Question

find the exact value of each expression.
(a) \\( \cos 75 ^ { \circ } + \cos 15 ^ { \circ } = \\)
(b) \\( \cos 67.5 ^ { \circ } \sin 22.5 ^ { \circ } = \\)

Explanation:

Step1: Use sum - to - product formula for \(a\)

The sum - to - product formula is \(\cos A+\cos B = 2\cos\frac{A + B}{2}\cos\frac{A - B}{2}\).
For \(A = 75^{\circ}\) and \(B=15^{\circ}\), we have \(\frac{A + B}{2}=\frac{75^{\circ}+15^{\circ}}{2}=45^{\circ}\) and \(\frac{A - B}{2}=\frac{75^{\circ}-15^{\circ}}{2}=30^{\circ}\).
So \(\cos75^{\circ}+\cos15^{\circ}=2\cos45^{\circ}\cos30^{\circ}\).
Since \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\) and \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), then \(2\times\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}=\frac{\sqrt{6}}{2}\).

Step2: Use product - to - sum formula for \(b\)

The product - to - sum formula \(\cos\alpha\sin\beta=\frac{1}{2}[\sin(\alpha+\beta)-\sin(\alpha - \beta)]\).
For \(\alpha = 67.5^{\circ}\) and \(\beta = 22.5^{\circ}\), \(\alpha+\beta=90^{\circ}\) and \(\alpha - \beta = 45^{\circ}\).
So \(\cos67.5^{\circ}\sin22.5^{\circ}=\frac{1}{2}(\sin90^{\circ}-\sin45^{\circ})\).
Since \(\sin90^{\circ}=1\) and \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\), then \(\frac{1}{2}(1 - \frac{\sqrt{2}}{2})=\frac{2-\sqrt{2}}{4}\).

Answer:

(a) \(\frac{\sqrt{6}}{2}\)
(b) \(\frac{2 - \sqrt{2}}{4}\)