Problem 63 - Algebra and Trigonometry

Question

Write the first expression in terms of the second if the terminal point determined by \(t\) is in the given quadrant. \(\tan ^{2} t, \sin t ; \quad\) any quadrant

Step-by-Step Solution

Answer

\(\tan^2 t = \frac{\sin^2 t}{1 - \sin^2 t}\).

Step 1: Understand the Relationship

The relationship between \(\tan^2 t\) and \(\sin t\) involves using the Pythagorean identity for tangent and sine. We know that \(\tan^2 t = \frac{\sin^2 t}{\cos^2 t}\) and that \(\sin^2 t + \cos^2 t = 1\). These identities will help us express \(\tan^2 t\) in terms of \(\sin t\).

Step 2: Express \(\cos^2 t\) in Terms of \(\sin t\)

Starting with \(\sin^2 t + \cos^2 t = 1\), we can express \(\cos^2 t\) as \(1 - \sin^2 t\). This substitution will help us in rewriting \(\tan^2 t\).

Step 3: Rewrite \(\tan^2 t\) using the Identities

Substitute \(\cos^2 t = 1 - \sin^2 t\) into the expression for \(\tan^2 t\):\[\tan^2 t = \frac{\sin^2 t}{\cos^2 t} = \frac{\sin^2 t}{1 - \sin^2 t}\].

Step 4: Final Expression

Thus, the expression for \(\tan^2 t\) in terms of \(\sin t\) is:\[\tan^2 t = \frac{\sin^2 t}{1 - \sin^2 t}\].

Key Concepts

Pythagorean identity

In trigonometry, the Pythagorean identity is one of the fundamental relationships. It connects the squares of the sine and cosine functions. Specifically, the identity is given by:

• \( \sin^2 t + \cos^2 t = 1 \)

This equation tells us that if we know the sine or cosine of an angle, we can easily determine the other. When working through problems like finding \( \tan^2 t \) in terms of \( \sin t \), this identity is tremendously helpful.
For instance, if you're asked to find the cosine given a sine value, you would rearrange the identity to solve for \( \cos^2 t \):
\[ \cos^2 t = 1 - \sin^2 t \]
This allows us to swap variables and navigate complex trigonometric expressions.

tangent in terms of sine

Tangent is another crucial trigonometric function. It can be expressed in terms of sine and cosine as \( \tan t = \frac{\sin t}{\cos t} \). When squared, this relationship becomes \( \tan^2 t = \frac{\sin^2 t}{\cos^2 t} \).

In any trigonometric problem, expressing tangent in terms of sine is very useful. One application is rewriting \( \tan^2 t \) using the Pythagorean identity, which provides \( \cos^2 t = 1 - \sin^2 t \). By substituting \( \cos^2 t \) in \( \tan^2 t \), it converts into:
\[ \tan^2 t = \frac{\sin^2 t}{1 - \sin^2 t} \]
This formula ties the tangent directly to the sine, sidestepping direct reference to cosine. It simplifies calculations when cosine is not directly needed.

quadrantal angles

Quadrantal angles are angles that lie on the axes of the Cartesian plane, such as 0°, 90°, 180°, and 270°. These angles have special properties:

• At 0° and 180°: \( \sin t = 0 \), \( \cos t = \pm 1 \)
• At 90° and 270°: \( \sin t = \pm 1 \), \( \cos t = 0 \)

Certain angles result in undefined tangent values due to division by zero in \( \tan t = \frac{\sin t}{\cos t} \). For example, at 90° and 270°, cosine is 0, making tangent undefined.

Understanding quadrantal angles is crucial, especially when determining the value of trigonometric functions. These angles influence the sign and sometimes the definition of trigonometric expressions, like your task of expressing \( \tan^2 t \) without potential undefined behavior.