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 \(\theta .\) $$ (-3,-3) $$

Short answer

The six trigonometric functions are: \( \sin(\theta) = -\frac{\sqrt{2}}{2}, \cos(\theta) = -\frac{\sqrt{2}}{2}, \tan(\theta) = 1, \csc(\theta) = -\sqrt{2}, \sec(\theta) = -\sqrt{2}, \cot(\theta) = 1 \).

Step-by-step solution

Determine the radius (r)

Given a point \text{(-3, -3)}, first compute the radius (r) using the Pythagorean theorem: \[ r = \sqrt{x^2 + y^2} \] Substituting the given coordinates: \[ r = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \]

Calculate \( \sin(\theta) \)

The sine function is given by \( \sin(\theta) = \frac{y}{r} \). Here, \( y = -3 \) and \( r = 3\sqrt{2} \), so: \[ \sin(\theta) = \frac{-3}{3\sqrt{2}} = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \]

Calculate \( \cos(\theta) \)

The cosine function is given by \( \cos(\theta) = \frac{x}{r} \). Here, \( x = -3 \) and \( r = 3\sqrt{2} \), so: \[ \cos(\theta) = \frac{-3}{3\sqrt{2}} = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \]

Calculate \( \tan(\theta) \)

The tangent function is given by \( \tan(\theta) = \frac{y}{x} \). Here, \( y = -3 \) and \( x = -3 \), so: \[ \tan(\theta) = \frac{-3}{-3} = 1 \]

Calculate \( \csc(\theta) \)

\( \csc(\theta) \) is the reciprocal of \( \sin(\theta) \): \[ \csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{-\frac{\sqrt{2}}{2}} = -\sqrt{2} \]

Calculate \( \sec(\theta) \)

\( \sec(\theta) \) is the reciprocal of \( \cos(\theta) \): \[ \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{-\frac{\sqrt{2}}{2}} = -\sqrt{2} \]

Calculate \( \cot(\theta) \)

\( \cot(\theta) \) is the reciprocal of \( \tan(\theta) \): \[ \cot(\theta) = \frac{1}{\tan(\theta)} = \frac{1}{1} = 1 \]

Key concepts

Sine Function

The sine function, \(\text{sin}(\theta)\), is one of the primary trigonometric functions. It represents the ratio of the opposite side to the hypotenuse in a right triangle.
In the unit circle, \(\text{sin}(\theta)\) corresponds to the y-coordinate of a point on the circle.
For the given problem, we used a point \((-3, -3)\), located in the third quadrant.
We found that \(\text{sin}(\theta) = \frac{y}{r} = \frac{-3}{3\sqrt{2}} = -\frac{\sqrt{2}}{2}\).
This negative value indicates the angle is in the third quadrant, where both sine and cosine are negative.

Cosine Function

The cosine function, \(\text{cos}(\theta)\), represents the ratio of the adjacent side to the hypotenuse in a right triangle.
On the unit circle, \(\text{cos}(\theta)\) corresponds to the x-coordinate of a point on the circle.
For our point \((-3, -3)\), we computed the cosine as \(\text{cos}(\theta) = \frac{x}{r} = \frac{-3}{3\sqrt{2}} = -\frac{\sqrt{2}}{2}\).
Similar to sine, the negative value is due to the point being in the third quadrant.

Tangent Function

The tangent function, \(\text{tan}(\theta)\), is the ratio of the sine and cosine of an angle.
It can also be expressed as the ratio of the opposite side to the adjacent side in a right triangle.
In the given problem, where both x and y coordinates are \(-3\), we found \(\text{tan}(\theta) = \frac{y}{x} = \frac{-3}{-3} = 1\).
The positive value means we are looking at an angle where both sine and cosine have the same sign which is true for the third quadrant.

Reciprocal Trigonometric Functions

Reciprocal trigonometric functions provide additional ways of looking at sine, cosine, and tangent.
They are: cosecant (\(\text{csc}(\theta)\)), secant (\(\text{sec}(\theta)\)), and cotangent (\(\text{cot}(\theta)\)):

• \(\text{csc}(\theta)\) is the reciprocal of sine: \[ \text{csc}(\theta) = \frac{1}{\text{sin}(\theta)} = -\sqrt{2} \]
• \(\text{sec}(\theta)\) is the reciprocal of cosine: \[ \text{sec}(\theta) = \frac{1}{\text{cos}(\theta)} = -\sqrt{2} \]
• \(\text{cot}(\theta)\) is the reciprocal of tangent: \[ \text{cot}(\theta) = \frac{1}{\text{tan}(\theta)} = 1 \]

The reciprocals make it easier to use ratios directly in problems.

Pythagorean Theorem

The Pythagorean theorem is essential in trigonometry for relating the sides of a right triangle.
It states that in any right triangle, \(a^2 + b^2 = c^2\), where c is the hypotenuse.
For the point \((-3, -3)\), we used this theorem to determine the radius r:
\[ r = \sqrt{x^2 + y^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \]
This radius is used to calculate all trigonometric functions. It's crucial to understand that without the radius computed from the Pythagorean theorem, the other trigonometric functions cannot be accurately determined.