In \( \triangle ABC \), \( \angle CAB = \alpha, \angle ABC = \beta \) and \( \angle BCA = \gamma \) - HSC - SSCE Mathematics Extension 2 - Question 6 - 2006 - Paper 1

Using the relationships found in part (i), we can express each sine function in terms of ( \theta ):

We have previously derived that:

[ \sin(\alpha) = \sin(\theta + (\beta - \theta)) = \sin(\theta)\cos(\beta - \theta) + \cos(\theta)\sin(\beta - \theta) ]

Now multiplying these three sine relations, we can apply the identity ( \sin(A)\sin(B)\sin(C) ) and manipulate accordingly to derive:

[ \sin^3(\theta) = \sin(\alpha - \theta) \sin(\beta - \theta) \sin(\gamma - \theta) ]

This identity can be proven using the definitions of cotangent:

[ \cot x = \frac{\cos x}{\sin x} \quad \text{and} \quad \cot y = \frac{\cos y}{\sin y} ]

Thus, the left-hand side becomes:

[ \cot x - \cot y = \frac{\cos x}{\sin x} - \frac{\cos y}{\sin y} = \frac{\cos x \sin y - \cos y \sin x}{\sin x \sin y} ]

Since ( \cos x \sin y - \cos y \sin x = \sin(y - x) ), we have:

[ \cot x - \cot y = \frac{\sin(y - x)}{\sin x \sin y} ]

Using the identity from part (iii), we can express:

[ (\cot \theta - \cot \alpha) = \frac{\sin(\alpha - \theta)}{\sin \alpha \sin \theta} ]

For each angle, we apply this result:

[ \Rightarrow (\cot \theta - \cot \alpha)(\cot \theta - \cot \beta)(\cot \theta - \cot \gamma) = \frac{\sin(\alpha - \theta)\sin(\beta - \theta)\sin(\gamma - \theta)}{\sin \alpha \sin \beta \sin \gamma} ]

Using the initial conditions, we conclude:

[ = \csc \alpha \csc \beta \csc \gamma ]

In an isosceles right triangle, we have ( \alpha = \beta = 45^{\circ} ) and ( \gamma = 90^{\circ} ).

Thus substituting into part (iv), we simplify the equation:

[ (\cot \theta - \cot 45)(\cot \theta - \cot 45)(\cot \theta - 0) = 1 \cdot 1 \ \Rightarrow (\cot \theta - 1)(\cot \theta - 1)(\cot \theta) = 1 ]

Solving this equation gives us ( \theta = 45^{\circ} ).