Problem 12 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the exact value of each expression. (a) \(\cos \left(\frac{\pi}{4}+\frac{\pi}{3}\right)\) (b) \(\cos \frac{\pi}{4}+\cos \frac{\pi}{3}\)

Step-by-Step Solution

Answer

(a) \( \frac{\sqrt{2} - \sqrt{6}}{4} \), (b) \( \frac{\sqrt{2} + 1}{2} \)

Step 1: Identify Known Values

From the unit circle, we know that \( \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \) and \( \cos \frac{\pi}{3} = \frac{1}{2} \). We'll be using these values in our calculations.

Step 2: Calculation of Expression (a)

We know that the formula for cosine of sum of two angles, \( \cos(a + b) \), is \( \cos a \cos b - \sin a \sin b \). Applying this formula here, \( \cos \left(\frac{\pi}{4}+\frac{\pi}{3}\right) = \cos \frac{\pi}{4} \cos \frac{\pi}{3} - \sin \frac{\pi}{4} \sin \frac{\pi}{3} \). We know the values of the cosine and sine functions at \( \frac{\pi}{4} \) and \( \frac{\pi}{3} \), hence we can substitute those in to obtain \( \frac{\sqrt{2}}{2} \cdot \frac{1}{2} - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4} \).

Step 3: Calculation of Expression (b)

This calculation is more straightforward since we're simply adding the cosine functions for \( \frac{\pi}{4} \) and \( \frac{\pi}{3} \). Using the known values, \( \cos \frac{\pi}{4}+\cos \frac{\pi}{3} = \frac{\sqrt{2}}{2} + \frac{1}{2} = \frac{\sqrt{2} + 1}{2} \).