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Find the exact value of the indicated function (no decimals). Note that since the degree sign is not used, the angle is assumed to be in radians. $$\tan \frac{\pi}{6}$$

Short Answer

Expert verified
\(\tan\frac{\pi}{6} = \frac{\sqrt{3}}{3}\)

Step by step solution

01

Understanding the Tangent Function

The tangent of an angle in a right triangle is the ratio of the length of the opposite side to the length of the adjacent side. For angles in radians, we often use unit circle properties to find these values.
02

Use the Unit Circle and Special Triangles

In a unit circle, the angle \(\frac{\pi}{6}\) radians corresponds to 30 degrees. This angle forms a special 30-60-90 triangle where the ratios of the sides are 1:\(\sqrt{3}\):2. The tangent is the ratio of the side opposite the 30-degree angle (which is 1) to the side adjacent to the 30-degree angle (which is \(\sqrt{3}\)).
03

Calculating the Tangent Value

Therefore, the tangent of \(\frac{\pi}{6}\) is the ratio \(\frac{1}{\sqrt{3}}\). To rationalize the denominator, multiply the numerator and the denominator by \(\sqrt{3}\) to get \(\frac{\sqrt{3}}{3}\).

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Key Concepts

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

Unit Circle
The unit circle is a fundamental concept in trigonometry and an essential tool for solving problems involving trigonometric functions. It is a circle with a radius of 1 unit centered at the origin of a coordinate plane. Points on the unit circle correspond to angles, and for any angle, there's a unique point on the perimeter of the unit circle. This defines the coordinates of the point, \( (x, y) \), where \( x \) and \( y \) are the cosine and sine of the angle, respectively.

When dealing with the tangent function, the unit circle helps us visualize the angle and the ratio that defines the tangent. Especially for angles like \( \frac{\pi}{6} \), or 30 degrees, the unit circle is helpful because it allows us to draw a special kind of triangle (specifically, the 30-60-90 triangle) and use it to find the exact value of the tangent function. Remember, the tangent of an angle in the unit circle is equivalent to \( \frac{y}{x} \) — the ratio between the y-coordinate (opposite side) and the x-coordinate (adjacent side) of the corresponding point on the circle.
Special Triangles
Understanding special triangles is crucial when working with trigonometric functions, such as the tangent. These are triangles with angles and side lengths that follow a specific pattern, making it easier to remember and calculate their trigonometric ratios. One of the most common special triangles is the 30-60-90 triangle.

30-60-90 Triangle

In this type of right triangle, the angles are always 30 degrees, 60 degrees, and 90 degrees. The sides opposite these angles have specific ratios: if the shortest side (opposite the 30-degree angle) is 1 unit, then the side opposite the 60-degree angle is \( \sqrt{3} \) units, and the hypotenuse is 2 units long. These ratios simplify the process of finding the sine, cosine, and tangent of the angles involved. Specifically, for the tangent of a 30-degree angle (or \( \frac{\pi}{6} \) radians), the ratio involves only the side opposite the 30-degree angle and the side opposite the 60-degree angle. This knowledge directly helps to solve the exercise by providing a clear understanding of the relationships between the sides and the angles.
Radians
Radians are an alternative way of measuring angles to degrees, and they are the standard unit of angular measure used in many areas of mathematics. One complete revolution around a circle is \( 2\pi \) radians, equivalent to 360 degrees. Using radians can often simplify calculations in trigonometry because it directly relates the angle size to the arc length on a unit circle.

Radians are especially important when working with trigonometric functions because they provide a more natural way to describe the relationships between angles and the lengths of arcs. For instance, the angle \( \frac{\pi}{6} \) radians points to using a pi-based system which matches with the unit circle's circumference. This change from degrees to radians often provides an immediate visual cue for students that a unit circle or a special triangle relationship might be useful in finding the solution, thus improving their ability to translate between different angle measurements and apply appropriate trigonometric concepts.

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Most popular questions from this chapter

Find the exact value of the indicated function (no decimals). Note that since the degree sign is not used, the angle is assumed to be in radians. $$\sin \frac{\pi}{3}$$

Steamboat Problem: Mark Twain sat on the deck of a river steamboat. As the paddle wheel turned, a point on the paddle blade moved so that its distance, \(d\), from the water's surface was a sinusoidal function of time. When Twain's stopwatch read 4 sec, the point was at its highest, \(16 \mathrm{ft}\) above the water's surface. The wheel's diameter was \(18 \mathrm{ft},\) and it completed a revolution every \(10 \mathrm{sec}\) a. Sketch the graph of the sinusoid. b. What is the lowest the point goes? Why is it reasonable for this value to be negative? c. Find a particular equation for distance as a function of time. (IMAGE CANNOT COPY) d. How far above the surface was the point when Mark's stopwatch read 17 sec? e. What is the first positive value of \(t\) at which the point was at the water's surface? At that time, was the point going into or coming out of the water? How can you tell? f. "Mark Twain" is a pen name used by Samuel Clemens. What is the origin of that pen name? Give the source of your information.

Sketch the sinusoid described and write a particular equation for it. Check the equation on your grapher to make sure it produces the graph you sketched. The period equals \(72^{\circ},\) amplitude is 3 units, phase displacement (for \(y=\cos \theta\) ) equals \(6^{\circ}\) and the sinusoidal axis is at \(y=4\) units.

Sketch one cycle of the graph of the parent sinusoid \(y=\cos \theta,\) starting at \(\theta=0^{\circ} .\) What is he amplitude of this graph?

Inflection Point Problem: Sketch the graph of a function that has high and low critical points. On the sketch, show a. A point of inflection b. A region where the graph is concave up c. A region where the graph is concave down

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