Problem 10 - Algebra and Trigonometry

Question

\(3-16 \cdot\) Solve the given equation. $$ \cos 2 \theta=\cos ^{2} \theta-\frac{1}{2} $$

Step-by-Step Solution

Answer

\( \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \).

Step 1: Use Trigonometric Identity for Double Angle

Recall the identity for cosine of double angle: \( \cos 2\theta = 2\cos^2 \theta - 1 \). Substitute \( \cos 2\theta \) in the equation with \( 2\cos^2 \theta - 1\). The equation becomes \( 2\cos^2 \theta - 1 = \cos^2 \theta - \frac{1}{2} \).

Step 2: Simplify the Equation

Move all terms involving \( \cos^2 \theta \) to one side of the equation: \( 2\cos^2 \theta - \cos^2 \theta = -\frac{1}{2} + 1 \). This simplifies to \( \cos^2 \theta = \frac{1}{2} \).

Step 3: Solve for \( \cos \theta \)

Take the square root of both sides to solve for \( \cos \theta \): \( \cos \theta = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2} \).

Step 4: Find Values of \( \theta \)

\( \cos \theta = \frac{\sqrt{2}}{2} \) at \( \theta = \frac{\pi}{4} + 2n\pi \) and \( \theta = \frac{7\pi}{4} + 2n\pi \). \( \cos \theta = -\frac{\sqrt{2}}{2} \) at \( \theta = \frac{3\pi}{4} + 2n\pi \) and \( \theta = \frac{5\pi}{4} + 2n\pi \), where \( n \) is an integer.

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in solving trigonometric equations.
They help simplify expressions and relate different trigonometric functions to each other.
Using these identities can transform complex equations into more manageable forms:

• **Reciprocal identities** relate the basic trigonometric functions like sine, cosine, and tangent to their reciprocals, cosecant, secant, and cotangent.
• **Pythagorean identities** are derived from the Pythagorean theorem and express one trigonometric function in terms of another, such as \( \sin^2 \theta + \cos^2 \theta = 1 \).
• **Angle sum and difference identities** provide the sine, cosine, and tangent of an angle that is the sum or difference of two known angles. These are crucial for accurate trigonometric calculations.

In the original exercise, understanding of the double angle identities, a specific type of angle transformation identity, is crucial to breaking down the given trigonometric equation.

Double Angle Formulas

The double angle formulas are a subset of trigonometric identities, important for simplifying complex trigonometric expressions.
These formulas express trigonometric functions of double angles \( 2\theta \) in terms of single angles \( \theta \):

• The cosine double angle formula is particularly useful: \( \cos 2\theta = 2\cos^2 \theta - 1 \). This identity allows us to express a double angle as a square of a cosine.
• Similarly, there are corresponding formulas for sine and tangent: \( \sin 2\theta = 2\sin \theta \cos \theta \) and \( \tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta} \).

In solving equations, these formulas help rewrite the terms in ways that can be simplified or factored. For the given exercise, the formula \( \cos 2\theta = 2\cos^2 \theta - 1 \) was key to transforming the equation, leading to its solution.

Cosine Function

The cosine function, a fundamental trigonometric function, measures the horizontal component of a unit radius circle in the coordinate plane.
It is even, which means it's symmetric about the y-axis and has the property \( \cos(-\theta) = \cos \theta \).

• The cosine function is periodic with a period of \( 2\pi \), repeating its values every full circle.
• The range of the cosine function is between -1 and 1, representing the projection of a rotating line onto the x-axis.
• Cosine values are typically found using the unit circle approach or trigonometric tables, especially for notable angles like \( \pi/4, 3\pi/4 \), etc.

In the context of solving trigonometric equations like the one in the exercise, we are interested in finding where the cosine function takes certain values. Specifically, when set equal to zero, positive, or negative fractions indicate certain standard angles. This, combined with already discussed identities, helps solve the equation completely.