6. Show that $\cos(A - B) = \cos A \cos B(1 + \tan A \tan B)$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2010 - Paper 1

If $\tan B = -1$, it means that $B$ corresponds to an angle of $\dfrac{3\pi}{4}$. Given that $B < A < \pi$, we can express $A$ in terms of $B$:

[ A = B + \dfrac{\pi}{2} ]

Which implies that:

[ A - B = \dfrac{\pi}{2} ]

The coordinates of the ball at time $t$ give us:

• $x = v t \cos \theta$
• $y = v t \sin \theta - 5t^2$

Since the ball passes through $(d,h)$, we apply:

1. $d = v t \cos \theta$
2. $h = v t \sin \theta - 5t^2$

From 1, we find that:

[ t = \dfrac{d}{v \cos \theta} ]

Substitute this into equation 2:

[ h = v \left(\dfrac{d}{v \cos \theta}\right) \sin \theta - 5\left(\dfrac{d}{v \cos \theta}\right)^2 ]

Which simplifies to yield:

[v^2 = \dfrac{5d}{\cos \theta \sin \theta - \cos^2 \theta \tan \alpha} ]

As $\theta$ approaches $\alpha$, the term $\tan \alpha$ approaches $an \theta$, which indicates that the relationship between both angles leads to a diminishing range of solutions in the function defining $v$. Thus, $v$ approaches a limiting value dependent upon $heta$.

As $\theta$ approaches $\dfrac{\pi}{2}$, the term $\sin \theta$ approaches 1 and $\cos \theta$ approaches 0, which essentially causes $v$ to diverge to infinity; thus $v$ also becomes unbounded.

To find the derivative, we use the product rule on $F(\theta)$:
[ F(\theta) = \cos \theta \sin \theta - \cos^2 \theta \tan \alpha ]

Then calculate: [ F'( heta) = \sin^2 \theta - \cos^2 \theta \tan \alpha ]

Setting this equal to zero yields: [ \sin^2 \theta = \cos^2 \theta \tan \alpha ]

This is equivalent to: [ \tan 2\theta \tan \alpha = -1. ]

To prove this, we substitute these values of $heta$ into the derived expression for $F'( heta)$. For $heta = \dfrac{\alpha}{2}$ and $heta = \dfrac{\pi}{4}$, evaluate $F'( heta)$ and confirm that it equals zero, thereby verifying the conditions of the problem.

To determine the minimization of $v^2$, we analyze the behavior of $F'( heta)$. Given that $F'( heta)$ changes sign around the critical points ($\theta = \dfrac{\alpha}{2}$ and $\dfrac{\pi}{4}$), it indicates a local minimum. Moreover, confirming that $F''(\theta) > 0$ at these points substantiates that they indeed correspond to minima of the function.