Problem 46 - Algebra 2

Question

ACT/SAT If \(\sin x=m\) and \(0

Step-by-Step Solution

Answer

C: \( \tan x = \frac{m}{\sqrt{1-m^2}} \)

Step 1: Understand the trigonometric identity

To solve for \( \tan x \) given \( \sin x = m \), we need to use trigonometric identities. We know that for any angle \( x \), \( \tan x = \frac{\sin x}{\cos x} \).

Step 2: Find \( \cos x \) in terms of \( m \)

Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we have \( m^2 + \cos^2 x = 1 \). Solving for \( \cos x \), we get \( \cos x = \sqrt{1 - m^2} \) since \( 0 < x < 90^{\circ} \), and cosine is positive in this range.

Step 3: Substitute into the tangent formula

Now substitute the expressions for \( \sin x \) and \( \cos x \) into the tangent formula: \(\tan x = \frac{\sin x}{\cos x} = \frac{m}{\sqrt{1-m^2}} \).

Step 4: Match with the given options

The expression \( \frac{m}{\sqrt{1-m^2}} \) matches option C. Therefore, \( \tan x = \frac{m}{\sqrt{1-m^2}} \) is the correct answer.

Key Concepts

Pythagorean identity

The Pythagorean identity is a cornerstone in trigonometry. It ties together the sine and cosine functions in a very fundamental way. This identity states that for any angle \( x \), the equation \( \sin^2 x + \cos^2 x = 1 \) always holds true. This equation can be seen as a reflection of the Pythagorean Theorem in a circle of radius 1, also known as the unit circle. Here, \( \sin x \) and \( \cos x \) represent the lengths of the legs of a right triangle, while 1 is the hypotenuse.

In practice:

• When you know one of the trigonometric functions, like sine or cosine, you can use this identity to find the other.
• It helps us convert between trigonometric functions, which is useful for solving various problems.
• In our case, given \( \sin x = m \), the identity helps us express \( \cos x \) as \( \cos x = \sqrt{1 - m^2} \) assuming \( 0 < x < 90^\circ \), which ensures both sine and cosine are positive.

sine and cosine functions

The sine and cosine functions are essential trigonometric functions that describe the relationships between the angles and sides of right triangles. Depending on the angle \( x \), they can change values smoothly, and are periodic functions with a cycle every \( 360^\circ \) or \( 2\pi \) radians.

Understanding sine and cosine:

• \( \sin x \) is the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle.
• \( \cos x \) is the ratio of the length of the adjacent side to the hypotenuse.
• They oscillate between -1 and 1, but for \( 0 < x < 90^\circ \), both functions remain positive.

In combination, they can describe not only the length of triangle sides but also as components of circular motion, where \( \cos x \) represents the horizontal component and \( \sin x \) represents the vertical component.

tangent function

The tangent function is another fundamental trigonometric function, which is derived from the sine and cosine functions. For a given angle \( x \), the tangent is defined as the ratio of sine to cosine:\( \tan x = \frac{\sin x}{\cos x} \). This makes the tangent function particularly useful because it directly links two already important functions: sine and cosine.

Key points about tangent:

• \( \tan x \) can be interpreted as the slope of the line that represents the angle \( x \) in the coordinate plane.
• As a ratio, it's undefined where \( \cos x = 0 \) (e.g., \( x = 90^\circ \), \( 270^\circ \)) because division by zero is undefined.
• In our problem, knowing \( \sin x = m \) adds specific meaning: substituting \( m \) and our expression for \( \cos x \), we use \( \tan x = \frac{m}{\sqrt{1-m^2}} \).
• This derivation allows us to visualize \( \tan x \) as dependent on both \( m \) and the relationship defined by the Pythagorean identity.

Understanding the fundamental nature of \( \tan x \) enables solving complex problems involving triangle properties and circular functions.