Problem 90 - Precalculus : Building Concepts and Connections

Question

Find the exact value of each expression without using a calculator. $$\sin \frac{\pi}{6}+\cot \frac{\pi}{6}$$

Step-by-Step Solution

Answer

The exact value of the expression \( \sin \frac{\pi}{6}+\cot \frac{\pi}{6} \) is \( \frac{1}{2} + \sqrt{3} \).

Step 1: Identify Known Values

From trigonometric identities, it is known that \( \sin \frac{\pi}{6} \) equals to \(\frac{1}{2}\), and \( \tan \frac{\pi}{6} \) equals to \(\frac{\sqrt{3}}{3}\).

Step 2: Calculate Reciprocal of Tangent

Since \( \cot \) is the reciprocal of \( \tan \), calculate the reciprocal of \( \tan \frac{\pi}{6} \), which will give \( \cot \frac{\pi}{6} = \frac{3}{\sqrt{3}}\). To rationalise the denominator, multiply the numerator and denominator by \( \sqrt{3} \), which gives \( \cot \frac{\pi}{6} = \sqrt{3}\).

Step 3: Sum up the Values

Now, add the values calculated \( \sin \frac{\pi}{6} + \cot \frac{\pi}{6} = \frac{1}{2} + \sqrt{3}\)

Key Concepts

Sine of Pi/6

The sine function measures the vertical component of an angle in the unit circle. For an angle of \frac{\text{π}}{6}\radians or 30 degrees, the exact value of \text{sin} \( \frac{\text{π}}{6} \) is well-known in trigonometry as \( \frac{1}{2} \). This value can be visualized in the special right triangle, where the opposite side is half the length of the hypotenuse. Understanding this fundamental value is crucial for solving a variety of trigonometric problems.

To remember this value, think of a 30°-60°-90° right triangle. The side opposite the 30° angle is half the length of the hypotenuse, illustrating why the sine of 30° or \( \frac{\text{π}}{6} \) is \( \frac{1}{2} \).

Cotangent of Pi/6

The cotangent function, expressed as \( \text{cot} \) is the reciprocal of the tangent function. In the case of \( \text{cot} \left( \frac{\text{π}}{6} \right) \), we begin by finding the tangent of the angle \( \tcancel{ \frac{\text{π}}{6} } \), which is known to be \( \frac{\text{√}3}{3} \). Therefore, to find the cotangent, you simply take the reciprocal of this value. When we flip \( \frac{\text{√}3}{3} \), we get \( \frac{3}{\text{√}3} \), but in trigonometry, it is standard to present values without radicals in the denominator, which leads us to the concept of rationalizing the denominator.

After rationalizing, the exact value of \( \text{cot} \left( \frac{\text{π}}{6} \right) \) is \( \text{√}3 \). This step might seem minor, but it’s important for maintaining consistency and clarity in mathematical communication.

Rationalizing the Denominator

Rationalizing the denominator is a technique used to eliminate radicals, such as square roots, from the bottom of a fraction. This is done for clarity and to ensure the result is in its simplest form. To rationalize a denominator, you multiply both the numerator and the denominator by the radical present in the denominator. For instance, to rationalize \( \frac{3}{\text{√}3} \), multiply the top and bottom by \( \text{√}3 \) to get \( \frac{3\text{√}3}{3} \) which simplifies to \( \text{√}3 \).

This method not only simplifies the expression but also helps when adding or comparing fractions with radicals, making it an essential skill in mathematics, especially in trigonometry.

Reciprocal Identities

Reciprocal identities are fundamental relationships in trigonometry that relate pairs of trigonometric functions. They express that the sine, cosine, tangent, cotangent, secant, and cosecant functions are reciprocals of each other in specific pairings. For example, the reciprocal of \( \text{sin} \) is \( \text{csc} \) (cosecant), the reciprocal of \( \text{tan} \) is \( \text{cot} \) (cotangent), and the reciprocal of \( \text{sec} \) is \( \text{cos} \) (cosine).

When given one function value, such as \( \text{tan} \left( \frac{\text{π}}{6} \right) \) in this case, you can find its reciprocal identity \( \text{cot} \left( \frac{\text{π}}{6} \right) \) easily. This illustrates the elegance and interconnectedness of trigonometric functions and their aliases in mathematical equations.