use a half - angle identity to find the exact value. \ncos 15°\n\ncos 15° = \n(simplify your answer…

Explanation:

Step1: Choose the half - angle identity

The half - angle identity for cosine is $\cos\frac{\alpha}{2}=\pm\sqrt{\frac{1 + \cos\alpha}{2}}$. Since $15^{\circ}=\frac{30^{\circ}}{2}$ and $15^{\circ}$ is in the first quadrant (where cosine is positive), we use $\cos\frac{\alpha}{2}=\sqrt{\frac{1+\cos\alpha}{2}}$ with $\alpha = 30^{\circ}$.

Step2: Substitute $\alpha = 30^{\circ}$ into the formula

We know that $\cos30^{\circ}=\frac{\sqrt{3}}{2}$. Substitute into the formula: $$\cos15^{\circ}=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}$$

Step3: Simplify the expression inside the square root

First, simplify the fraction inside the square root: $$\frac{1+\frac{\sqrt{3}}{2}}{2}=\frac{\frac{2 + \sqrt{3}}{2}}{2}=\frac{2+\sqrt{3}}{4}$$ So, $\cos15^{\circ}=\sqrt{\frac{2+\sqrt{3}}{4}}$.

Step4: Simplify the square root

Since $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ ($a = 2+\sqrt{3}$, $b = 4$), we have $\cos15^{\circ}=\frac{\sqrt{2+\sqrt{3}}}{2}$. Another form (by rationalizing and using double - angle identities in reverse) is $\frac{\sqrt{6}+\sqrt{2}}{4}$.

Answer:

$\frac{\sqrt{6}+\sqrt{2}}{4}$