4. If \( \cos 2 \mathrm{~A}=\frac{3}{5} \) and \( 0^{\circ} \leq \mathrm{A} \leq 90^{\circ} \), determine, without using a calculator, the value of each of the following in its \( \operatorname{simplest} \) form: \( \begin{array}{lll}\text { (a) } \cos \mathrm{A} & \text { (b) } \sin \mathrm{A} & \text { (c) } \cos \left(\mathrm{A}+45^{\circ}\right)\end{array} \)

To solve the problem, we start with the known condition: \[ \cos 2A = \frac{3}{5} \] We can use the double angle identity for cosine: \[ \cos 2A = 2\cos^2 A - 1 \] Setting this equal to the given value: \[ 2\cos^2 A - 1 = \frac{3}{5} \] Now, we will solve for \(\cos A\). First, we add 1 to both sides: \[ 2\cos^2 A = \frac{3}{5} + 1 \] Converting 1 to a fraction with a denominator of 5: \[ 2\cos^2 A = \frac{3}{5} + \frac{5}{5} = \frac{8}{5} \] Next, we divide both sides by 2: \[ \cos^2 A = \frac{8}{10} = \frac{4}{5} \] Now, we take the square root to find \(\cos A\): \[ \cos A = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \] Since \(0^{\circ} \leq A \leq 90^{\circ}\), \(\cos A\) is positive. ### (a) Value of \(\cos A\) \[ \cos A = \frac{2\sqrt{5}}{5} \] ### (b) Value of \(\sin A\) We can find \(\sin A\) using the Pythagorean identity: \[ \sin^2 A + \cos^2 A = 1 \] Substituting \(\cos^2 A\): \[ \sin^2 A + \frac{4}{5} = 1 \] Subtracting \(\frac{4}{5}\) from both sides: \[ \sin^2 A = 1 - \frac{4}{5} = \frac{1}{5} \] Taking the square root: \[ \sin A = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \] ### (c) Value of \(\cos(A + 45^{\circ})\) Using the cosine addition formula: \[ \cos(A + 45^{\circ}) = \cos A \cos 45^{\circ} - \sin A \sin 45^{\circ} \] We know that \(\cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2}\). Substituting the values we found: \[ \cos(A + 45^{\circ}) = \left(\frac{2\sqrt{5}}{5}\right) \left(\frac{\sqrt{2}}{2}\right) - \left(\frac{\sqrt{5}}{5}\right) \left(\frac{\sqrt{2}}{2}\right) \] Factoring out \(\frac{\sqrt{2}}{2}\): \[ \cos(A + 45^{\circ}) = \frac{\sqrt{2}}{2} \left(\frac{2\sqrt{5}}{5} - \frac{\sqrt{5}}{5}\right) \] Simplifying inside the parentheses: \[ \cos(A + 45^{\circ}) = \frac{\sqrt{2}}{2} \left(\frac{2\sqrt{5} - \sqrt{5}}{5}\right) = \frac{\sqrt{2}}{2} \left(\frac{\sqrt{5}}{5}\right) \] Thus, we have: \[ \cos(A + 45^{\circ}) = \frac{\sqrt{10}}{10} \] ### Final Answers \[ \begin{array}{lll} \text{(a) } \cos A = \frac{2\sqrt{5}}{5} & \text{(b) } \sin A = \frac{\sqrt{5}}{5} & \text{(c) } \cos(A + 45^{\circ}) = \frac{\sqrt{10}}{10} \end{array} \]