(b) (1) Using the fact that \( 1=\frac{\sqrt{a}}{\sqrt{a}} \) where \( a>0 \), show that \( \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2} \) (2) Hence show that: (i) \( \cos 45^{\circ}=\frac{\sqrt{2}}{2} \) (ii) \( \tan 30^{\circ}=\frac{\sqrt{3}}{3} \) (iii) \( \tan 60^{\circ}=\frac{3}{\sqrt{3}} \) (iv) \( \sin 60^{\circ}=\frac{3}{2 \sqrt{3}} \) Without using a calculator, solve the following equations, where the angles are acu (1) \( \sin x=\frac{1}{2} \) (2) \( \cos x=\frac{1}{2} \) (3) \( 2 \sin x=\sqrt{3} \) (4) \( \sqrt{2} \cos x=1 \) (5) \( \quad \tan x=\sqrt{3} \) (6) \( \sqrt{3} \tan x-1=0 \) (7) \( \tan x=1 \) (8) \( \sin 2 x=0,5 \) (9) \( \cos \left(3 x-15^{\circ}\right) \)

To prove that \( \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \), we start with the fact that \( 1 = \frac{\sqrt{a}}{\sqrt{a}} \) for \( a = 2 \). We can multiply both sides by \( \frac{1}{\sqrt{2}} \), giving us \( \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \). It’s a neat manipulation that connects a simple fraction to a fundamental ratio found in trigonometry! Now, let's move onto the relationships in trigonometry! For \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \), this follows from the fact that in a 45-45-90 triangle, the legs are equal, and hence the cosine of 45 degrees yields this ratio. As for \( \tan 30^{\circ} = \frac{\sqrt{3}}{3} \), we can think of the 30-60-90 triangle where the opposite side (1) compared to the adjacent side (\(\sqrt{3}\)) gives us this charismatic ratio. Fun fact – knowing just these fundamental angles can open the door to solving many trigonometric puzzles!