Evaluate $$\int_0^{\frac{\pi}{2}} \cos x \sin^2 x \, dx.$$ By making the substitution $t = \tan \frac{\theta}{2}$ or otherwise, show that $$\csc \theta + \cot \theta = \cot \frac{\theta}{2}.$$ The points $P(2ap^2, ap^2)$ and $Q(2aq, aq^2)$ lie on the parabola $x^2 = 4ay$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2005 - Paper 1

Using the substitution $t = \tan \frac{\theta}{2}$, we have the relationships:

$\sin \theta = \frac{2t}{1 + t^2}, \quad \cos \theta = \frac{1 - t^2}{1 + t^2}.$

Now,[ \csc \theta = \frac{1}{\sin \theta} = \frac{1 + t^2}{2t}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{(1 - t^2) \cdot 2t}{2t} = \frac{1 - t^2}{t} .] Thus, $\csc \theta + \cot \theta = \frac{1 + t^2}{2t} + \frac{1 - t^2}{t} = \frac{1 + t^2 + 2(1 - t^2)}{2t} = \frac{3 - t^2}{2t}.$

To show that this equals $\cot \frac{\theta}{2}$, we rewrite it in terms of $t$: $\cot \frac{\theta}{2} = \frac{1}{t}.$

Therefore, $\csc \theta + \cot \theta = \frac{3 - t^2}{2t} = \cot \frac{\theta}{2}$, which is what we set out to show.

To find the intersection point of the normals at points $P$ and $Q$, we set their equations equal: $x + py = 2ap^2 + ap^3$ $x + qy = 2aq + aq^3.$

Subtract these equations: $(p - q)y = (2ap^2 + ap^3) - (2aq + aq^3).$

Solving for $y$, we can substitute back to find $x$ and determine coordinates for $P$ and $Q$ as defined in the texts.

The equation of the chord can be analyzed by substituting the point $(0, a)$ into the line equation: $y = \frac{1}{2}(p + q)(0) - apq = a.$

This gives us $-apq = a$, leading to $pq = -1$.

We already have the coordinates of $R$. By using the relationships from our previous steps, we can derive that the locus will take the form of an equation in $p$ and $q$. By tracing values of $p$ and $q$, we can express this locus in Cartesian coordinates.

To prove this statement using induction, we first check the base case for $n=2$: $4^2 - 1 - 7(2) = 16 - 1 - 14 = 1 > 0.$

Now assume it holds for $n=k$, i.e., $4^k - 1 - 7k > 0$. We must show it holds for $n=k+1$: $4^{k+1} - 1 - 7(k+1) = 4 \cdot (4^k) - 1 - 7k - 7 = 4(4^k - 1 - 7k) + 3 > 0.$

Thus by the principle of induction, the statement holds for all $n \geq 2$.