Problem 14 - Precalculus with Calculus Previews

Question

Find all solutions of the given trigonometric equation if \(\theta\) represents an angle measured in degrees. $$ 2 \sin \theta=\sqrt{2} $$

Step-by-Step Solution

Answer

\(\theta = 45^\circ + 360^\circ n\) and \(\theta = 135^\circ + 360^\circ n\), for any integer \(n\).

Step 1: Isolate the Trigonometric Function

The given equation is \(2 \sin \theta = \sqrt{2}\). To isolate \(\sin \theta\), divide both sides by 2: \[ \sin \theta = \frac{\sqrt{2}}{2} \]

Step 2: Identify General Solutions for Sine

Recall that the sine of an angle equals \(\frac{\sqrt{2}}{2}\) at specific reference angles. In the unit circle, these angles are \(45^\circ\) and \(135^\circ\) in the first and second quadrants, respectively.

Step 3: Determine All Solutions for \(\theta\) in Degrees

Since sine is periodic with a period of \(360^\circ\), the general solutions for \(\theta\) are: \[ \theta = 45^\circ + n \times 360^\circ \] and \[ \theta = 135^\circ + n \times 360^\circ \] where \(n\) is any integer.

Step 4: Verify Solutions and Write Final Answer

Verify both solutions by plugging them back into the original equation: For \(\theta = 45^\circ\), \(2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}\), which is true. For \(\theta = 135^\circ\), \(2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}\), which is also true. Thus, the solutions \(\theta = 45^\circ + 360^\circ n\) and \(\theta = 135^\circ + 360^\circ n\) are verified.

Key Concepts

sin function

The sine function, often written as \( \sin(\theta) \), is a fundamental building block in trigonometry. It relates a specific angle in a right triangle to the ratio of the length of the opposite side to the hypotenuse. This trigonometric function is periodic, meaning it repeats its values in regular intervals. For sine, this interval or period is \(360^\circ\) (or \(2\pi\) if measured in radians).

• The sine function is an odd function, which means \( \sin(-\theta) = -\sin(\theta) \).
• Its range is confined between -1 and 1, as it is a ratio of side lengths in a triangle.

In the context of the given equation \(2 \sin \theta = \sqrt{2}\), isolating \(\sin \theta\) by dividing by 2 gives \(\sin \theta = \frac{\sqrt{2}}{2}\). This equation shows that the sine function's value at angles \(45^\circ\) and \(135^\circ\) is precisely \(\frac{\sqrt{2}}{2}\), highlighting the significance of reference angles in solving trigonometric equations.

unit circle

The unit circle is an essential concept in trigonometry, serving as a tool to visualize and solve trigonometric equations. Centered at the origin of the coordinate system, the unit circle has a radius of 1.

• Every point on the unit circle corresponds to an angle \(\theta\) and has coordinates \((\cos(\theta), \sin(\theta))\).
• The sine of an angle \(\theta\) is simply the y-coordinate of the corresponding point on the circle.

This circle helps in understanding the periodic nature of trigonometric functions. For instance, in the exercise, finding all solutions involves recognizing-angles where \(\sin \theta = \frac{\sqrt{2}}{2}\). On the unit circle, these specific angles are \(45^\circ\) and \(135^\circ\), due to their corresponding y-coordinates meeting the equation \(\sin \theta = \frac{\sqrt{2}}{2}\).

reference angles

Reference angles are crucial in understanding trigonometric functions and solving equations like the one in our exercise. A reference angle is the acute angle that \(\theta\) makes with the x-axis.

• They simplify the task of finding sine, cosine, and tangent values by providing a base reference in the first quadrant.
• All angles in standard position have the same sine or cosine value as their reference angles, with possibly a sign change depending on the quadrant.

For example, from the unit circle, we know that the sine function has a value of \(\frac{\sqrt{2}}{2}\) at \(45^\circ\), making \(\theta = 45^\circ\) its own reference angle. The angle \(135^\circ\) in the second quadrant also uses \(45^\circ\) as its reference angle. This symmetry allows us to set \(\theta = 135^\circ\) implying that it shares the same sine value as \(45^\circ\). Understanding reference angles allows you to quickly locate solutions for trigonometric equations across different quadrants.