Problem 6 - Algebra 2 and Trigonometry

Question

In \(3-14,\) find the exact values of \(\theta\) in the interval \(0^{\circ} \leq \theta < 360^{\circ}\) that satisfy each equation. $$ 2 \sin \theta+1=\csc \theta $$

Step-by-Step Solution

Answer

The solutions are \(\theta = 30^\circ\), \(150^\circ\), and \(270^\circ\).

Step 1: Transform the Equation

The equation to solve is \(2 \sin \theta + 1 = \csc \theta\). Begin by expressing \(\csc \theta\) in terms of \(\sin \theta\):\[ \csc \theta = \frac{1}{\sin \theta} \]Substitute this identity into the equation:\[ 2 \sin \theta + 1 = \frac{1}{\sin \theta} \]

Step 2: Create a Common Equation

Multiply both sides by \(\sin \theta\) to eliminate the fraction:\[ 2 \sin^2 \theta + \sin \theta = 1 \]

Step 3: Rearrange the Equation

Reorganize the equation into a standard quadratic form:\[ 2 \sin^2 \theta + \sin \theta - 1 = 0 \]

Step 4: Factor the Quadratic

Attempt to factor the quadratic equation:\[ (2 \sin \theta - 1)(\sin \theta + 1) = 0 \]This gives us potential solutions for \(\sin \theta\):1. \(2 \sin \theta - 1 = 0\) 2. \(\sin \theta + 1 = 0\)

Step 5: Solve for Sine Values

Solve each of the factored equations for \(\sin \theta\):- From \(2 \sin \theta - 1 = 0\), we have: \[ \sin \theta = \frac{1}{2} \]- From \(\sin \theta + 1 = 0\), we have:\[ \sin \theta = -1 \]

Step 6: Find Angles for \(\sin \theta = \frac{1}{2}\)

For \(\sin \theta = \frac{1}{2}\), the angles in the interval \(0^\circ \leq \theta < 360^\circ\) are \(\theta = 30^\circ\) and \(\theta = 150^\circ\).

Step 7: Find Angle for \(\sin \theta = -1\)

For \(\sin \theta = -1\), the angle in the interval \(0^\circ \leq \theta < 360^\circ\) is \(\theta = 270^\circ\).

Step 8: Conclusion

The angles that satisfy the original equation \(2 \sin \theta + 1 = \csc \theta\) within the interval are \(\theta = 30^\circ\), \(\theta = 150^\circ\), and \(\theta = 270^\circ\).

Key Concepts

Sine Function

The sine function, denoted as \( \sin \theta \), is a fundamental trigonometric function derived from the opposite side's length over the hypotenuse in a right triangle. It plays a vital role in connecting angles to ratios of side lengths.

Key characteristics of the sine function:

• *Range*: The sine function values range from -1 to 1.
• *Periodicity*: It has a periodic nature, repeating every 360° or \(2\pi\) radians.
• *Symmetry*: It is an odd function, which means \( \sin(-\theta) = -\sin(\theta) \).

Understanding the sine function is crucial for solving equations involving trigonometric terms, like \( 2 \sin \theta + 1 = \csc \theta \). By using identities and properties of the sine function, we can transform and solve trigonometric equations effectively.

Cosecant Function

The cosecant function, denoted as \( \csc \theta \), is the reciprocal of the sine function. This means \( \csc \theta = \frac{1}{\sin \theta} \). It represents the ratio of the hypotenuse to the opposite side in a right-angled triangle.

Key features of the cosecant function include:

• *Range*: Since it is the reciprocal of sine, it does not exist where \( \sin \theta = 0 \). The range is \(( -\infty, -1] \cup [1, \infty )\).
• *Periodicity*: Like sine, the cosecant function is periodic with a period of 360° or \(2\pi\) radians.
• *Vertical asymptotes*: Occur at integer multiples of \( \pi \) where the sine function equals zero.

Utilizing the reciprocal identity of cosecant helps simplify trigonometric equations through substitution, as seen when transforming \( 2 \sin \theta + 1 = \csc \theta \) into a solvable equation involving sine only.

Quadratic Equation

A quadratic equation is any equation that can be arranged in the form \( ax^2 + bx + c = 0 \). In the context of trigonometric functions, it often arises when rearranging and simplifying expressions with squared trigonometrics, such as \( 2 \sin^2 \theta + \sin \theta - 1 = 0 \).

Solving a quadratic equation involves:

• *Factoring*: Expressing the quadratic as a product of two binomials. For \( (2 \sin \theta - 1)(\sin \theta + 1) = 0 \), we have simplified the equation down.
• *Quadratic formula*: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), is another method if factoring is not feasible.

Once a quadratic equation is in its factored or solved form, finding its roots enables the identification of potential solutions for the trigonometric equation, such as which angles satisfy \( \sin \theta \).

Angle Solution

Finding the angle solutions to trigonometric equations involves determining the specific angles that satisfy the equation within a given interval, typically \(0^{\circ} \leq \theta < 360^{\circ}\).

To solve for angles:

• *Determine possible sine values*: From the solutions derived for \( \sin \theta \). In our example, these values were \( \sin \theta = \frac{1}{2} \) and \( \sin \theta = -1 \).
• *Reference the unit circle*: Identify angles where these sine values occur. \( \sin \theta = \frac{1}{2} \) corresponds to \(30^{\circ}\) and \(150^{\circ}\), while \( \sin \theta = -1 \) corresponds to \(270^{\circ}\).

Utilizing the unit circle to find these angles ensures solutions fall within the designated range, confirming they satisfy the original trigonometric equation.