Problem 71
Question
Verify the identity. $$\tan ^{2} u-\sin ^{2} u=\tan ^{2} u \sin ^{2} u$$
Step-by-Step Solution
Verified Answer
The identity is verified as both sides simplify to \( \frac{\sin^4 u}{\cos^2 u} \).
1Step 1: Express Both Sides Using Sine and Cosine
Start by writing all trigonometric functions, including the left-hand side (LHS) and the right-hand side (RHS), in terms of sine and cosine. Note: \( \tan u = \frac{\sin u}{\cos u} \).For the LHS:\[ \tan^2 u - \sin^2 u = \left(\frac{\sin^2 u}{\cos^2 u}\right) - \sin^2 u \]For the RHS:\[ \tan^2 u \sin^2 u = \left(\frac{\sin^2 u}{\cos^2 u}\right) \sin^2 u = \frac{\sin^4 u}{\cos^2 u} \]
2Step 2: Simplify the Left-Hand Side
For LHS:Multiply both terms over a common denominator of \( \cos^2 u \):\[ \frac{\sin^2 u}{\cos^2 u} - \sin^2 u = \frac{\sin^2 u}{\cos^2 u} - \frac{\sin^2 u \cdot \cos^2 u}{\cos^2 u} \]Simplify the expression:\[ = \frac{\sin^2 u - \sin^2 u \cos^2 u}{\cos^2 u} = \frac{\sin^2 u (1 - \cos^2 u)}{\cos^2 u} \]
3Step 3: Recognize and Use Trigonometric Identity
Recall the Pythagorean identity: \( \sin^2 u + \cos^2 u = 1 \). From this, \( 1 - \cos^2 u = \sin^2 u \).Replace \( 1 - \cos^2 u \) in the expression:\[ \frac{\sin^2 u (\sin^2 u)}{\cos^2 u} = \frac{\sin^4 u}{\cos^2 u} \]
4Step 4: Compare Simplified Left-Hand Side and Right-Hand Side
Observe that the LHS simplifies to:\[ \frac{\sin^4 u}{\cos^2 u} \]The RHS is already given as:\[ \frac{\sin^4 u}{\cos^2 u} \]Since both expressions are the same, the identity is verified.
Key Concepts
Pythagorean identitySimplifying trigonometric expressionsSine and cosine definitionsVerifying identities
Pythagorean identity
The Pythagorean identity is one of the most fundamental relationships in trigonometry. It states that for any angle \( u \), the sum of the squares of sine and cosine equals 1, which can be written as:
In the given exercise, this identity is particularly useful to simplify expressions. When we encounter terms like \( 1 - \cos^2 u \), we can confidently replace them with \( \sin^2 u \) thanks to the Pythagorean identity. This is a handy tool that simplifies complex trigonometric expressions easily and efficiently.
- \( \sin^2 u + \cos^2 u = 1 \)
In the given exercise, this identity is particularly useful to simplify expressions. When we encounter terms like \( 1 - \cos^2 u \), we can confidently replace them with \( \sin^2 u \) thanks to the Pythagorean identity. This is a handy tool that simplifies complex trigonometric expressions easily and efficiently.
Simplifying trigonometric expressions
Simplifying trigonometric expressions involves reducing them to a more manageable form. This often includes expressing trigonometric functions using a common denominator or substituting equivalent expressions using identities.
For example, in the original exercise, we transformed both sides of the identity using sine and cosine to find a common basis for comparison. By expressing both \( \tan u \) as \( \frac{\sin u}{\cos u} \) and \( \tan^2 u \) as \( \frac{\sin^2 u}{\cos^2 u} \), we created an opportunity to simplify terms directly.
Arranging under a common denominator and using algebraic manipulations allow students to see more clearly how the expressions relate to each other, facilitating further simplification and verification.
For example, in the original exercise, we transformed both sides of the identity using sine and cosine to find a common basis for comparison. By expressing both \( \tan u \) as \( \frac{\sin u}{\cos u} \) and \( \tan^2 u \) as \( \frac{\sin^2 u}{\cos^2 u} \), we created an opportunity to simplify terms directly.
Arranging under a common denominator and using algebraic manipulations allow students to see more clearly how the expressions relate to each other, facilitating further simplification and verification.
Sine and cosine definitions
The sine and cosine functions are the core of trigonometry. Their definitions on the unit circle are straightforward:
This means that any point \((\cos u, \sin u)\) on the unit circle directly correlates to the angle \( u \). These definitions are essential, especially when simplifying and verifying trigonometric identities, as they allow us to replace trigonometric functions with their algebraic equivalents like we did in the exercise.
- \( \sin u \) is the y-coordinate of the point on the unit circle.
- \( \cos u \) is the x-coordinate of the point on the unit circle.
This means that any point \((\cos u, \sin u)\) on the unit circle directly correlates to the angle \( u \). These definitions are essential, especially when simplifying and verifying trigonometric identities, as they allow us to replace trigonometric functions with their algebraic equivalents like we did in the exercise.
Verifying identities
Verifying trigonometric identities is a process where you show that both sides of an equation are equivalent for all values of the variables involved. It involves transforming one or both sides of an equation using known identities and simplifications until they match exactly.
In the provided exercise, we verified the identity by working through each side of the equation separately, simplifying them, and finally using the Pythagorean identity to reach a common form.
In the provided exercise, we verified the identity by working through each side of the equation separately, simplifying them, and finally using the Pythagorean identity to reach a common form.
- Start by expressing all trigonometric functions using sine and cosine.
- Use algebraic manipulations such as factoring, expanding, or simplifying fractions.
- Apply relevant trigonometric identities, such as the Pythagorean identity, where applicable.
Other exercises in this chapter
Problem 71
Find the value of the product or sum. $$\cos 255^{\circ}-\cos 195^{\circ}$$
View solution Problem 71
Addition Formula for Tangent Use the Addition Formulas for cosine and sine to prove the Addition Formula for Tangent. \([\)Hint: Use $$ \tan (s+t)=\frac{\sin (s
View solution Problem 72
Find the value of the product or sum. $$\cos \frac{\pi}{12}+\cos \frac{5 \pi}{12}$$
View solution Problem 72
Verify the identity. $$\frac{\tan v \sin v}{\tan v+\sin v}=\frac{\tan v-\sin v}{\tan v \sin v}$$
View solution