Problem 19
Question
Show that \(1+\cos \theta\) and \(\frac{\sin ^{2} \theta}{1-\cos \theta}\) form an identity.
Step-by-Step Solution
Verified Answer
\(1 + \cos \theta = \frac{\sin^{2} \theta}{1-\cos \theta}\) is an identity after simplification.
1Step 1: Express the Right-Hand Side
First, examine the expression \( \frac{\sin ^{2} \theta}{1-\cos \theta} \). To simplify this, use the Pythagorean identity \( \sin^2 \theta = 1 - \cos^2 \theta \). Substitute this identity into the expression: \[ \frac{1 - \cos^2 \theta}{1 - \cos \theta} \].
2Step 2: Simplify the Fraction
Notice that the numerator \( 1 - \cos^2 \theta \) can be factored using the difference of squares: \( (1 - \cos \theta)(1 + \cos \theta) \). Substitute this into the expression: \[ \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 - \cos \theta} \].
3Step 3: Cancel Common Terms
Now, observe that the terms \( 1 - \cos \theta \) in the numerator and the denominator are the same and can be canceled out, provided \( 1 - \cos \theta eq 0 \). This simplifies the expression to: \[ 1 + \cos \theta \].
4Step 4: Conclude Identity
At this point, both sides of the expression are \( 1 + \cos \theta \), thus confirming that the initial expression \( 1 + \cos \theta = \frac{\sin^2 \theta}{1 - \cos \theta} \) is an identity.
Key Concepts
Pythagorean IdentityDifference of SquaresAlgebraic Simplification
Pythagorean Identity
In trigonometry, the Pythagorean identity is a fundamental relation that connects sine and cosine functions. The most familiar form of this identity is: - \( \sin^2 \theta + \cos^2 \theta = 1 \) This particular relation emerges directly from the Pythagorean theorem and holds for any angle \( \theta \) in the unit circle. We can derive various expressions and use this identity for substitution purposes in complex equations. For example, we can rearrange it to state: - \( \sin^2 \theta = 1 - \cos^2 \theta \) This form is particularly useful when we need to simplify expressions involving trigonometric functions, such as in the given exercise where we simplify \( \frac{\sin^2 \theta}{1-\cos \theta} \) by substituting with \( 1-\cos^2 \theta \). Using Pythagorean identities provides a reliable means to simplify and solve trigonometric expressions by expressing one function in terms of another. This process is a crucial step in proving identities and simplifying complex trigonometric functions.
Difference of Squares
The difference of squares is a powerful algebraic tool used to factor certain types of expressions. It states that: - \( a^2 - b^2 = (a-b)(a+b) \) This pattern arises because the middle terms cancel each other out when you expand the product on the right-hand side. In the context of our trigonometric problem, the expression \( 1 - \cos^2 \theta \) is a difference of squares because it can be rewritten as: - \( (1 - \cos \theta)(1 + \cos \theta) \) Factoring in this manner allows us to simplify expressions significantly because it can lead to canceling out terms when placed over a common denominator. When you encounter expressions resembling \( a^2 - b^2 \), utilizing the difference of squares can lead to easier and more elegant simplifications, by reducing complexity and emphasizing common factors.
Algebraic Simplification
Algebraic simplification involves using mathematical properties to condense and make expressions more manageable. This process often involves canceling out common terms, factoring, or applying algebraic identities. In our exercise, after substituting the Pythagorean identity, the expression \( \frac{1 - \cos^2 \theta}{1 - \cos \theta} \) is simplified by factoring the numerator using the difference of squares. The goal is to identify and eliminate common terms whenever possible, as we do with \( (1 - \cos \theta) \) from both the numerator and the denominator. Simplification is not only about making the expression smaller, but also about making it equivalent to another, ideally simpler expression, which is exactly what happens here. When done correctly, algebraic simplification can transform intricate expressions into easy-to-understand forms, aiding in the recognition of underlying identities or simpler equivalent forms.
Other exercises in this chapter
Problem 19
Find the exact value of each expression by using the half-angle formulas. \(\sin 345^{\circ}\)
View solution Problem 19
Find the exact value of each expression. \(\sin \left(-15^{\circ}\right)\)
View solution Problem 19
Simplify each expression. \(\tan \theta \cot \theta\)
View solution Problem 19
State the amplitude, period, and phase shift for each function. Then graph the function. $$ y=\cos \left(\theta+\frac{\pi}{3}\right) $$
View solution