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

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})