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Chapter 5: Problem 28

Write the expression in algebraic form. \(\cos \left(\arcsin \frac{x-h}{r}\right)\)

Short Answer

Expert verified
The algebraic form of the given expression is \( \frac{\sqrt{r^2 - x^2 + 2hx - h^2}}{r} \)

Step by step solution

01

Interpretation of the Expression

The expression given is \( \cos \left(\arcsin \frac{x-h}{r}\right) \). This is based on the property of the basic trigonometric function that \( \cos(\arcsin x) = \sqrt{1 - x^2} \). Here, \(x = \frac{x-h}{r}\)
02

Substituting the Value of \(x\)

Substitute \( x = \frac{x-h}{r} \) in the expression \( \cos(\arcsin x) = \sqrt{1 - x^2} \) to get the required algebraic form. So, it will be \( \sqrt{1 - \left(\frac{x-h}{r}\right)^2} \)
03

Simplification

Simplify the above equation to get the final algebraic form: \( \sqrt{1 - \left(\frac{x^2 - 2hx + h^2 }{r^2}\right)} = \frac{\sqrt{r^2 - x^2 + 2hx - h^2}}{r} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Algebraic Form
In mathematics, expressing a trigonometric function like cosine or sine in algebraic form means representing it as a polynomial or a combination of the basic arithmetic operations: addition, subtraction, multiplication, division, and root extraction, without trigonometric functions involved.

For instance, when we have a trigonometric expression involving an inverse operation like arcsin, rewriting it algebraically allows us to simplify complex expressions into forms that are often more understandable and easier to work with. This process may involve the use of Pythagorean theorem, which relates the sides of a right triangle to its hypotenuse, as well as other algebraic techniques for manipulating expressions.
Inverse Trigonometric Functions
Inverse trigonometric functions, often denoted as arcfunctions such as arcsin, arccos, and arctan, are used to find the angle that corresponds to a given trigonometric ratio. The arcsin function, for example, can be thought of as the inverse operation of the sine function.

It's important to remember that these functions return an angle whose sine, cosine, or tangent is the given number. However, since trigonometric functions are not one-to-one, we generally restrict the domain to ensure that each value corresponds to a unique angle.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for all admissible values. They are essential tools for simplifying trigonometric expressions and solving trigonometric equations.

One of the most fundamental identities is the Pythagorean identity, which states that for any angle \theta, \( \cos^2\theta + \sin^2\theta = 1 \). It's this identity that, in part, allows us to associate the trigonometric functions with the sides of a right triangle and ultimately to express trigonometric functions in algebraic terms using the inverse functions.
Simplifying Expressions
The process of simplifying expressions often involves reducing complex algebraic expressions to a simpler or more readable form without changing their value. This can be done by using arithmetic operations, factoring, expanding, and by employing various mathematical identities and properties.

In our example, simplifying the expression involves both algebraic manipulation through the expansion of a binomial and then grouping like terms. Also, it leverages the Pythagorean identity when dealing with the squares of the functions involved. Through simplification, we can obtain a clearer understanding of the relationship between the variables in an expression and more easily compute its value for specific instances.

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Most popular questions from this chapter

An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result in part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$ \frac{d v}{d t}=-\left(32+k v^{2}\right) $$ where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Find the velocity as a function of time by solving the equation $$ \int \frac{d v}{32+k v^{2}}=-\int d t $$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part (c) if \(k=0.001\). Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral \(\int_{0}^{t_{0}} v(t) d t\) where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. Let \(f\) be twice-differentiable and one-to-one on an open interval \(I\). Show that its inverse function \(g\) satisfies \(g^{\prime \prime}(x)=-\frac{f^{\prime \prime}(g(x))}{\left[f^{\prime}(g(x))\right]^{3}}\) If \(f\) is increasing and concave downward, what is the concavity of \(f^{-1}=g\) ?

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(f(x)=x^{n}\) where \(n\) is odd, then \(f^{-1}\) exists.

Find any relative extrema of the function. Use a graphing utility to confirm your result. $$ f(x)=x \sinh (x-1)-\cosh (x-1) $$

Find \(d y / d x\) at the given point for the equation. \(x=2 \ln \left(y^{2}-3\right), \quad(0,4)\)
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