Solving equations Solve each equation.


sin² 2Θ = 1/2, -π/2 ≤ Θ ≤ π/2

Verified step by step guidance

1. Step 1: Recognize that the equation is in the form of \( \sin^2(2\Theta) = \frac{1}{2} \). To solve for \( \Theta \), first take the square root of both sides to get \( \sin(2\Theta) = \pm \frac{1}{\sqrt{2}} \).
2. Step 2: Recall that \( \sin(2\Theta) = \pm \frac{1}{\sqrt{2}} \) corresponds to angles where the sine function equals \( \frac{1}{\sqrt{2}} \) or \( -\frac{1}{\sqrt{2}} \). These angles are \( \frac{\pi}{4}, \frac{3\pi}{4}, -\frac{\pi}{4}, \) and \( -\frac{3\pi}{4} \).
3. Step 3: Set \( 2\Theta = \frac{\pi}{4}, \frac{3\pi}{4}, -\frac{\pi}{4}, \) and \( -\frac{3\pi}{4} \) to find the possible values of \( 2\Theta \).
4. Step 4: Solve for \( \Theta \) by dividing each angle by 2. This gives \( \Theta = \frac{\pi}{8}, \frac{3\pi}{8}, -\frac{\pi}{8}, \) and \( -\frac{3\pi}{8} \).
5. Step 5: Verify that each solution for \( \Theta \) falls within the given interval \( -\frac{\pi}{2} \leq \Theta \leq \frac{\pi}{2} \).

Key Concepts

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, relate angles to ratios of sides in right triangles. The sine function, specifically, gives the ratio of the length of the opposite side to the hypotenuse. Understanding these functions is crucial for solving equations involving angles, as they provide the foundational relationships needed to manipulate and solve trigonometric equations.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as arcsin, arccos, and arctan, are used to find angles when the values of the trigonometric functions are known. For example, if sin(Θ) = 1/2, then Θ can be found using arcsin(1/2). These functions are essential for solving equations where the angle is the unknown, allowing us to determine the angle that corresponds to a given sine, cosine, or tangent value.

Periodic Nature of Trigonometric Functions

Trigonometric functions are periodic, meaning they repeat their values in regular intervals. For sine, the period is 2π, which implies that sin(Θ) = sin(Θ + 2πk) for any integer k. This periodicity is important when solving trigonometric equations, as it allows for multiple solutions within a specified range, such as the interval -π/2 ≤ Θ ≤ π/2 in the given problem.