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 \(\theta .\) $$ \left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right) $$

Short answer

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

Step-by-step solution

Identify the Coordinates

The given point on the terminal side of the angle \(\theta\) is \(\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)\). Let \(x = \frac{\sqrt{2}}{2}\) and \(y = -\frac{\sqrt{2}}{2}\).

Calculate the Radius (r)

The radius \(r\) can be found using the Pythagorean theorem: \[ r = \sqrt{x^2 + y^2} \] Plug in the values of \(x\) and \(y\): \[ r = \sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{2}{4} + \frac{2}{4}} = \sqrt{1} = 1 \]

Find Sine and Cosine

The sine and cosine functions are defined as: \[ \sin(\theta) = \frac{y}{r} \] \[ \cos(\theta) = \frac{x}{r} \] Since \(r = 1\), we get: \[ \sin(\theta) = -\frac{\sqrt{2}}{2} \] \[ \cos(\theta) = \frac{\sqrt{2}}{2} \]

Find Tangent

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

Find Cosecant

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

Find Secant

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

Find Cotangent

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

Key concepts

Angle in Standard Position

An angle is in its standard position when its vertex is at the origin of the coordinate system and its initial side lies along the positive x-axis. The angle then sweeps counterclockwise to its terminal side. For instance, given the point \(\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)\), the terminal side of the angle intercepts the x-axis and y-axis forming an angle \(\theta\).
The position of the point helps determine the quadrant where \(\theta\) lies. Here, the coordinates imply that the terminal side of \(\theta\) is in the fourth quadrant, where the x-values are positive, and y-values are negative.

Pythagorean Theorem

The Pythagorean theorem is essential in trigonometry for finding the length of the hypotenuse, given the lengths of the other two sides in a right triangle. It is stated as \[a^2 + b^2 = c^2\].
In this context, the theorem is used to find the radius (r), which is the hypotenuse of a triangle formed by the x and y coordinates.
Using the given point \(\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)\), we calculate the radius as: \[ r = \sqrt{x^2 + y^2} = \sqrt{\left(\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{2}{4} + \frac{2}{4}} = \sqrt{1} = 1 \].
This radius 'r' plays a crucial role in defining trigonometric functions.

Reciprocal Trigonometric Functions

Reciprocal trigonometric functions are the reciprocals of the basic trigonometric functions. They are defined as:

• \textbf{Cosecant (csc)} — the reciprocal of sine. \[ \csc(\theta) = \frac{1}{\sin(\theta)} \]
• \textbf{Secant (sec)} — the reciprocal of cosine. \[ \sec(\theta) = \frac{1}{\cos(\theta)} \]
• \textbf{Cotangent (cot)} — the reciprocal of tangent. \[ \cot(\theta) = \frac{1}{\tan(\theta)} \]

For the given angle \(\theta\) with coordinates \(\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)\), we have found these values:
\textbf{Cosecant:} \[ \csc(\theta) = \frac{1}{-\frac{\sqrt{2}}{2}} = -\sqrt{2} \]
\textbf{Secant:} \[ \sec(\theta) = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \]
\textbf{Cotangent:} \[ \cot(\theta) = \frac{1}{-1} = -1 \]
These functions can help in solving various trigonometric equations and identities.

Sine and Cosine

Sine and cosine are fundamental trigonometric functions that relate the angles of a right triangle to the lengths of its sides.

• \textbf{Sine (sin)} is the ratio of the side opposite the angle to the hypotenuse. \[ \sin(\theta) = \frac{y}{r} \]
• \textbf{Cosine (cos)} is the ratio of the adjacent side to the hypotenuse. \[ \cos(\theta) = \frac{x}{r} \]

In the given exercise, the point \(\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)\) allows us to find:
\textbf{Sine:} \[ \sin(\theta) = \frac{-\frac{\sqrt{2}}{2}}{1} = -\frac{\sqrt{2}}{2} \]
\textbf{Cosine:} \[ \cos(\theta) = \frac{\frac{\sqrt{2}}{2}}{1} = \frac{\sqrt{2}}{2} \]
Understanding these functions is crucial for studying waves, oscillations, and many other phenomena.

Tangent and Cotangent

Tangent and cotangent functions are derived from sine and cosine and are pivotal in trigonometry.

• \textbf{Tangent (tan)} is defined as the ratio of sine to cosine. \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{y}{x} \]
• \textbf{Cotangent (cot)} is the reciprocal of tangent. \[ \cot(\theta) = \frac{1}{\tan(\theta)} = \frac{x}{y} \]

For the point \(\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)\), we get:
\textbf{Tangent:} \[ \tan(\theta) = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1 \]
\textbf{Cotangent:} \[ \cot(\theta) = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 \]
These functions are essential in various areas, including calculus and engineering.