4. Find the value of the ratios below in terms of \( \sqrt{2} \) and \( \sqrt{3} \) using the sum or difference of two special angles. \( \begin{array}{llll}4.1 \cos 15^{\circ} & \left.\text { [Hint: } 15^{\circ}=60^{\circ}-45^{\circ}\right] & 4.2 & \sin 75^{\circ} \\ 4.3 \cos 105^{\circ} & 4.4 & \sin 195^{\circ} \\ 4.5 \sin 165^{\circ} & 4.6 & \tan 15^{\circ}\end{array} \)
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To find the requested trigonometric values involving special angles, we can use the angle subtraction or addition formulas. To solve \( \cos 15^\circ \) which is \( \cos(60^\circ - 45^\circ) \), we use the formula \(\cos(A - B) = \cos A \cos B + \sin A \sin B\). Thus, we have: \[ \cos 15^\circ = \cos 60^\circ \cos 45^\circ + \sin 60^\circ \sin 45^\circ = \left(\frac{1}{2} \cdot \frac{\sqrt{2}}{2}\right) + \left(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} = \frac{\sqrt{2} + \sqrt{6}}{4}. \] Next, to find \( \sin 75^\circ \), we note that \( 75^\circ = 45^\circ + 30^\circ \). Using the addition formula \(\sin(A + B) = \sin A \cos B + \cos A \sin B\), we calculate: \[ \sin 75^\circ = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ = \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}. \] For \( \cos 105^\circ \), we rewrite it using \( 180^\circ - 75^\circ \): \[ \cos 105^\circ = -\cos 75^\circ = -\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) = -\frac{\sqrt{6} + \sqrt{2}}{4}. \] Now, \( \sin 195^\circ \) can be expressed as \( \sin(180^\circ + 15^\circ) \): \[ \sin 195^\circ = -\sin 15^\circ = -\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right) = -\frac{\sqrt{6} - \sqrt{2}}{4}. \] For \( \sin 165^\circ \), observe that \( 165^\circ = 180^\circ - 15^\circ \): \[ \sin 165^\circ = \sin 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}. \] Finally, for \( \tan 15^\circ \): Using the formula \( \tan A = \frac{\sin A}{\cos A} \): \[ \tan 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ} = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}. \] Thus, summarizing the results: \[ \begin{align*} 4.1 & : \cos 15^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}, \\ 4.2 & : \sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}, \\ 4.3 & : \cos 105^\circ = -\frac{\sqrt{6} + \sqrt{2}}{4}, \\ 4.4 & : \sin 195^\circ = -\frac{\sqrt{6} - \sqrt{2}}{4}, \\ 4.5 & : \sin 165^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}, \\ 4.6 & : \tan 15^\circ = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}. \end{align*} \]
