Using sum and difference identities, what is the exact value of $\cos (15°)$?

• A. $\frac{\sqrt{2}}{2}$
• B. $\frac{\sqrt{6}+\sqrt{2}}{4}$
• C. $\frac{1}{2}$
• D. $\frac{\sqrt{6}-\sqrt{2}}{4}$

Answer: (B)

Verified step by step guidance

1. Recognize that 15° can be expressed as the difference of two special angles whose cosine and sine values are well known, for example, 45° and 30°. So, write 15° as 45° - 30°.
2. Recall the cosine difference identity: \(\cos(A - B) = \cos A \cos B + \sin A \sin B\). Here, let \(A = 45^\circ\) and \(B = 30^\circ\).
3. Substitute the known values for \(\cos 45^\circ\), \(\cos 30^\circ\), \(\sin 45^\circ\), and \(\sin 30^\circ\) into the identity. These are \(\cos 45^\circ = \frac{\sqrt{2}}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\sin 45^\circ = \frac{\sqrt{2}}{2}\), and \(\sin 30^\circ = \frac{1}{2}\).
4. Calculate the expression by multiplying and adding the terms according to the identity: \(\cos 15^\circ = \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right)\).
5. Simplify the expression by combining the fractions and radicals to get the exact value of \(\cos 15^\circ\) in simplest radical form.