Problem 21 - Precalculus Essentials

Question

In Exercises \(15-22,\) write each expression as the sine, cosine, or tangent of a double angle. Then find the exact value of the expression. $$\frac{2 \tan \frac{\pi}{12}}{1-\tan ^{2} \frac{\pi}{12}}$$

Step-by-Step Solution

Answer

The expression \(\frac{2 \tan \frac{\pi}{12}}{1-\tan ^{2} \frac{\pi}{12}}\) is equivalent to \(\tan \frac{\pi}{6}\) and its exact value is \(\sqrt{3}/3\).

Step 1: Identify the relevant double angle formula

Given the expression \(\frac{2 \tan \frac{\pi}{12}}{1-\tan ^{2} \frac{\pi}{12}}\), it is in the format of double angle formula for tangent which is \(\tan 2\theta = \frac{2 \tan \theta}{1- \tan^2 \theta}\). Hence, the expression is equivalent to \(\tan 2\theta\), where \(\theta\) is \(\frac{\pi}{12}\)

Step 2: Substitution

Substitute the value of \(theta\) into \(2 \theta\). Hence, we get \(\tan 2 * \frac{\pi}{12} = \tan \frac{\pi}{6}\)

Step 3: Find the exact value of the expression

The exact value of \(\tan \frac{\pi}{6}\) can be found using the standard trigonometric values, which is \(\sqrt{3}/3\).

Key Concepts

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables involved. They are an essential part of solving many types of trigonometric problems, including simplifying expressions, proving other identities, and solving equations.

One of the most valuable sets of trigonometric identities are those for double angles. The double angle formulas express trigonometric functions of double angles – that is angles of the form 2θ – in terms of functions of the original angle θ. For example, the double angle formula for sine is
\[\sin(2\theta) = 2\sin(\theta)\cos(\theta)\]
and for cosine, we have \[\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)\]
or alternatively,
\[\cos(2\theta) = 2\cos^2(\theta) - 1 = 1 - 2\sin^2(\theta)\]
The double angle formula for tangent, which was used in the given exercise, is
\[\tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)}\]
Having these identities at hand enables us to tackle more sophisticated problems easily.

Exact Trigonometric Values

Having a solid understanding of exact trigonometric values is crucial in solving trigonometric problems without a calculator. These values are based on the unit circle and special triangles, such as the 45°-45°-90° triangle and the 30°-60°-90° triangle. The angles associated with these triangles, which are multiples of \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), and \(\frac{\pi}{3}\), have trigonometric values that can be expressed as simple fractions or radical expressions.

For example, some exact values include:

• \(\sin(\frac{\pi}{6}) = \frac{1}{2}\)
• \(\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\)
• \(\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\)

These values are based on the symmetric properties of the unit circle and can be used for finding the sine, cosine, and tangent of any angle that is a multiple of these special angles. In our exercise, the exact value of \(\tan(\frac{\pi}{6})\) turns out to be \(\frac{\sqrt{3}}{3}\), based on its relation to the 30°-60°-90° triangle.

Solving Trigonometric Equations

Solving trigonometric equations involves finding the angles that satisfy a given trigonometric statement. This often requires the use of trigonometric identities and known exact trigonometric values, as seen in the given exercise. The process typically involves isolating the trigonometric function and using algebraic techniques to solve for the variable angle.

Here’s a typical approach to solving a trigonometric equation:

1. First, simplify the trigonometric equation using known identities.
2. If needed, apply inverse trigonometric functions to both sides to find the angle solutions.
3. Check if there are additional solutions within a given interval based on the periodicity of the trigonometric functions.
4. Verify the solutions using the exact trigonometric values or a unit circle, if necessary.

Finding Additional Solutions:

In some cases, there are multiple angles that satisfy a given trigonometric equation due to the periodic nature of trigonometric functions. For instance, a function like sine or cosine repeats every \(2\pi\) units, so if \(\theta\) is a solution, then \(\theta + 2k\pi\), where \(k\) is an integer, will also be a solution.

In the context of the given exercise, we translated the given tangent expression into a double angle formula and then used the exact value of \(\tan(\frac{\pi}{6})\) to find the exact value. Similar strategies can be adapted to different forms of trigonometric equations.