Question

use a half-angle identity to find the exact value.
cos 195°
cos 195° =
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Express 195° as half of 390°

Since \(195^{\circ}=\frac{390^{\circ}}{2}\), we can use the half - angle identity for cosine: \(\cos\frac{\theta}{2}=\pm\sqrt{\frac{1 + \cos\theta}{2}}\). First, we need to determine the sign of \(\cos195^{\circ}\). The angle \(195^{\circ}\) is in the third quadrant, where cosine is negative. So we will use the negative sign in the half - angle formula.

Step2: Identify \(\theta\) for the half - angle formula

Let \(\theta = 390^{\circ}\). We know that the cosine function has a period of \(360^{\circ}\), so \(\cos390^{\circ}=\cos(360^{\circ}+ 30^{\circ})=\cos30^{\circ}=\frac{\sqrt{3}}{2}\) (because \(\cos(A + 360^{\circ}n)=\cos A\) for any integer \(n\) and \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\)).

Step3: Substitute into the half - angle formula

Using the half - angle formula \(\cos\frac{\theta}{2}=-\sqrt{\frac{1+\cos\theta}{2}}\) with \(\theta = 390^{\circ}\) (and thus \(\frac{\theta}{2}=195^{\circ}\)):

$$
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We can also rationalize or rewrite \(\sqrt{2+\sqrt{3}}\) as \(\frac{\sqrt{3}+ 1}{\sqrt{2}}\) (by noting that \((\frac{\sqrt{3}+1}{\sqrt{2}})^2=\frac{3 + 2\sqrt{3}+1}{2}=\frac{4 + 2\sqrt{3}}{2}=2+\sqrt{3}\)). So \(-\frac{\sqrt{2+\sqrt{3}}}{2}=-\frac{\sqrt{3}+1}{2\sqrt{2}}\times\sqrt{2}\times\sqrt{2}=-\frac{\sqrt{6}+\sqrt{2}}{4}\) (after rationalizing the denominator: \(\frac{\sqrt{2+\sqrt{3}}}{2}=\frac{\sqrt{\frac{4 + 2\sqrt{3}}{2}}}{2}=\frac{\sqrt{(\sqrt{3}+1)^2/2}}{2}=\frac{\sqrt{3}+1}{2\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4}\), and since we have a negative sign, \(\cos195^{\circ}=-\frac{\sqrt{6}+\sqrt{2}}{4}\))

Answer:

\(-\frac{\sqrt{6}+\sqrt{2}}{4}\)