Problem 109 - A Graphical Approach to Precalculus with Limits

Question

Write each expression as an algebraic expression in \(u, u>0\). $$\tan \left(\sin ^{-1} \frac{u}{\sqrt{u^{2}+2}}\right)$$

Step-by-Step Solution

Answer

The expression simplifies to \( \frac{u}{\sqrt{2}} \).

Step 1: Understanding the Expression

We need to simplify \( \tan(\sin^{-1} y) \). The given expression is \( \tan \left(\sin^{-1} \frac{u}{\sqrt{u^{2}+2}}\right) \). We need to find \( y = \frac{u}{\sqrt{u^{2}+2}} \) first.

Step 2: Identify y in terms of u

Here, the value of \( y \) is \( \frac{u}{\sqrt{u^2+2}} \). This expression is the sine value in the inverse sine function.

Step 3: Use a Right Triangle to Find the Sides

To find \( \tan(\sin^{-1} y) \), visualize a right triangle where the angle \( \theta = \sin^{-1} y \). Then the opposite side is \( u \), the hypotenuse is \( \sqrt{u^2+2} \), and we need to find the adjacent side.

Step 4: Calculate the adjacent side

Using the Pythagorean theorem, adjacent side = \( \sqrt{(\sqrt{u^2+2})^2 - u^2} = \sqrt{u^2+2 - u^2} = \sqrt{2} \).

Step 5: Calculate \( \tan(\theta) \)

\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{u}{\sqrt{2}} \].

Step 6: Write the Expression

Therefore, \( \tan \left(\sin^{-1} \frac{u}{\sqrt{u^2+2}}\right) = \frac{u}{\sqrt{2}} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find the angle that corresponds to a given trigonometric value. While trigonometric functions like sine, cosine, and tangent take an angle and return a ratio, inverse trigonometric functions take a ratio and return the angle. These functions are especially handy in real-world situations where the angle is the missing piece, not the side lengths.

• For example, \( \sin^{-1}(x) \) gives us the angle whose sine is \( x \).
• With \( \tan(\sin^{-1} y) \), the \( \sin^{-1} \) portion determines the angle where the sine of that angle equals \( y \).

In our exercise, we're working with \( \tan(\sin^{-1} \frac{u}{\sqrt{u^2 + 2}}) \). Here, the key step is finding what angle gives us the sine ratio of \( \frac{u}{\sqrt{u^2 + 2}} \). This angle then feeds into the tangent function.

Right Triangle Relationships

Right triangles are a primary way to understand trigonometric functions and their inverses. In our scenario, when you see \( \sin^{-1} \frac{u}{\sqrt{u^2+2}} \), visualize this as the angle \( \theta \) in a right triangle.

• The opposite side length relative to \( \theta \) is \( u \).
• The hypotenuse is \( \sqrt{u^2+2} \).

To find \( \tan(\theta) \), we must first know all sides of the triangle. As per the Pythagorean theorem, if you know two sides, you can find the third. Here, we determine the length of the adjacent side as \( \sqrt{2} \) using this relationship:

• The adjacent side is \( \sqrt{\text{hypotenuse}^2 - \text{opposite}^2} \).
• This simplifies to \( \sqrt{u^2+2 - u^2} = \sqrt{2} \).

This relationship provides us with the necessary information to resolve \( \tan(\theta) \).

Algebraic Simplification

Algebraic simplification involves taking complex expressions and reducing them to simpler forms. It's like uncluttering a messy desk, where cleaning it up makes the workspace clearer and tasks more manageable.
In the problem, you start with a complex expression: \( \tan(\sin^{-1} \frac{u}{\sqrt{u^2+2}}) \). To simplify, you used right triangle relationships to resolve the trigonometric functions into a straightforward ratio. This process included:

• Identifying the geometric relationships between sides of a triangle based on given trigonometric values.
• Finding the tangent of the angle, obtained using inverse sine, via the formula \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \).

After the simplification, the expression \( \tan(\sin^{-1} \frac{u}{\sqrt{u^2+2}}) \) becomes \( \frac{u}{\sqrt{2}} \), a manageable algebraic expression in terms of \( u \). This makes computations simpler and shows the power of combining trigonometric and algebraic concepts.