Problem 125 - Precalculus

Question

For the following exercises, find the requested value. If \(\cos (t)=\frac{2}{9}\) and \(t\) is in the \(1^{\text { st }}\) quadrant, find \(\sin (t)\)

Step-by-Step Solution

Answer

\( \sin(t) = \frac{\sqrt{77}}{9} \)

Step 1: Understanding the Problem

We need to find the value of \( \sin(t) \) given that \( \cos(t) = \frac{2}{9} \) and \( t \) is in the 1st quadrant.

Step 2: Use Pythagorean Identity

Recall the Pythagorean identity: \[ \sin^2(t) + \cos^2(t) = 1 \]Substitute \( \cos(t) = \frac{2}{9} \) into the identity.

Step 3: Substitute Cosine Value

Substitute the value into the equation:\[ \sin^2(t) + \left(\frac{2}{9}\right)^2 = 1 \]\[ \sin^2(t) + \frac{4}{81} = 1 \]

Step 4: Solve for \( \sin^2(t) \)

Rearrange the equation to solve for \( \sin^2(t) \):\[ \sin^2(t) = 1 - \frac{4}{81} \]\[ \sin^2(t) = \frac{81}{81} - \frac{4}{81} = \frac{77}{81} \]

Step 5: Solve for \( \sin(t) \)

Since \( t \) is in the 1st quadrant, \( \sin(t) \) is positive. So:\[ \sin(t) = \sqrt{\frac{77}{81}} \]\[ \sin(t) = \frac{\sqrt{77}}{9} \]

Key Concepts

Pythagorean identity

The Pythagorean identity is a fundamental concept in trigonometry, often used to relate the basic trigonometric functions of sine and cosine. It is derived from the Pythagorean theorem and is presented as:

• \( \sin^2(t) + \cos^2(t) = 1 \)

This identity is crucial when you have the value of either the sine or cosine function and need to find the other.
In the original exercise, for example, we know \( \cos(t) = \frac{2}{9} \) and need to find \( \sin(t) \). By substituting the given cosine value into the identity:

• \( \sin^2(t) + \left(\frac{2}{9}\right)^2 = 1 \)

we can solve for \( \sin^2(t) \). This simplifies to \( \sin^2(t) = 1 - \frac{4}{81} \), and further simplifies to \( \frac{77}{81} \).
This allows us to find \( \sin(t) \) as the square root of \( \frac{77}{81} \), which is \( \frac{\sqrt{77}}{9} \).
The Pythagorean identity simplifies problems of finding missing trigonometric values when the angle position is considered.

trigonometric functions

Trigonometric functions are essential in mathematics, especially for studying triangles and modeling periodic phenomena. The primary trigonometric functions are sine, cosine, and tangent. Each function relates the angles of a right triangle to the ratio of its sides.

• Sine (\( \sin \)): It is defined as the ratio of the opposite side to the hypotenuse of a right triangle.
• Cosine (\( \cos \)): It is the ratio of the adjacent side to the hypotenuse.
• Tangent (\( \tan \)): It is the ratio of the sine to the cosine, or opposite to adjacent side.

Trigonometric functions also have identities that help us solve various problems, such as the Pythagorean identity. In solving for \( \sin(t) \), knowing \( \cos(t) \) allows us to use this identity to find missing values oscillating with the same regular interval.
These functions are periodic, meaning they repeat values over regular intervals, specifically multiples of \( 2\pi \) for sine and cosine.
Understanding these functions and their relationships is necessary for areas like geometry, physics, and engineering.

first quadrant angles

The first quadrant of the unit circle is where the angles range from \(0\) to \(\frac{\pi}{2}\) radians, or from \(0\) to \(90^\circ\). In this quadrant, both sine and cosine functions have positive values. This is because:

• The x-coordinate on the unit circle (representing cosine) is positive
• The y-coordinate (representing sine) is also positive

These properties are what allow us to confidently state that \( \sin(t) \) and \( \cos(t) \) are positive in this region.
In the given problem, we know that \(t\) is in the first quadrant from the start, which directly affects the algebra: once we solve \( \sin^2(t) = \frac{77}{81} \), the positive nature of \( \sin(t) \) in this quadrant dictates its value as the positive square root: \( \frac{\sqrt{77}}{9} \).
Recognizing the sign of functions based on quadrant helps avoid mistakes and complex calculations when assigning proper values during problem-solving.