Problem 84 - A Graphical Approach to Precalculus with Limits

Question

Use the identity \(\cos ^{2} s+\sin ^{2} s=1\) to find the value of \(x\) or \(y,\) as appropriate. Then, assuming that \(s\) corresponds to the given point on the unit circle, find the six circular function values for \(s\). $$\left(-\frac{1}{4}, y\right), y<0$$

Step-by-Step Solution

Answer

\(y = -\frac{\sqrt{15}}{4}\). Function values: \(\cos s = -\frac{1}{4}\), \(\sin s = -\frac{\sqrt{15}}{4}\), \(\tan s = \sqrt{15}\), \(\csc s = -\frac{4\sqrt{15}}{15}\), \(\sec s = -4\), \(\cot s = \frac{\sqrt{15}}{15}\).

Step 1: Use Trigonometric Identity

We use the identity \(\cos^2 s + \sin^2 s = 1\). Since the x-coordinate represents \(\cos s\), it is given by \(-\frac{1}{4}\). Therefore, the equation becomes:\[\left(-\frac{1}{4}\right)^2 + \sin^2 s = 1\]

Step 2: Calculate \( \sin^2 s \)

Compute \(\left(-\frac{1}{4}\right)^2\) which equals \(\frac{1}{16}\). The equation now is:\[\frac{1}{16} + \sin^2 s = 1\]Subtract \(\frac{1}{16}\) from both sides:\[\sin^2 s = 1 - \frac{1}{16} = \frac{15}{16}\]

Step 3: Solve for \( \sin s \)

Take the square root of both sides. Remember that since \(y < 0\), \(\sin s\) must be negative:\[\sin s = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}\]So, \(y\) is \(-\frac{\sqrt{15}}{4}\).

Step 4: Determine Remaining Function Values

For the remaining trigonometric function values, use:- \( \tan s = \frac{\sin s}{\cos s} = \frac{-\frac{\sqrt{15}}{4}}{-\frac{1}{4}} = \sqrt{15} \)- \( \csc s = \frac{1}{\sin s} = -\frac{4}{\sqrt{15}} = -\frac{4\sqrt{15}}{15} \) after rationalizing- \( \sec s = \frac{1}{\cos s} = -4 \)- \( \cot s = \frac{\cos s}{\sin s} = \frac{-\frac{1}{4}}{-\frac{\sqrt{15}}{4}} = \frac{1}{\sqrt{15}} = \frac{\sqrt{15}}{15} \) after rationalizing

Key Concepts

Unit Circle

The unit circle is a fundamental concept in trigonometry. It's a circle with a radius of 1, centered at the origin of the coordinate plane. This circle is extensively used to define the trigonometric functions for all real numbers.

• On the unit circle, the coordinates of any point \( (x, y) \) can be interpreted as \( (\cos s, \sin s) \) for an angle \( s \).
• The unit circle helps in visualizing circular functions, especially with angles beyond \( 0 \) to \( 2\pi \) approximately \( 360^{\circ} \).
• For \((-\frac{1}{4}, y)\), the \( x\) -coordinate represents \( \cos s\).

In this particular problem, knowing the position of the point relative to the unit circle allows us to use the trigonometric identities to find the corresponding \( y \) value, which represents \( \sin s \).Given the condition \( y < 0 \), we find that \( \sin s \) must be negative.

Circular Functions

Circular functions, also known as trigonometric functions, are derived from the geometry of the unit circle. Each function represents a distinct ratio or characteristic based on the coordinates of a point on the circle.

• The six fundamental circular functions are sine ( \(\sin s\) ), cosine ( \(\cos s\) ), tangent ( \(\tan s\) ), cosecant ( \(\csc s\) ), secant ( \(\sec s\) ), and cotangent ( \(\cot s\) ).
• Sine ( \(\sin s\) ) and cosine ( \(\cos s\) ) are the fundamental values derived directly from the unit circle coordinates.
• The other four functions are "quotients" or "reciprocals" of these basic functions, such as:
  • Tangent, which is the ratio \(\frac{\sin s}{\cos s}\)
  • Cosecant, the reciprocal of sine \(\frac{1}{\sin s}\)
  • Secant, the reciprocal of cosine \(\frac{1}{\cos s}\)
  • Cotangent, the reciprocal of tangent or \(\frac{\cos s}{\sin s}\)

Using these, we find function values based on the position \((-\frac{1}{4}, -\frac{\sqrt{15}}{4})\) given from the exercise, calculating each function's exact value.

Pythagorean Identity

The Pythagorean Identity is a central equation in trigonometry relating the sine and cosine of an angle. It translates the Pythagorean Theorem to trigonometric functions:\[\cos^2 s + \sin^2 s = 1\]

• This equation holds true for any angle \( s \) making it incredibly useful to solve trigonometric equations.
• In our task, \(\cos s\) is given as \(-\frac{1}{4} \), enabling us to solve for \(\sin s\) using: \[ (-\frac{1}{4})^2 + \sin^2 s = 1 \]
• Solving the equation allows us to deduce\(\sin s = -\frac{\sqrt{15}}{4}\) by substituting and rearranging the terms to focus on \( \sin^2 s\).

Employing the identity efficiently, especially with restricted conditions like \( y < 0 \), allows easy calculation of \(y\) which is critical in determining all trigonometric function values.

Trigonometric Ratios

Trigonometric ratios are the functions that depict the relationship between angles and sides of a right triangle, or geometrically on the unit circle.

• These ratios include sine, cosine, and tangent, which can be visually understood through unit circle coordinates as \((x, y) = (\cos s, \sin s)\).
• Ratios provide significant insight when solving problems concerning angles and lengths in trigonometry.
• On the unit circle, as given in the exercise:
  • \(\sin s = -\frac{\sqrt{15}}{4}\)
  • \(\cos s = -\frac{1}{4}\)
  • \(\tan s = \frac{\sin s}{\cos s} = \sqrt{15}\)

Understanding trigonometric ratios is fundamental for deriving additional values like \(\csc, \sec, \cot\), and extensively unit circle trigonometry, making it easier to solve broader mathematical problems involving circular functions.