text{cos}(\theta)} \)
• **Cotangent (cot):** Reciprocal of tangent, \( \text{cot}(\theta) = \frac{1}{\text{tan}(\theta)} \)

Using our initial calculations: \[ \text{csc}(\theta) = \frac{1}{\frac{\text{√}3}{2}} = \frac{2\text{√}3}{3} \] \[ \text{sec}(\theta) = \frac{1}{-\frac{1}{2}} = -2 \] \[ \text{cot}(\theta) = \frac{1}{-\text{√}3} = -\frac{\text{√}3}{3} \] These functions help mathematicians and scientists switch between forms and solve various complex problems easily.