Tan (Theta - Pie) Answer Explained

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SUMMARY

The discussion clarifies the mathematical expression Tan(Theta - Pi) and confirms that it equals Tan(Theta). The tangent function has a periodicity of π radians, which leads to the identity Tan(Theta + Pi) = Tan(Theta). The confusion arises from the misinterpretation of "pie" as a dessert instead of "pi," the mathematical constant. The sum/difference identity for tangent is also referenced to further validate the conclusion.

PREREQUISITES
  • Understanding of trigonometric functions, specifically tangent.
  • Familiarity with the concept of periodicity in trigonometric functions.
  • Knowledge of mathematical identities, particularly the sum/difference identity for tangent.
  • Basic understanding of radians and their application in trigonometry.
NEXT STEPS
  • Study the properties of the tangent function, focusing on its periodicity.
  • Learn about the sum and difference identities for trigonometric functions.
  • Explore the implications of radians versus degrees in trigonometric calculations.
  • Practice solving trigonometric equations involving periodic functions.
USEFUL FOR

Students of mathematics, educators teaching trigonometry, and anyone interested in understanding the properties of the tangent function and its identities.

captainnumber36
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What does Tan (Theta - Pie) = ?

I know Tan (theta + pie) = tan (theta).

They say the answer is tan (theta), but I think it's some kind of typo.
 
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captainnumber36 said:
What does Tan (Theta - Pie) = ?

I know Tan (theta + pie) = tan (theta).

They say the answer is tan (theta), but I think it's some kind of typo.

Pie is a dessert, while pi is a Greek letter used to represent the ratio of a circle's circumference to its diameter. Having said that, the period of the tangent function is $\pi$ radians, which means:

$$\tan(\theta+\pi k)=\tan(\theta)$$ where $k\in\mathbb{Z}$ (this means $k$ can be any integer, even negative ones)

So, it's not a typo, your book is correct.
 
captainnumber36 said:
What does Tan (Theta - Pie) = ?

I know Tan (theta + pie) = tan (theta).

They say the answer is tan (theta), but I think it's some kind of typo.

sum/difference identity for tangent ...

$\tan(a \pm b) = \dfrac{\tan{a} \pm \tan{b}}{1 \mp \tan{a} \cdot \tan{b}}$

now, let $a = \theta$ and $b = \pi$ ... substitute & evaluate
 

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