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 the problems step by step, we will use known trigonometric identities and values. ### 5.2.1: Calculate \( \sin 15^{\circ} \) Using the sine subtraction formula: \[ \sin(15^{\circ}) = \sin(45^{\circ} - 30^{\circ}) = \sin(45^{\circ})\cos(30^{\circ}) - \cos(45^{\circ})\sin(30^{\circ}) \] We know: - \( \sin(45^{\circ}) = \frac{\sqrt{2}}{2} \) - \( \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \) - \( \cos(45^{\circ}) = \frac{\sqrt{2}}{2} \) - \( \sin(30^{\circ}) = \frac{1}{2} \) Substituting these values: \[ \sin(15^{\circ}) = \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4} \] ### 5.2.2: Calculate \( \sin 75^{\circ} \) Using the sine addition formula: \[ \sin(75^{\circ}) = \sin(45^{\circ} + 30^{\circ}) = \sin(45^{\circ})\cos(30^{\circ}) + \cos(45^{\circ})\sin(30^{\circ}) \] Substituting the known values: \[ \sin(75^{\circ}) = \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} \] ### 5.2.3: Calculate \( 1 - \tan^{2} 15^{\circ} \) Using the identity \( \tan^{2} \theta = \frac{\sin^{2} \theta}{\cos^{2} \theta} \): \[ \tan(15^{\circ}) = \frac{\sin(15^{\circ})}{\cos(15^{\circ})} \] We can find \( \cos(15^{\circ}) \) using the cosine subtraction formula: \[ \cos(15^{\circ}) = \cos(45^{\circ} - 30^{\circ}) = \cos(45^{\circ})\cos(30^{\circ}) + \sin(45^{\circ})\sin(30^{\circ}) \] Substituting the known values: \[ \cos(15^{\circ}) = \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} \] Now, we can find \( \tan^{2}(15^{\circ}) \): \[ \tan^{2}(15^{\circ}) = \frac{\sin^{2}(15^{\circ})}{\cos^{2}(15^{\circ})} = \frac{\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^{2}}{\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^{2}} = \frac{(\sqrt{6} - \sqrt{2})^{2}}{(\sqrt{6} + \sqrt{2})^{2}} \] Calculating the squares: \[ (\sqrt{6} - \sqrt{2})^{2} = 6 - 2\sqrt{12} + 2 = 8 - 2\sqrt{12} \] \[ (\sqrt{6} + \sqrt{2})^{2} = 6 + 2\sqrt{12} + 2 = 8 + 2\sqrt{12} \] Thus, \[ \tan^{2}(15^{\circ}) = \frac{8 - 2\sqrt{12}}{8 + 2\sqrt{12}} \] Now, we can find \( 1 - \tan^{2}(15^{\circ}) \): \[ 1 - \tan^{2}(15^{\circ}) = 1 - \frac{8 - 2\sqrt{12}}{8 + 2\sqrt{12}} = \frac{(8 + 2\sqrt{12}) - (8 - 2\sqrt{12})}{8 + 2\sqrt{12}} = \frac{4\sqrt{12}}{8 + 2\sqrt{12}} = \frac{2\sqrt{12}}{4 + \sqrt{12}} \] Thus, the answers are: - \( 5.2.1: \frac{\sqrt{6} - \sqrt{2}}{4} \) - \( 5.2.2: \frac{\sqrt{6} + \sqrt{2}}{4} \) - \( 5.2.3: \frac{2\sqrt{12}}{4 + \sqrt{12}} \)