Problem 136 - Precalculus

Question

For the following exercises, find the exact value, if possible, without a calculator. If it is not possible, explain why. $$ \tan ^{-1}\left(\sin \left(\frac{-5 \pi}{2}\right)\right) $$

Step-by-Step Solution

Answer

The exact value is \(-\frac{\pi}{4}\).

Step 1: Simplify the Angle in Sine

The first step is to simplify the angle \( \frac{-5\pi}{2} \) used in the sine function. Since \( \sin(\theta) \) has a period of \( 2\pi \), we can add \( 2\pi \) repeatedly until the angle is within the range \([0, 2\pi)\).\\[\frac{-5\pi}{2} + 2\pi = \frac{-\pi}{2}\]. Since \(-\pi/2\) is within the range \([-\pi, \pi]\), we can use this value directly.

Step 2: Evaluate the Sine Function

Evaluate \( \sin(\frac{-\pi}{2}) \). From the unit circle, \( \sin(\frac{-\pi}{2}) = -1 \). Therefore, \( \sin(\frac{-5\pi}{2}) = -1 \).

Step 3: Apply the Inverse Tangent Function

Now, we need to find \( \tan^{-1}(-1) \). The inverse tangent function \( \tan^{-1}(x) \) returns values in the range \((-\frac{\pi}{2}, \frac{\pi}{2})\). We know that \( \tan(-\frac{\pi}{4}) = -1 \), so \( \tan^{-1}(-1) = -\frac{\pi}{4} \).

Key Concepts

Understanding Trigonometric Functions

Trigonometric functions are fundamental in mathematics, particularly in dealing with angles and triangles. They help us describe the relationships between the angles and sides of triangles, and they also extend to circles. The main trigonometric functions include sine (\(\sin\)), cosine (\(\cos\)), and tangent (\(\tan\)). These functions are defined as ratios:

• The sine of an angle in a right triangle is the opposite side divided by the hypotenuse.
• The cosine is the adjacent side divided by the hypotenuse.
• The tangent is the opposite side divided by the adjacent side.

In the problem given, we are primarily focused on the sine function and the inverse tangent function.
The sine function repeats its values every \(2\pi\) radians due to the circle's symmetry. This periodic behavior allows us to simplify angles outside the \([0, 2\pi)\) range. On the other hand, inverse functions, like \(\tan^{-1}\), return an angle whose tangent is the given value.
They "undo" the original trigonometric function by providing the angle rather than the ratio.

Exploring the Unit Circle

The unit circle is a powerful tool for understanding trigonometric functions and their inverses. It is a circle with a radius of 1, centered at the origin of a coordinate plane. Key angles on the unit circle are measured in radians, which are a more natural angle measure for calculus and higher mathematics.
On the unit circle:

• The x-coordinate represents the cosine of the angle.
• The y-coordinate represents the sine of the angle.

In our exercise, the unit circle is particularly useful for evaluating \(\sin\left(\frac{-5\pi}{2}\right)\). This angle is not immediately on the unit circle, but since \(\sin\) has a period of \(2\pi\), we can find an equivalent angle:
Adding \(2\pi\) to \(\frac{-5\pi}{2}\) simplifies to \(\frac{-\pi}{2}\). This simplification tells us to look at the angle corresponding to this new position, which we find using the unit circle to be the point where y-coordinate \(-1\) is achieved. Subsequently, the sine of \(\frac{-\pi}{2}\) is \(-1\).

Concept of Angle Simplification

Angle simplification is the key to solving many trigonometric problems, especially when dealing with non-standard angles. Since trigonometric functions are periodic, they repeat their values over a specified interval. For sine and cosine, this interval is \(2\pi\), while for tangent, it is \(\pi\).
When simplifying an angle like \(\frac{-5\pi}{2}\), the goal is to bring it into a common reference interval where the function's values are well known, typically \([0, 2\pi)\) or \([-\pi, \pi]\) for sine. By repeatedly adding or subtracting the function’s period, we aim to transform a complex angle into one that is more manageable. This way, you only need to know the trigonometric values of angles within this simpler interval.
In our example, after simplifying \(\sin\left(\frac{-5\pi}{2}\right)\) to \(\sin\left(\frac{-\pi}{2}\right)\), the next step is direct and relies on memorization or familiarity with the unit circle to know that this is \(-1\).