Problem 48 - Precalculus with Limits

Question

In Exercises 9-50, verify the identity \( \cos(\sin^{-1} x) = \sqrt{1- x^2} \)

Step-by-Step Solution

Answer

The exercise is solved by making use of the definition \( y = \sin^{-1}{x} \) which implies \( x = \sin{y} \) , and then substituting it into the Pythagorean identity, which gives us \( \cos{\theta} = \sqrt{1 - x^{2}} \) . Afterwards, \( \theta \) is replaced by \( \sin^{-1}{x} \) to verify the identity \( \cos(\sin^{-1} x) = \sqrt{1- x^2} \).

Step 1: Understand the Relationship between \( \sin^{-1}{x} \) and \( x \)

If we let the angle \( \theta = \sin^{-1}{x} \), we know from the properties of inverse trigonometric functions that \( x = \sin{ \theta} \) .

Step 2: Apply the Pythagorean identity

In any right triangle, the Pythagorean theorem states that the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. This translates neatly into trigonometry as having \( \cos^{2}{\theta} = 1 - \sin^{2}{ \theta} \) . Now, we've shown in step 1 that \( \sin{\theta} = x \). Substituting into this identity, we get \( \cos^{2}{ \theta} = 1 - x^{2} \). Taking square root on both sides give us \( \cos{\theta} = \sqrt{1 - x^{2}} \) . Confused about whether it's positive or negative root? Remember that we have stated initially that \( \theta = \sin^{-1}{x} \) , which means \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \). In this range, cosine function is always nonnegative, so we only take the positive root..

Step 3: Replace \( \theta \)

Finally, recall that we originally defined \( \theta = \sin^{-1}{x} \). So replace \( \theta \) with \( \sin^{-1}{x} \) in \( \cos{\theta} = \sqrt{1 - x^{2}} \). Therefore, this proves that \( \cos(\sin^{-1} x) = \sqrt{1- x^2} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential in mathematics. They help us convert a trigonometric ratio back into an angle. Consider the sine function. The inverse sine, denoted as \( \sin^{-1} \), is used to find the angle whose sine is a given number. If we say \( \theta = \sin^{-1} x \), it means \( \theta \) is the angle such that \( \sin \theta = x \).

Inverse trigonometric functions are particularly useful in situations where we know the values of a trigonometric ratio and want to determine the corresponding angle.

• \( \sin^{-1} x \) maps from \([-1, 1] \) to \([-\frac{\pi}{2}, \frac{\pi}{2}] \).
• Since \( \theta \) is defined as \( \sin^{-1} x \), \( x \) must be within the range \([-1, 1] \).
• This function helps unravel trigonometric equations, converting them into angle measurements.

By understanding the domain and range of these functions, you can solve equations and understand trigonometric identities better.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry and trigonometry. It's the basis for understanding the relationship between the sides of right triangles. According to the theorem, in a right triangle, the square of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the other two sides. This can be expressed as:

• \( a^2 + b^2 = c^2 \)

In trigonometry, this theorem manifests through identities. For instance, \( \cos^2 \theta + \sin^2 \theta = 1 \).

When \( \theta = \sin^{-1} x \), and understood that \( \sin \theta = x \), we can derive:

• \( \cos^2 \theta = 1 - \sin^2 \theta = 1 - x^2 \)
• Therefore, \( \cos \theta = \sqrt{1 - x^2} \)

Remember, cosine is non-negative in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\), confirming why we select the positive square root.

Trigonometric Functions

Trigonometric functions are some of the foundational functions in mathematics, especially in the study of angles and triangles. The main trigonometric functions are sine, cosine, and tangent. These functions derive from the relationships between the angles and sides of a right triangle.

• Sine (\( \sin \)): Opposite side over hypotenuse.
• Cosine (\( \cos \)): Adjacent side over hypotenuse.
• Tangent (\( \tan \)): Opposite side over adjacent side.

The beauty of trigonometric functions lies in their periodicity and symmetry. This is what makes them so useful in modeling cyclic phenomena.

For example, in verifying identities such as \( \cos(\sin^{-1} x) = \sqrt{1- x^2} \), we see how these functions interplay with each other. You understand that:

• The value of \( \sin \) directly affects \( \cos \), as shown in the identity transformation using the Pythagorean identity.
• Angle determination through inverse functions highlights these relationships effectively.

Mastering these functions allows for a deeper understanding of how angles and distances relate in various fields, from physics to engineering.