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Chapter 6: Problem 73

Show that $$ \cos (2 \theta) \leq \cos ^{2} \theta $$ for every angle \(\theta\).

Short Answer

Expert verified
We can prove this inequality by using trigonometric identities and inequalities. First, rewrite \(\cos(2\theta)\) as \(2\cos^2(\theta) - 1\). Then, rearrange the inequality to \(\cos^2(\theta) - 1 \leq 0\). Use the Pythagorean identity \(\sin^2(\theta) = 1 - \cos^2(\theta)\) and substitute it back into the inequality, simplifying to \(-\cos^2(\theta) \leq 0\). Since \(\cos^2(\theta)\) is always non-negative for all angles \(\theta\), the inequality holds true, and we have shown that \(\cos(2\theta) \leq \cos^2(\theta)\) for every angle \(\theta\).

Step by step solution

01

Rewrite \(\cos(2\theta)\) using trigonometric identity

We can use a well-known trigonometric identity to express the left side of the inequality in terms of \(\cos(\theta)\). The identity states that: \[ \cos(2\theta) = 2\cos^2(\theta) - 1 \] This allows us to rewrite the inequality as: \[ 2\cos^2(\theta) - 1 \leq \cos^2(\theta) \]
02

Rearrange the inequality

We want to get the inequality in the form of a single expression that is greater than or equal to zero. To do this, we can subtract \(\cos^2(\theta)\) from both sides of the inequality and rewrite it as follows: \[ 2\cos^2(\theta) - 1 - \cos^2(\theta) \leq 0 \] Simplifying, we get: \[ \cos^2(\theta) - 1 \leq 0 \]
03

Use Pythagorean identity

To further simplify the inequality, we can make use of the Pythagorean identity, which states that for any angle \(\theta\): \[ \sin^2(\theta) + \cos^2(\theta) = 1 \] From this identity, we can isolate \(\sin^2(\theta)\) as follows: \[ \sin^2(\theta) = 1 - \cos^2(\theta) \]
04

Substitute back into the inequality

Now, we can substitute this expression for \(\sin^2(\theta)\) back into the inequality: \[ 1 - \cos^2(\theta) - 1 \leq 0 \] This simplifies to: \[ -\cos^2(\theta) \leq 0 \]
05

Analyze the inequality for all values of theta

Since \(\cos^2(\theta)\) is always positive or zero for all values of \(\theta\), we can say that \(-\cos^2(\theta)\) is always non-positive, which means that the inequality holds true for all values of \(\theta\): \[ -\cos^2(\theta) \leq 0 \] Since the inequality holds for every angle \(\theta\), we have shown that: \[ \cos(2\theta) \leq \cos^2(\theta) \] for every angle \(\theta\).

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