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Chapter 6: Problem 93

Suppose \(\theta\) is an angle such that \(\cos \theta\) is rational. Explain why \(\cos (2 \theta)\) is rational.

Short Answer

Expert verified
Recall the double angle formula for cosine: \(\cos (2\theta) = 2 \cos^2 (\theta) - 1\). Given \(\cos \theta\) is rational, let \(\cos \theta = \frac{a}{b}\), with \(a\) and \(b\) as integers and \(b\neq0\). Substituting this value, we get \(\cos (2\theta) = 2 \cdot \frac{a^2}{b^2} - 1\). Rewrite this expression as a rational number: \(\cos (2\theta) = \frac{2a^2 - b^2}{b^2}\). Since both the numerator and denominator are integers, \(\cos (2\theta)\) is rational.

Step by step solution

01

Recall the double angle formula for cosine

The double angle formula for cosine is given by: \[ \cos (2\theta) = 2 \cos^2 (\theta) - 1 \] This formula will help us find \(\cos (2\theta)\) using \(\cos \theta\), which is given to be a rational number.
02

Substitute the given rational value of \(\cos \theta\)

Since \(\cos \theta\) is a rational number, let's say that \(\cos \theta = \frac{a}{b}\), where \(a\) and \(b\) are integers and \(b\neq0\). Using the double angle formula, we'll substitute this value in the expression of \(\cos (2\theta)\): \[ \cos (2\theta) = 2 \left(\frac{a}{b}\right)^2 - 1\]
03

Simplify the expression

Now, let's simplify the expression: \[ \cos (2\theta) = 2 \cdot \frac{a^2}{b^2} - 1 \]
04

Make the expression look like a rational number

We need to rewrite the expression in such a way that it looks like a rational number, i.e., in the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q\neq0\). To do that, we can rewrite the above expression as: \[ \cos (2\theta) = \frac{2a^2}{b^2} - \frac{b^2}{b^2}\]
05

Combine the two fractions

Next, combine the two fractions: \[ \cos (2\theta) = \frac{2a^2 - b^2}{b^2} \] Given that \(2a^2 - b^2\) is an integer and \(b^2\) is an integer (\(b\neq0\)), we can now conclude that \(\cos (2\theta)\) is a rational number.

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