3.1 If \( \sin 28^{\circ}=p \), write the following in terms of \( p \). 3.1.1 \( \tan 28^{\circ} \) (2) 3.1.2 \( \cos 208^{\circ} \) (2) \( 3.1 .3 \sin 388^{\circ} \) (2) \( 3.1 .4 \cdot \tan \left(-332^{\circ}\right) \) (2) 3.2 It is given that \( 37 \sin \alpha+35=0 \) and \( \tan \alpha>0 \). With the sid of a diagram and without the use of a calculator, determine th: value of : \( 3.2 .1 \cos \alpha \) (4) \( 3.2 .2 \cos \left(-270^{\circ}+\alpha\right) \) (2) \( 3.2 .3 \frac{\sin \left(90^{\circ}+\alpha\right)}{\sin \left(540^{\circ}-\alpha\right)} \) \( 3.2 .4 \sin \alpha \) if \( a+\beta=900^{\circ} \) (3)

To find \( \tan 28^{\circ} \) in terms of \( p \): Using the identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), and knowing \( \sin 28^{\circ} = p \), we can find \( \cos 28^{\circ} \) using the Pythagorean identity: \[ \cos 28^{\circ} = \sqrt{1 - p^2} \] So, \[ \tan 28^{\circ} = \frac{p}{\sqrt{1 - p^2}} \] For \( \cos 208^{\circ} \): 208° is in the third quadrant where cosine is negative. We can express it as: \[ \cos 208^{\circ} = \cos(180^{\circ} + 28^{\circ}) = -\cos 28^{\circ} \] Since \( \cos 28^{\circ} = \sqrt{1 - p^2} \), we find: \[ \cos 208^{\circ} = -\sqrt{1 - p^2} \] For \( \sin 388^{\circ} \): This angle can be simplified as: \[ 388^{\circ} - 360^{\circ} = 28^{\circ} \] Thus, \[ \sin 388^{\circ} = \sin 28^{\circ} = p \] For \( \tan(-332^{\circ}) \): First, reduce the angle by adding 360°: \[ -332^{\circ} + 360^{\circ} = 28^{\circ} \] Therefore, \[ \tan(-332^{\circ}) = \tan(28^{\circ}) = \frac{p}{\sqrt{1 - p^2}} \] Now, let’s tackle the given equations involving \( \alpha \): With the equation \( 37 \sin \alpha + 35 = 0 \): This gives us \( \sin \alpha = -\frac{35}{37} \). Given that \( \tan \alpha > 0 \), we know that \( \alpha \) is in the third quadrant where cosine is also negative. Thus: \[ \cos \alpha = -\sqrt{1 - \left(-\frac{35}{37}\right)^2} = -\sqrt{1 - \frac{1225}{1369}} = -\sqrt{\frac{144}{1369}} = -\frac{12}{37} \] For \( \cos(-270^{\circ} + \alpha) \): Using the cosine addition formula: \[ \cos(-270^{\circ} + \alpha) = \cos(-270^{\circ}) \cos(\alpha) + \sin(-270^{\circ}) \sin(\alpha) \] This translates to: \[ \cos(-270^{\circ}) = 0, \quad \sin(-270^{\circ}) = -1 \] Therefore: \[ \cos(-270^{\circ} + \alpha) = 0 \cdot \cos(\alpha) - 1 \cdot \sin(\alpha) = \frac{35}{37} \] Finally, for \( \frac{\sin(90^{\circ} + \alpha)}{\sin(540^{\circ} - \alpha)} \): We can simplify this as follows: \[ \sin(90^{\circ} + \alpha) = \cos \alpha = -\frac{12}{37} \] And: \[ 540^{\circ} - \alpha = 360^{\circ} + (180^{\circ} - \alpha) = \sin(180^{\circ} - \alpha) = \sin \alpha = -\frac{35}{37} \] Thus: \[ \frac{\sin(90^{\circ} + \alpha)}{\sin(540^{\circ} - \alpha)} = \frac{-\frac{12}{37}}{-\frac{35}{37}} = \frac{12}{35} \] With the information gathered, we know: \[ \sin \alpha = -\frac{35}{37} \] And finally, if \( a + \beta = 900^{\circ} \), since \( \beta \) is derived from our conditions, we can find \( \sin \alpha \) remains at \( -\frac{35}{37} \).