Question

use a double - angle identity to find the exact value of the expression. $\cos ^{2}90^{\circ }-\sin ^{2}90^{\circ }$ $\cos ^{2}90^{\circ }-\sin ^{2}90^{\circ }=$ (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Recall the double - angle identity

The double - angle identity for cosine is \(\cos2\alpha=\cos^{2}\alpha - \sin^{2}\alpha\).

Step2: Substitute \(\alpha = 90^{\circ}\)

When \(\alpha = 90^{\circ}\), we have \(\cos^{2}90^{\circ}-\sin^{2}90^{\circ}=\cos(2\times90^{\circ})\).

Step3: Calculate \(\cos(2\times90^{\circ})\)

Since \(2\times90^{\circ}=180^{\circ}\), and \(\cos180^{\circ}=- 1\).

Answer:

\(-1\)