Problem 72 - Algebra 2

Question

Solve each trigonometric equation for \(\theta\) in terms of \(y\) Sample $$\begin{aligned} y &=2 \sin 3 \theta+4 \\ \sin 3 \theta &=\frac{y-4}{2} \\\ 3 \theta &=\sin ^{-1}\left(\frac{y-4}{2}\right) \\ \theta &=\frac{1}{3} \sin ^{-1}\left(\frac{y-4}{2}\right) \end{aligned}$$ $$ y=2 \cos \pi \theta+1 $$

Step-by-Step Solution

Answer

\(\theta = \frac{1}{3} \sin^{-1}\left(\frac{y - 4}{2}\right)\) and \( \theta = \frac{1}{\pi} \cos^{-1}\left(\frac{y - 1}{2}\right) \)

Step 1: Isolate the trigonometric term

The first step is to isolate the trigonometric term in each equation. In the first equation \( y = 2 \sin 3 \theta + 4 \), to get the \(\sin 3 \theta\) term on one side, subtract 4 from both sides. This gives \(\sin 3\theta = \frac{y - 4}{2}\)

Step 2: Find theta

Now that \( \sin 3\theta \) has been isolated, one can solve for \( \theta \) using the inverse sine function. Thus, \( 3\theta = \sin^{-1}\left(\frac{y - 4}{2}\right) \). Dividing by 3 on both sides gives \( \theta = \frac{1}{3} \sin^{-1}\left(\frac{y - 4}{2}\right) \)

Step 3: Solve the second equation

Follow similar steps for the second equation \( y = 2 \cos \pi \theta + 1 \). First isolate the cosine term by subtracting 1 from both sides, then divide by 2 gives \( \cos \pi\theta = \frac{y - 1}{2} \)

Step 4: Find theta for second equation

Now, using inverse cosine, \( \pi\theta = \cos^{-1}\left(\frac{y - 1}{2}\right) \). To find \( \theta \), divide by \( \pi \) on both sides, which results in \( \theta = \frac{1}{\pi} \cos^{-1}\left(\frac{y - 1}{2}\right) \)

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions help us find angles when we know the value of a trigonometric function. They are essential in solving trigonometric equations where the angle needs to be expressed in terms of other variables like 'y'. For example, if we know that \( \sin \theta = x \), then the inverse sine function \( \sin^{-1} \) allows us to find \( \theta \). So, \( \theta = \sin^{-1} (x) \). This is particularly useful in mathematical problems that require us to determine the angle from a sine, cosine, or other trigonometric values.

Let's take a deeper look into how they are utilized in solving equations. When you have an equation like \( y = 2 \sin 3 \theta + 4 \), and you isolate \( \sin 3 \theta \) to \( \frac{y-4}{2} \), you use \( \sin^{-1} \) to find \( 3\theta \). This step involves reversing the effect of the sine function by using its inverse. Finally, you isolate \( \theta \) by dividing the result by 3, which gives you \( \theta = \frac{1}{3} \sin^{-1}\left(\frac{y-4}{2}\right) \). The inverse functions of cosine, tangent, etc., work in a similar way, adapting for each specific trigonometric function.

Inverse trigonometric functions are often confined to specific ranges which define the principal values to assure a unique solution. It's essential to be mindful of these restrictions during problem-solving to find solutions within accepted bounds.

Sine and Cosine Functions

The sine and cosine functions are foundational concepts in trigonometry, both representing relationships within right-angled triangles and are functions of an angle. The sine of an angle in a right triangle is the ratio of the length of the opposite side to the hypotenuse. Similarly, the cosine function is the ratio of the adjacent side to the hypotenuse.

These functions can also be represented as wave functions when dealing with circular motion or oscillatory motions such as light or sound waves. Sine and cosine functions are periodic, meaning they repeat at regular intervals, specifically every \(2\pi\) radians or \(360\) degrees. This periodic nature is crucial when solving equations, because if a solution for \( \theta \) fits an equation, then \( \theta + 2n\pi \) and \( \theta - 2n\pi \) (where \( n \) is an integer) are also solutions.

In our example, the equation is reformulated into terms such as \(2\sin 3\theta\) or \(2\cos \pi\theta\) by isolating and breaking down these wave forms into manageable expressions. Understanding these functions and their properties, such as amplitude and phase shift, assists greatly in graphing and translates into solving more complex trigonometric expressions.

Equation Solving Steps

Solving trigonometric equations involves several key steps that ensure you arrive at the correct solution systematically. Let’s list them for clarity:

• Isolating the Trigonometric Function: To isolate, you need to manipulate the given equation so that the trigonometric part stands alone. This often involves subtracting or dividing constants from both sides of the equation, as seen when we move terms away from sine or cosine terms.
• Using Inverse Functions: After isolating, use the corresponding inverse trigonometric function to find the angle. For example, if you have \( \sin 3\theta = \frac{y-4}{2} \), you'd use \( \sin^{-1} \) to find \( 3\theta \) and subsequently uncover \( \theta \).
• Dividing by Coefficients: If your variable like \( \theta \) is multiplied by a coefficient (for instance, 3), divide it out to solve for the single variable \( \theta \).

This methodology not only helps tackle the equation in a logical sequence but aids in breaking down complex expressions into simpler, solvable parts. In the cases where the coefficient is a transcendental or unusual number like \( \pi \), such steps remain applicable but require attention to units and potential simplifications.