Question

use a double - angle identity to find the exact value of the expression.
(cos ^{2} 180^{circ}-sin ^{2} 180^{circ})
(cos ^{2} 180^{circ}-sin ^{2} 180^{circ}=square)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Apply double - angle identity

The double - angle identity for cosine is \(\cos2\alpha=\cos^{2}\alpha - \sin^{2}\alpha\).
For \(\alpha = 180^{\circ}\), the given expression \(\cos^{2}180^{\circ}-\sin^{2}180^{\circ}\) can be written as \(\cos(2\times180^{\circ})\).

Step2: Simplify the angle

Calculate \(2\times180^{\circ}=360^{\circ}\). So, the expression becomes \(\cos360^{\circ}\).

Step3: Evaluate the cosine function

We know that \(\cos360^{\circ}=\cos(0^{\circ}+ 360^{\circ})\). Since the cosine function has a period of \(360^{\circ}\), \(\cos(360^{\circ}+\theta)=\cos\theta\). So, \(\cos360^{\circ}=\cos0^{\circ}\). And \(\cos0^{\circ}=1\).

Answer:

\(1\)