Problem 16 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 9-16: Let \(P\) be the point on the unit circle \(U\) that corresponds to \(t\). Find the coordinates of \(P\) and the exact values of the trigonometric functions of \(t\), whenever possible. (a) \(7 \pi / 4\) (b) \(-3 \pi / 4\)

Step-by-Step Solution

Answer

(a) Coordinates: \( \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \), Trig: \( \cos t = \frac{\sqrt{2}}{2}, \sin t = -\frac{\sqrt{2}}{2}, \tan t = -1 \). (b) Coordinates: \( \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \), Trig: \( \cos t = -\frac{\sqrt{2}}{2}, \sin t = -\frac{\sqrt{2}}{2}, \tan t = 1 \).

Step 1: Understand the Unit Circle

The unit circle is a circle with a radius of 1, centered at the origin of the coordinate plane. Any point \( P \) on this circle can be represented by the coordinates \( (\cos(t), \sin(t)) \), where \( t \) is the angle in radians from the positive x-axis.

Step 2: Calculate for t = \(\frac{7\pi}{4}\)

The angle \( t = \frac{7\pi}{4} \) is in the fourth quadrant of the unit circle. In this quadrant, the cosine value is positive, and the sine value is negative. From reference angles, we know that \( \frac{7\pi}{4} \) corresponds to \( 360° - 45° = 315° \). Therefore, \( \cos(315°) = \frac{\sqrt{2}}{2} \) and \( \sin(315°) = -\frac{\sqrt{2}}{2} \). Thus, the coordinates of \( P \) are \( \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \).

Step 3: Trigonometric Functions for \( t = \frac{7\pi}{4} \)

The trigonometric functions are calculated as follows: \( \cos(\frac{7\pi}{4}) = \frac{\sqrt{2}}{2} \), \( \sin(\frac{7\pi}{4}) = -\frac{\sqrt{2}}{2} \), \( \tan(\frac{7\pi}{4}) = \frac{\sin(\frac{7\pi}{4})}{\cos(\frac{7\pi}{4})} = -1 \). The reciprocal functions are: \( \csc(\frac{7\pi}{4}) = -\sqrt{2} \), \( \sec(\frac{7\pi}{4}) = \sqrt{2} \), and \( \cot(\frac{7\pi}{4}) = -1 \).

Step 4: Calculate for t = \(-\frac{3\pi}{4}\)

The angle \( t = -\frac{3\pi}{4} \) implies a clockwise rotation equivalent to \( 315° \), landing in the third quadrant. In the third quadrant, both cosine and sine values are negative. Hence, \( \cos(-135°) = -\frac{\sqrt{2}}{2} \) and \( \sin(-135°) = -\frac{\sqrt{2}}{2} \). Thus, the coordinates of \( P \) are \( \left( -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \).

Step 5: Trigonometric Functions for \( t = -\frac{3\pi}{4} \)

The trigonometric functions are calculated as follows: \( \cos(-\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} \), \( \sin(-\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} \), \( \tan(-\frac{3\pi}{4}) = 1 \). The reciprocal functions are: \( \csc(-\frac{3\pi}{4}) = -\sqrt{2} \), \( \sec(-\frac{3\pi}{4}) = -\sqrt{2} \), and \( \cot(-\frac{3\pi}{4}) = 1 \).

Key Concepts

Trigonometric Functions

Trigonometric functions are the heart of the unit circle concept. These functions include sine (sin), cosine (cos), and tangent (tan). Each function represents a specific ratio:

• **Sine (sin)**: It represents the ratio of the opposite side to the hypotenuse in a right triangle, or simply the y-coordinate of a point on the unit circle.
• **Cosine (cos)**: It represents the ratio of the adjacent side to the hypotenuse, or the x-coordinate of a point on the unit circle.
• **Tangent (tan)**: Calculated as the sine divided by the cosine, giving the gradient of the line.

When using the unit circle, these functions are simplified since the hypotenuse is 1. This simplifies calculations as all the points on the unit circle lie on a circle with radius 1. So, for any angle \(t\), the point \(P(\cos(t), \sin(t))\) provides the exact coordinates and directly gives the basic trigonometric functions.
Understanding how these functions work together can make solving circle problems much easier!

Coordinates

Coordinates on the unit circle are simply the values of the cosine and sine functions for a particular angle \(t\).
Every point \(P(x, y)\) on the unit circle represents \(x = \cos(t)\) and \(y = \sin(t)\).
It's like plotting the angle as if it's a map direction with terms of radians, moving counter-clockwise, starting from the positive x-axis.For example, given an angle \(\frac{7\pi}{4}\), you can find:

• The x-coordinate by calculating \(\cos(\frac{7\pi}{4}) = \frac{\sqrt{2}}{2}\)
• The y-coordinate by calculating \(\sin(\frac{7\pi}{4}) = -\frac{\sqrt{2}}{2}\)

The beauty of a unit circle is that it allows visualization of the trigonometric functions and helps determine sine, cosine, and tangent ratios easily using these coordinates, no matter the complexity of the angle.

Reciprocal Functions

Reciprocal functions are derived from the basic trigonometric functions. They involve flipping the numerator and denominator of trigonometric function ratios. These include:

• **Cosecant (csc)**: The reciprocal of sine, expressed as \(\csc(t) = \frac{1}{\sin(t)}\). It indicates the length of the hypotenuse (or line from the origin to the circle) divided by the opposite side (or the y-coordinate).
• **Secant (sec)**: The reciprocal of cosine, expressed as \(\sec(t) = \frac{1}{\cos(t)}\). It indicates the hypotenuse divided by the adjacent side (or the x-coordinate).
• **Cotangent (cot)**: The reciprocal of tangent, expressed as \(\cot(t) = \frac{1}{\tan(t)} = \frac{\cos(t)}{\sin(t)}\). This provides the ratio of the adjacent to the opposite side.

For any angle like \(\frac{7\pi}{4}\), you can find the reciprocal functions easily once you know sine and cosine:

• \(\csc(\frac{7\pi}{4}) = -\sqrt{2}\)
• \(\sec(\frac{7\pi}{4}) = \sqrt{2}\)
• \(\cot(\frac{7\pi}{4}) = -1\)

Understanding reciprocal functions is vital for more advanced trigonometry problems, giving deeper insight into angles and their properties.