Problem 35 - Calculus of a Single Variable

Question

Solve the equation for \(x\). \(\arcsin \sqrt{2 x}=\arccos \sqrt{x}\)

Step-by-Step Solution

Answer

\(x = 1/3\)

Step 1: Identify the ranges of the functions

The arcsine and arccosine functions have a domain in \([-1,1]\) and range in \([0,\pi]\) and \([0, \pi/2]\), respectively. Therefore, \(\sqrt{2x}\) and \(\sqrt{x}\) must be in \([-1,1]\). This gives us: \(0 \leq x \leq 0.5\) for \(\sqrt{2x}\) and \(0 \leq x \leq 1\) for \(\sqrt{x}\). Thus, \(x\) must be in the range \([0, 0.5]\).

Step 2: Solve the equation

Knowing the range of the sine and cosine functions allows us to transform the equation. Since \(\sin(\pi/2 - a) = \cos(a)\) for any angle \(a\), we can rewrite our equation: \(\arcsin \sqrt{2x} = \pi/2 - \arccos \sqrt{x}\). Then, take the sine of both sides, and we get: \(\sqrt{2x} = \sin(\pi/2 - \arccos \sqrt{x}) = \sqrt{1 - x}\), by applying \(\sin(\pi/2 - a) = \sqrt{1 - (\cos a)^2}\). Finally, square both sides to get the equation \((2x) = (1 - x)\), which simplifies to \(3x = 1\).

Step 3: Check the solution

Solving for \(x\) gives \(x = 1/3\). However, we need to check whether this solution falls within the initial range of \(x\) obtained in Step 1. Indeed, \(1/3\) is in the interval \([0, 0.5]\). Therefore, \(x = 1/3\) is a valid solution.

Key Concepts

Arcsine Function

The arcsine function, denoted by \(\arcsin\), is the inverse of the sine function. This means it is used to determine the angle whose sine is a given number. To grasp this concept, it's crucial to understand that the sine function takes an angle and gives a number, while arcsine takes a number and gives an angle.

• The domain of the arcsine function is \( [-1,1] \), meaning it only accepts inputs within this interval.
• The range is \( [0,\pi] \), which corresponds to the angles measured in radians from \(\frac{\radians}{2}}\) for \(\frac{\radians}{2}}\).

For example, if you want to find the angle whose sine is \(\frac{1}{2}\), you would calculate \(\frac{1}{2}\), you would calculate \(\frac{1}{2}\).

Arccosine Function

Moving on to the arccosine function, or \(\arccos\), it serves as the inverse of the cosine function. This function provides the answer to questions like 'For which angle is the cosine of that angle equal to a certain number?'.

• Its domain is identical to the arcsine, \( [-1,1] \).
• However, the range differs slightly, spanning from \( [0, \frac{\radians}{2}}\).

So, to obtain the angle with a cosine of \(\frac{\radians}{2}}\) you'd use \(\frac{\radians}{2}}\).Keeping these parameters in mind is essential when solving trigonometric equations like in our exercise.

Domains and Ranges of Trigonometric Functions

Grasping the domains and ranges of trigonometric functions is the cornerstone of solving trigonometric equations. Each trigonometric function has its specific domain and range:

• For sine and cosine, the domain is \( [-1,1] \) but they differ in ranges where sine's range is \( [-\frac{\radians}{2}}\) and cosine's is \( [0, \frac{\radians}{2}}\).
• For other trigonometric functions like tangent and cotangent, the domain excludes certain points where these functions are undefined.

Understanding these constraints allows one to find valid solutions within the correct intervals, such as ensuring the solution to an arcsine or arccosine falls within their respective ranges.

Inverse Trigonometric Functions

Inverse trigonometric functions are the reverse of traditional trigonometric functions, providing an angle as output for a given real number input. They include \(\arcsin\), \(\arccos\), and \(\arctan\), among others.

• These functions are critical for solving trigonometric equations where an angle must be found from a known sine, cosine, or tangent value.
• The key is to remember that their outputs are angles and to ensure these angles are within the correct range to be physically or geometrically meaningful.

The use of inverse trigonometric functions is evident in our initial problem, where both \(\arcsin\) and \(\arccos\) are utilized to find the value of \(x\) that satisfies the given equation.

Sine and Cosine Relationship

The relationship between sine and cosine is foundational in trigonometry and is particularly useful when dealing with equations involving both functions.

• Two key identities reflect their relationship: \(\sin(a) = \cos(\frac{\radians}{2}}}}\) \(\cos(a) \) and \(\sin(a) = \sqrt{1 - (\cos(a))^2}\).
• These identities provide alternate expressions for sine or cosine, which is critical for transforming and ultimately solving equations like the one in our exercise.
• Moreover, knowing that sine and cosine are essentially shifts of each other on the unit circle can often simplify complex trigonometric problems into solvable forms.

This interplay between sine and cosine ensures that if we know one, we can often determine the other, as shown step-by-step in transforming and solving the given equation.