Determine the following in terms of \( t \), without the use of a calculato \( 5.2 .1 \sin 15^{\circ} \) \( 5.2 .2 \sin 75^{\circ} \) 5.2 .3 \( 1-\tan ^{2} 15^{\circ} \) (Give the answer as a single fraction.)

To solve these problems, let's break them down one by one. ### \( 5.2 .1 \sin 15^{\circ} \) The angle \( 15^{\circ} \) can be expressed using the sine subtraction formula: \[ \sin 15^{\circ} = \sin(45^{\circ} - 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} \] Using known values: \(\sin 45^{\circ} = \frac{\sqrt{2}}{2}, \cos 30^{\circ} = \frac{\sqrt{3}}{2}, \cos 45^{\circ} = \frac{\sqrt{2}}{2}, \sin 30^{\circ} = \frac{1}{2}\): \[ \sin 15^{\circ} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4} \] ### \( 5.2 .2 \sin 75^{\circ} \) Using the sine addition formula, we find: \[ \sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ} \] Plugging in the same known values: \[ \sin 75^{\circ} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} \] ### \( 5.2 .3 1 - \tan^2 15^{\circ} \) Using the identity \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \), we start with \( \tan 15^{\circ} = \frac{\sin 15^{\circ}}{\cos 15^{\circ}} \). Using the cosine subtraction formula: \[ \cos 15^{\circ} = \cos(45^{\circ} - 30^{\circ}) = \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} \] Thus: \[ \tan 15^{\circ} = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}} \] \( \tan^2 15^{\circ} = \left(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}\right)^2 = \frac{(6 - 2 - 2\sqrt{12})}{(6 + 2 + 2\sqrt{12})} = \frac{4 - 2\sqrt{12}}{8 + 2\sqrt{12}} \) Now plug this into \(1 - \tan^2 15^{\circ}\) to give \( 1 - \frac{(4 - 2\sqrt{12})}{(8 + 2\sqrt{12})} = \frac{(8 + 2\sqrt{12}) - (4 - 2\sqrt{12})}{8 + 2\sqrt{12}} = \frac{4 + 4\sqrt{12}}{8 + 2\sqrt{12}} \) This