Problem 8 - Algebra and Trigonometry Real Mathematics, Real People

Question

Value is a solution of the equation. \(2 \cos ^{2} 4 x-1=0\) (a) \(x=\frac{\pi}{16}\) (b) \(x=\frac{3 \pi}{16}\)

Step-by-Step Solution

Answer

Both \(x=\frac{\pi}{16}\) and \(x=\frac{3 \pi}{16}\) are solutions to the equation.

Step 1: Substitute the value

First substitute \(x=\frac{\pi}{16}\) into the equation. We get \(2 \cos^2{4 \times \frac{\pi}{16}} - 1 = 2 \cos^2{\frac{\pi}{4}} - 1\).

Step 2: Calculate the solution

The value of \(\cos^2{\frac{\pi}{4}}\) is \(\frac{1}{2}\). So the equation becomes \(2 \times \frac{1}{2} - 1) = 0\).

Step 3: Verify the solution

The result from Step 2 is 0 which implies that \(x=\frac{\pi}{16}\) is a solution to the equation.

Step 4: Repeat for the second value

Now, repeat steps 1 to 3 with \(x=\frac{3 \pi}{16}\). Substitute \(x\) into the equation to get \(2 \cos^2{4 \times \frac{3 \pi}{16}} - 1 = 2 \cos^2{\frac{3 \pi}{4}} - 1\). The value of \(\cos^2{\frac{3 \pi}{4}}\) is also \(\frac{1}{2}\).

Step 5: Verify the second solution

Substitute \(\cos^2{\frac{3 \pi}{4}}\) into the equation to get \(2 \times \frac{1}{2} - 1 = 0\). The result is 0. So, \(x=\frac{3 \pi}{16}\) is also a solution to the equation.

Key Concepts

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for every value of the variable within a certain range. One fundamental identity is the Pythagorean Identity:

• \( \sin^{2} x + \cos^{2} x = 1 \)

This means that for any angle \( x \), the square of the sine of \( x \) plus the square of the cosine of \( x \) is always one.
Trigonometric identities are extremely useful when solving equations, simplifying expressions, and in fostering deeper understanding of trigonometric relationships.
In our exercise, identifying that \( 2\cos^2{4x} - 1 = 0 \) can be linked to the double angle identities doubles the usefulness. For instance, one manipulation gives us the identity \( 2\cos^2A-1 = \cos{2A} \); therefore, \( 2\cos^2{4x} - 1 \) can be rewritten as \( \cos{8x} \).

Cosine Function

The cosine function, one of the primary trigonometric functions, is crucial to understanding the set trigonometric equations covered in our exercise. Cosine measures the horizontal distance from the origin to the point on the unit circle corresponding to the angle. It is periodic with a period of \(2\pi\), meaning \(\cos(x) = \cos(x + 2\pi)\).
To master trigonometry, it's vital to remember key values of the cosine function:

• \( \cos(0) = 1 \)
• \( \cos(\frac{\pi}{2}) = 0 \)
• \( \cos(\pi) = -1 \)
• \( \cos(\frac{3\pi}{2}) = 0 \)
• \( \cos(2\pi) = 1 \)

These values are essential when calculating the solution to cosine equations, like finding \( \cos^2{\frac{\pi}{4}} = \frac{1}{2} \) used in our problem steps.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree, typically in the form \( ax^2+bx+c=0 \). Solving quadratic equations is a crucial algebraic skill since they appear frequently in trigonometry when dealing with squared trigonometric functions.
The solution process might involve:

• Factoring the equation when possible.
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)

In our trigonometric problem, the equation \( 2 \cos^2{4x} - 1 = 0 \) shares properties with a quadratic equation. Here, instead of \( x \), we have \( \cos{4x} \).
Replacing trigonometric terms with algebraic expressions like \( u = \cos{4x} \), transforms our equation into a quadratic form \( 2u^2 - 1 = 0 \) before solving.