Problem 29 - Precalculus Mathematics for Calculus

Question

Find all solutions of the given equation. $$5 \sin \theta-1=0$$

Step-by-Step Solution

Answer

The solutions are: \( \theta = 0.201 + 2k\pi \) and \( \theta = 2.940 + 2k\pi \) for any integer \(k\).

Step 1: Isolate the Trigonometric Function

Start by isolating the sine function in the equation: \[5 \sin \theta - 1 = 0\]To isolate \(\sin \theta\), add 1 to both sides:\[5 \sin \theta = 1\]Now, divide both sides by 5:\[\sin \theta = \frac{1}{5}\]

Step 2: Find the General Solution

Since \(\sin \theta = \frac{1}{5}\), we need to find the angle \(\theta\) such that the sine of the angle equals \(\frac{1}{5}\). The general solution for \(\theta\) when \(\sin \theta = a\) is:\[\theta = \arcsin(a) + 2k\pi \quad \text{and} \quad \theta = \pi - \arcsin(a) + 2k\pi\]where \(k\) is any integer. Let's compute:\[\theta = \arcsin\left(\frac{1}{5}\right) + 2k\pi \]and\[\theta = \pi - \arcsin\left(\frac{1}{5}\right) + 2k\pi\]

Step 3: Provide Numeric Approximation

Use a calculator to find the numerical approximation of \(\arcsin\left(\frac{1}{5}\right)\):\[\theta \approx \arcsin\left(\frac{1}{5}\right) \approx 0.201\text{ (in radians)}\]So, the two possible values for \(\theta\) from the general solutions are:\[\theta = 0.201 + 2k\pi \quad \text{and} \quad \theta = \pi - 0.201 + 2k\pi\]The second solution numerically is:\[\theta \approx 2.940 + 2k\pi\]where \(k\) is any integer.

Key Concepts

sine function

The sine function is a fundamental concept in trigonometry that describes the relationship between the angles and sides of triangles, particularly right-angled triangles. In the unit circle, the sine of an angle is defined as the y-coordinate of the point where the terminal side of the angle intersects the circle. This is because the unit circle has a radius of 1, and it significantly simplifies calculations in trigonometric concepts. The general sine function is expressed as \(\sin \theta\), where \(\theta\) is the angle in question. The sine function is periodic with a period of \(2\pi\), meaning it repeats its values every \(2\pi\). This characteristic is essential in finding solutions to trigonometric equations, as it ensures there are infinitely many angles corresponding to a particular sine value, each varying by some multiple of \(2\pi\). Some key features of the sine function include:

• Range: The sine function outputs values between -1 and 1 inclusive.
• Even-Odd Property: \(\sin(-\theta) = -\sin(\theta)\), showing it's an odd function.
• Symmetry: It’s symmetrical about the origin.

Understanding the general behavior and properties of the sine function helps in solving trigonometric equations efficiently.

general solution

In trigonometry, finding the general solution to an equation, like \(\sin \theta = \frac{1}{5}\), involves determining all possible angles \(\theta\) that satisfy the equation. For the sine function, because of its periodic nature, the general solution gives us all the angles spaced \(2k\pi\) apart, where \(k\) is any integer. The general solution for \(\sin \theta = a\) involves two key formulas:

• \(\theta = \arcsin(a) + 2k\pi\)
• \(\theta = \pi - \arcsin(a) + 2k\pi\)

These formulas account for all the angles in the unit circle that would yield the same sine value due to symmetry in the sine wave across the y-axis. When calculating these, remember that \(\arcsin\) is the inverse sine function, returning the primary angle whose sine is \(a\), and \(\pi - \arcsin(a)\) gives the second angle in the interval \([0, \pi]\) that also has the same sine value. This approach ensures you capture all potential solutions.

arc sine

The arc sine function, denoted as \(\arcsin\) or \(\sin^{-1}\), is the inverse of the sine function. It is used to find the angle whose sine is a given number. This makes \(\arcsin\) a critical tool in solving trigonometric equations where you need to determine the angle when the sine value is known. The domain of \(\arcsin(x)\) is limited to the interval \([-1, 1]\), as these are the possible outputs of the sine function. The range of \(\arcsin\), or the angles it can return, is \([-\frac{\pi}{2}, \frac{\pi}{2}]\). This is because \(\arcsin\) returns the principal value, meaning the smallest angle between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). Key points to remember about \(\arcsin\) are:

• It will always yield an angle in the first or fourth quadrant of the unit circle.
• Ensure the input value is within the valid domain \([-1, 1]\).
• When used in the general solution, \(\arcsin\) calculates the primary angle for one solution while \(\pi - \arcsin(x)\) provides another angle in the range of the sine function.

Using \(\arcsin\) in combination with the periodicity of the sine function allows you to determine all possible angles that can satisfy a particular sine equation.