Question

write the expression as the sine, cosine, or tangent of a double - angle. then find the exact value of the expression.
\\( \cos ^ { 2 } 180 ^ { \circ } - \sin ^ { 2 } 180 ^ { \circ } \\)
which expression is \\( \cos ^ { 2 } 180 ^ { \circ } - \sin ^ { 2 } 180 ^ { \circ } \\) equal to? select the correct choice below and fill in the answer box within
your choice.
(simplify your answer. type your answer in degrees. use integers or decimals for any numbers in the expression.)
\\( \bigcirc \\) a. \\( \cos ^ { 2 } 180 ^ { \circ } - \sin ^ { 2 } 180 ^ { \circ } = \sin \square ^ { \circ } \\)
\\( \bigcirc \\) b. \\( \cos ^ { 2 } 180 ^ { \circ } - \sin ^ { 2 } 180 ^ { \circ } = \cos \square ^ { \circ } \\)
\\( \bigcirc \\) c. \\( \cos ^ { 2 } 180 ^ { \circ } - \sin ^ { 2 } 180 ^ { \circ } = \tan \square ^ { \circ } \\)

Explanation:

Step1: Recall the double - angle formula for cosine

The double - angle formula for cosine is \(\cos2\alpha=\cos^{2}\alpha - \sin^{2}\alpha\). In the given expression \(\cos^{2}180^{\circ}-\sin^{2}180^{\circ}\), we have \(\alpha = 180^{\circ}\). So, \(\cos^{2}180^{\circ}-\sin^{2}180^{\circ}=\cos(2\times180^{\circ})\)

Step2: Calculate the value of the cosine function

We know that \(\cos(360^{\circ})=\cos(0^{\circ}) = 1\). Also, using the unit - circle definition, \(\cos180^{\circ}=- 1\) and \(\sin180^{\circ}=0\). Then \(\cos^{2}180^{\circ}-\sin^{2}180^{\circ}=(-1)^{2}-0^{2}=1\) and \(\cos(2\times180^{\circ})=\cos360^{\circ}=1\)

Answer:

B. \(\cos^{2}180^{\circ}-\sin^{2}180^{\circ}=\cos360^{\circ}\)