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Find exact values of the given trigonometric functions without the use of a calculator. $$\arctan 0$$

Short Answer

Expert verified
The exact value of \( \arctan 0 \) is 0.

Step by step solution

01

Understanding the Function

The function \( \arctan \) or inverse tangent function, undoes the operation of the tangent function. It gives the angle whose tangent is a given number. In other words, \( \arctan (x) \) is the angle whose tangent is \( x \).
02

Finding the Value

As per the definition, \( \arctan (0) \) asks for the angle whose tangent is 0. Looking at the properties of the unit circle and associated trigonometric functions, it's known that \( \tan 0 = 0 \), as the tangent of 0 degrees or 0 radians is 0. So, \( \arctan 0 = 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions
Trigonometric functions are essential in mathematics, especially when dealing with angles and lengths in triangles. These functions relate the angles of a triangle to the ratios of its sides. The primary trigonometric functions include sine (**sin**), cosine (**cos**), and tangent (**tan**). Understanding these functions allows us to explore more complex concepts, including their inverses.

Sine of an angle is the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. Cosine is the ratio of the length of the adjacent side to the hypotenuse, while tangent is the ratio of the sine to the cosine, or equivalently, the opposite side to the adjacent side. For example, for an angle \( \theta \), the formulas are:
  • \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \)
  • \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \)
  • \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \)
These basic relationships form the foundation for exploring other properties of circles and angles, especially when using the unit circle approach for trigonometric calculations.
Arctan
The function \( \arctan \), or the inverse tangent function, is pivotal when you reverse the effect of the tangent function. It helps find the angle for which a given tangent value exists. If you have any real number, \( \arctan(x) \) will give you the angle \( \theta \) such that \( \tan(\theta) = x \).

The range of \( \arctan \) is typically from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) radians or from \(-90^\circ\) to \(90^\circ\). This range ensures that each \( x \) has one unique corresponding angle. Practical uses of \( \arctan \) include problems where the inverse of an angle's tangent needs to be calculated.

With the specific problem of \( \arctan(0) \), it asks us, "Which angle's tangent gives us 0?" The solution lies in the unit circle approach, where \( \tan(0) = 0 \), making \( \arctan(0) = 0 \). Understanding how the tangent function behaves around zero is key to solving this inverse problem.
Unit Circle
The unit circle is a foundational concept for understanding trigonometric functions and their inverses. It is a circle centered at the origin with a radius of one unit in the Cartesian coordinate system. The unit circle allows the representation of angles in radians and the visualization of trigonometric functions.

When working with the unit circle, any point \( (x, y) \) on the circle corresponds to an angle \( \theta \), where \( x = \cos(\theta) \) and \( y = \sin(\theta) \). Consequently, the tangent of the angle \( \theta \) is the ratio \( \frac{y}{x} \), which directly corresponds to \( \tan(\theta) \).
  • At \( \theta = 0 \), or the angle made with the positive x-axis, the point on the unit circle is \( (1, 0) \). Here, \( \tan(0) = \frac{0}{1} = 0 \).
The unit circle provides a clear visual aid and a consistent method for determining the values of trigonometric functions at various angles, making it an invaluable tool for understanding the relationship between angles and their trigonometric values.

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Most popular questions from this chapter

This set of exercises will draw on the ideas presented in this section and your general math background. Does the equation \(\sin (t+\pi)=\sin t+\sin \pi\) hold for all \(t\) ? Explain.

Graph at least two cycles of the given functions. $$h(x)=2 \cos \left(2 x+\frac{\pi}{2}\right)-1$$

The base of a railing for a staircase makes an angle of \(x\) degrees with the horizontal. Let \(d(x)\) be the horizontal distance between the two ends of the base of the railing. If point \(L\) on the railing is 5 feet higher than point \(M,\) find the positive number \(A\) such that \(d(x)=A\) cot \(x .\) Then use your function to find the length of the base of the railing if \(x=35^{\circ}\).

This set of exercises will draw on the ideas presented in this section and your general math background. What are the vertical asymptotes of the graph of \(f(x)=\tan x+\cot x ?\)

For Exercises \(61-72,\) fill in the given table with the missing information. A pproximate all nonexact answers to four decimal places. $$ \begin{array}{|r|c|c|c|c|} \hline & \text { Quadrant } & \sin t & \cos t & \tan t \\ \hline 61 . & \mathrm{I} & \frac{1}{2} & & \\ \hline 62 . & \mathrm{IV} & & \frac{1}{2} & \\ \hline 63 . & \mathrm{III} & & & 1 \\ \hline 64 . & \mathrm{II} & & & -1 \\ \hline 65 . & \mathrm{II} & & -\frac{1}{2} & \\ \hline 66 . & \mathrm{II} & & -\frac{\sqrt{3}}{2} & \\ \hline 67 . & \mathrm{IV} & -0.6 & & \\ \hline 68 . & \mathrm{III} & -0.8 & & \\ \hline 69 . & \mathrm{II} & & -\frac{5}{13} & \\ \hline 70 . & \mathrm{IV} & & \frac{12}{13} & \\ \hline 71 . & \mathrm{IV} & & & -2 \\ \hline 72 . & \mathrm{II} & & & \\ \hline \end{array} $$

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