Problem 33
Question
Use the addition identity for the tangent to show that \(\tan (t+\pi)=\tan t\) for all \(t\) in the domain of \(\tan t\)
Step-by-Step Solution
Verified Answer
Using the addition identity for tangent, \( \tan(t + \pi) = \tan t \).
1Step 1: Use the Addition Identity for Tangent
Recall the identity for the tangent of a sum: \( \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \). We want to apply this identity with \( a = t \) and \( b = \pi \).
2Step 2: Calculate \( \tan(\pi) \)
We know that \( \tan(\pi) = 0 \) because the tangent of any multiple of \( \pi \) is zero, as \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \) and \( \sin(\pi) = 0 \), \( \cos(\pi) = -1 \).
3Step 3: Apply the Identity
Substitute into the addition identity: \( \tan(t + \pi) = \frac{\tan t + \tan(\pi)}{1 - \tan t \cdot \tan(\pi)} = \frac{\tan t + 0}{1 - \tan t \cdot 0} = \frac{\tan t}{1} = \tan t \).
4Step 4: Conclusion
We've shown that \( \tan(t + \pi) = \tan t \), hence the identity is proven for all \( t \) in the domain of \( \tan t \).
Key Concepts
Tangent Addition FormulaDomain of Tangent FunctionSine and Cosine Relationship
Tangent Addition Formula
The tangent addition formula is an essential trigonometric identity. It helps us find the tangent of the sum of two angles. This formula is given by:
Using sine and cosine addition formulas, we can express it as the above identity.
The tangent addition formula is useful for deriving other tangent identities and solving various trigonometric problems.
In our exercise, the formula helps show that \( \tan(t+\pi) = \tan t \). Applying this identity simplifies calculations and proves this trigonometric property.
- \( \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \)
Using sine and cosine addition formulas, we can express it as the above identity.
The tangent addition formula is useful for deriving other tangent identities and solving various trigonometric problems.
In our exercise, the formula helps show that \( \tan(t+\pi) = \tan t \). Applying this identity simplifies calculations and proves this trigonometric property.
Domain of Tangent Function
Understanding the domain of the tangent function is crucial in trigonometry. The function \( \tan \theta \) is defined as \( \frac{\sin \theta}{\cos \theta} \).
Thus, it is undefined when \( \cos \theta = 0 \), since division by zero is impossible.
This condition occurs at odd multiples of \( \frac{\pi}{2} \) (e.g., \( \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots \)).
Thus, it is undefined when \( \cos \theta = 0 \), since division by zero is impossible.
This condition occurs at odd multiples of \( \frac{\pi}{2} \) (e.g., \( \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots \)).
- The domain of \( \tan \theta \) excludes these points
- Symbolically, it is expressed as all real numbers except \( \theta = (2n+1)\frac{\pi}{2} \), where \( n \) is an integer
Sine and Cosine Relationship
The sine and cosine relationship lays the foundation for understanding trigonometric identities like tangent. Both functions describe the ratios of sides in a right triangle relative to the angle.
They are complementary functions:
The key relationships are the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), and their phase shifts.
These play a vital role in proof-based problems that utilize identities.
In the tangent addition formula, sine and cosine ties ensure any simplifications used in proofs of tangent properties, like \( \tan(t+\pi) = \tan t \), stem from these deep-seated relationships.
They are complementary functions:
- \( \sin(\theta) = \cos(\frac{\pi}{2} - \theta) \)
- \( \cos(\theta) = \sin(\frac{\pi}{2} - \theta) \)
The key relationships are the Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \), and their phase shifts.
These play a vital role in proof-based problems that utilize identities.
In the tangent addition formula, sine and cosine ties ensure any simplifications used in proofs of tangent properties, like \( \tan(t+\pi) = \tan t \), stem from these deep-seated relationships.
Other exercises in this chapter
Problem 33
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