Problem 66 - Precalculus Essentials

Question

Use a right triangle to write each expression as an algebraic expression. Assume that \(x\) is positive and that the given inverse trigonometric function is defined for the expression in \(x\). $$\sin \left(\cos ^{-1} 2 x\right)$$

Step-by-Step Solution

Answer

The algebraic expression for \( \sin(\cos^{-1}(2x)) \) is \( \sqrt{1 - 4x^2} \).

Step 1: Understand the expression and visualize a right triangle

Consider a right triangle ABC with angle BAC being \( \alpha = \cos^{-1}(2x) \). By definition, the length of the adjacent side (AB) to angle \( \alpha \) is \( 2x \) times the length of the hypotenuse (AC), then let AC = 1 and AB = \( 2x \). BC, the length of the opposite side can be found using the Pythagorean theorem.

Step 2: Apply the Pythagorean Theorem

The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. Hence, in triangle ABC, \( (BC)^2 = (AC)^2 - (AB)^2 = 1^2 - (2x)^2 = 1 - 4x^2. Thus, BC = \( \sqrt{1 - 4x^2} \). Since x is positive and \( \cos^{-1}(2x) \) is defined for the expression in x, BC is also positive.

Step 3: Find the desired expression

Finally, we need to find \( \sin(\cos^{-1}(2x)) \) which is the sine of angle \( \alpha \). In a right triangle, the sine of an angle is given by the ratio of the length of the side opposite the angle to the length of the hypotenuse. Therefore, in triangle ABC, \( \sin(\alpha) = \frac{BC}{AC} = \sqrt{1 - 4x^2} \). So the algebraic expression for \( \sin(\cos^{-1}(2x)) \) is \( \sqrt{1 - 4x^2} \).

Key Concepts

Right Triangle

A right triangle is a special type of triangle that has one angle measuring exactly 90 degrees. This angle is called the right angle and is often marked by a small square in diagrams. Right triangles have several unique properties that make them particularly useful in mathematics, especially in trigonometry.
When working with a right triangle, the side opposite the right angle is called the hypotenuse. It is always the longest side in the triangle. The other two sides are known as the adjacent side and the opposite side, relative to one of the non-right angles in the triangle.
These triangles are essential in trigonometry because they allow us to define the sine, cosine, and tangent functions. We can use right triangles to connect these trigonometric ratios with angles measured in radians and degrees. This connection is crucial when studying the inverse trigonometric functions, like in the given exercise.

Pythagorean Theorem

The Pythagorean theorem is a mathematical principle that applies to right triangles. It relates the lengths of the triangle's sides. Mathematically, it is expressed by the equation \((AC)^2 = (AB)^2 + (BC)^2\), where AC is the hypotenuse, and AB and BC are the other two sides of the triangle.
This theorem is incredibly useful when we need to find an unknown side of a right triangle, provided we have the lengths of the other two sides. For our current problem, the theorem helps us determine the length of the opposite side (BC) when we know the hypotenuse (AC) and the adjacent side (AB).
Let's see how the Pythagorean theorem works in our exercise. Using the given lengths, we calculated that \((BC)^2 = 1 - 4x^2\). Solving for BC, we find \(BC = \sqrt{1 - 4x^2}\). This calculation gives us the necessary length to find the sine of angle \(\alpha\).

Sine and Cosine Relationships

Sine and cosine are two of the primary trigonometric functions. They describe the rotation or angle of a right triangle's side ratios. The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side to the hypotenuse. Conversely, the cosine of an angle is the ratio of the adjacent side to the hypotenuse.
In dealing with inverse trigonometric functions, we work backwards. For example, if we have \(\cos^{-1}(2x)\), it means finding the angle whose cosine ratio is \(2x\). Once we determine this angle, we can then find the sine of this angle using the sine definition as explained above.
In our exercise, we needed \(\sin(\alpha)\), where \(\alpha = \cos^{-1}(2x)\). Given our right triangle setup, we found \(\sin(\alpha) = \frac{\sqrt{1-4x^2}}{1}\). This result shows how inverse trigonometric functions translate into other trigonometric relationships within a right triangle.