nd the value of the ratios below in terms of \( \sqrt{2} \) and \( \sqrt{3} \) using the sum or fference of two special angles. Simplify your answers. \( \begin{array}{llll}\cos 15^{\circ} \\ \left.\text { [Hint: } 15^{\circ}=60^{\circ}-45^{\circ}\right] & 4.2 & \sin 75^{\circ} \\ 3 \cos 105^{\circ} & 4.4 & \sin 195^{\circ} \\ 5 & \sin 165^{\circ} & 4.6 & \tan 15^{\circ}\end{array} \)

To find the values of the trigonometric functions in terms of \( \sqrt{2} \) and \( \sqrt{3} \), we will use the sum or difference of angles where applicable. Let's solve each part step by step. ### 1. \( \cos 15^{\circ} \) Using the hint \( 15^{\circ} = 60^{\circ} - 45^{\circ} \), we apply the cosine difference formula: \[ \cos(a - b) = \cos a \cos b + \sin a \sin b \] where \( a = 60^{\circ} \) and \( b = 45^{\circ} \). Calculating: - \( \cos 60^{\circ} = \frac{1}{2} \) - \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \) - \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \) - \( \sin 45^{\circ} = \frac{\sqrt{2}}{2} \) Now substituting into the formula: \[ \cos 15^{\circ} = \cos 60^{\circ} \cos 45^{\circ} + \sin 60^{\circ} \sin 45^{\circ} \] \[ = \left(\frac{1}{2} \cdot \frac{\sqrt{2}}{2}\right) + \left(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}\right) \] \[ = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{2} + \sqrt{6}}{4} \] ### 2. \( \sin 75^{\circ} \) Using the identity \( 75^{\circ} = 45^{\circ} + 30^{\circ} \): \[ \sin(a + b) = \sin a \cos b + \cos a \sin b \] where \( a = 45^{\circ} \) and \( b = 30^{\circ} \). Calculating: - \( \sin 45^{\circ} = \frac{\sqrt{2}}{2} \) - \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \) - \( \sin 30^{\circ} = \frac{1}{2} \) - \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \) Now substituting into the formula: \[ \sin 75^{\circ} = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ} \] \[ = \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right) \] \[ = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} \] ### 3. \( \cos 105^{\circ} \) Using \( 105^{\circ} = 180^{\circ} - 75^{\circ} \): \[ \cos(180^{\circ} - x) = -\cos x \] Thus, \[ \cos 105^{\circ} = -\cos 75^{\circ} \] From the previous calculation, we have: \[ \cos 75^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4} \] So, \[ \cos 105^{\circ} = -\frac{\sqrt{6} - \sqrt{2}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4} \] ### 4. \( \sin 195^{\circ} \) Using \( 195^{\circ} = 180^{\circ} + 15^{\circ} \): \[ \sin(180^{\circ} + x) = -\sin x \] Thus, \[ \sin 195^{\circ} = -\sin 15^{\circ} \] Using the sine difference formula for \( \sin 15^{\circ} \): \[ \sin 15^{\circ} = \sin(60^{\circ} - 45^{\circ}) = \sin 60^{\circ} \cos 45^{\circ} - \cos 60^{\circ} \sin 45^{\circ} \] \[ = \left(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}\right) - \left(\frac{1}{2} \cdot \frac{\sqrt{2}}{2}\right) \] \[ = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4} \] Thus, \[ \sin 195^{\circ} = -\frac{\sqrt{6} - \sqrt{2}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4} \] ### 5. \( \sin 165^{\circ} \) Using \( 165^{\circ} = 180^{\circ} - 15^{\circ} \): \[ \sin(180^{\circ} - x) = \sin x \] Thus, \[ \sin 165^{\circ} = \sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4} \] ### 6. \( \tan 15^{\circ} \) Using the identity \( \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \): \[ \tan 15^{\circ} = \tan(60^{\circ} - 45^{\circ}) = \frac{\tan 60^{\circ} - \tan 45^{\circ}}{1 + \tan 60^{\circ} \tan 45^{\circ}} \] Calculating: - \( \tan 60^{\circ} = \sqrt{3} \) - \( \tan 45^{\circ} = 1 \) Substituting: \[ \tan 15^{\circ} = \frac{\sqrt{3} - 1}{1 + \sqrt{3} \cdot 1} = \