10a. The gradient of a tangent to the curve y = 4x\(^3\) at points P and Q is 108. Find the coordinates of P and Q bi. Given \(\hat{A}\) = 45º, \(\hat{B}\) ...

10a. The equation of the curve is \( y = 4x^3 \).  
The gradient of the tangent at any point is given by the derivative:  
\(\frac{dy}{dx} = 12x^2.\)  
At points P and Q, the gradient is 108:  
\(12x^2 = 108 \implies x^2 = 9 \implies x = \pm 3.\)

When \( x = 3 \):  
\(y = 4(3)^3 = 4 \times 27 = 108.\)
So one point is 3, 108).  

When x = -3 :  
\(y = 4(-3)^3 = 4 \times (-27) = -108.\) 
So the other point is (-3, -108).  

Coordinates of P and Q:  (3, 108), and (-3, -108).

10bi.  We use the angle subtraction formulas:  
sin(A - B) = sin A cos B - cos A sin B,  
cos(A - B) = cos A cos B + sin A sin B,
where \( A = 45^\circ \) and \( B = 30^\circ \).  

Known values:  
\(\sin 45^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2}\)

Now,  
\(\sin 15^\circ = \sin(45^\circ - 30^\circ) = \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}.\)  

\(\cos 15^\circ = \cos(45^\circ - 30^\circ) = \left( \frac{\sqrt{2}}{2} \right) \left( \frac{\sqrt{3}}{2} \right) + \left( \frac{\sqrt{2}}{2} \right) \left( \frac{1}{2} \right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}.\) 

10bii.  \(\tan 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ} = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}.\) 

Rationalise the denominator by multiplying numerator and denominator by \(\sqrt{6} - \sqrt{2}\):  
Numerator: \((\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} + 2 = 8 - 4\sqrt{3}\).  
Denominator: \((\sqrt{6})^2 - (\sqrt{2})^2 = 6 - 2 = 4 \).  

Thus,  \(\tan 15^\circ = \frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}.\)