Problem 12 - Algebra 2 and Trigonometry

Question

In \(9-14,\) find the exact values for \(\theta\) in the interval \(0 \leq \theta<2 \pi\) $$ \sin \theta+\sqrt{2}=\frac{\sqrt{2}}{2} $$

Step-by-Step Solution

Answer

\( \theta = \frac{5\pi}{4} \) and \( \theta = \frac{7\pi}{4} \).

Step 1: Isolate the Sine Function

Start by isolating the sine function in the equation. We have \( \sin \theta + \sqrt{2} = \frac{\sqrt{2}}{2} \). Subtract \( \sqrt{2} \) from both sides: \( \sin \theta = \frac{\sqrt{2}}{2} - \sqrt{2} \).

Step 2: Simplify the Right Side

Simplify \( \frac{\sqrt{2}}{2} - \sqrt{2} \) by expressing \( \sqrt{2} \) as \( \frac{2\sqrt{2}}{2} \) to get a common denominator: \( \sin \theta = \frac{\sqrt{2} - 2\sqrt{2}}{2} = \frac{-\sqrt{2}}{2} \).

Step 3: Find Reference Angle

Determine the reference angle \( \theta \) where \( \sin \theta = -\frac{\sqrt{2}}{2} \). The reference angle is \( \frac{\pi}{4} \) because \( \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).

Step 4: Determine Quadrants

Since \( \sin \theta \) is negative, \( \theta \) must be in the third or fourth quadrant. For the third quadrant, \( \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \). For the fourth quadrant, \( \theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \).

Key Concepts

Sine Function

The sine function is a fundamental concept in trigonometry, representing the y-coordinate of a point on the unit circle.
It is an important periodic function that oscillates between -1 and 1.
In the context of right triangles, the sine of an angle is defined as the ratio of the length of the opposite side to the hypotenuse.

• Expressed mathematically, for a right triangle with an angle \(\theta\), sine is given by \( \sin \theta = \frac{opposite}{hypotenuse} \).
• In the unit circle, because the hypotenuse is equal to the radius (which is 1), the sine of angle \(\theta\) is simply the y-coordinate of the point.

When solving trigonometric equations, like the one in this exercise, the goal is to isolate the sine function, making it easier to identify angles that satisfy the equation.

Reference Angle

Reference angles are a crucial tool in solving trigonometric equations. They are the positive acute angle that a terminal side makes with the x-axis, regardless of the quadrant.
These angles help us understand trigonometric values in all four quadrants by relating them back to the basic angles of \( \frac{\pi}{6} \), \( \frac{\pi}{4} \), and \( \frac{\pi}{3} \).
However, in this exercise, the reference angle \( \theta = \frac{\pi}{4} \) plays a key role because:

• The sine of \( \frac{\pi}{4} \) is known to be \( \frac{\sqrt{2}}{2} \).
• To find angles in the third and fourth quadrants that correspond to a negative sine value, we subtract from \( \pi \) or \( 2\pi \), respectively.

Understanding reference angles allows us to determine the signs and magnitudes of sine (and other trigonometric) functions beyond the first quadrant. They make it possible to apply knowledge of basic angles to more complex situations.

Unit Circle

The unit circle is a crucial concept to understand when solving trigonometric equations.
It is defined as a circle with a radius of exactly 1 unit, centered at the origin of a coordinate plane.
This circle is instrumental in defining the trigonometric functions for all real numbers. As a foundational tool in trigonometry, it helps to:

• Visualize angles and their terminal sides, allowing us to determine sine, cosine, and tangent values easily.
• Understand the periodic nature of trigonometric functions, as these repeat values every \(2\pi\) radians (or 360 degrees).

In the exercise, the unit circle helps us find the angles \( \frac{5\pi}{4} \) and \( \frac{7\pi}{4} \) because it shows:

• Where the line intersects the circle in the third and fourth quadrants.
• The values of coordinates at these intersection points, which correspond to specific sine values.

By effectively using the unit circle, students can accurately solve trigonometric equations and develop a deeper understanding of the relationships between angles and trigonometric ratios.