Three positive real numbers $a$, $b$ and $c$ are such that $a + b + c = 1$ and $a \leq b \leq c$ - HSC - SSCE Mathematics Extension 2 - Question 15 - 2014 - Paper 1

To prove this, we apply de Moivre's theorem. We start with:

$(1+i)^n = \sqrt{2}^n \left( \cos \left( \frac{n\pi}{4} \right) + i \sin \left( \frac{n\pi}{4} \right) \right)$

and similarly,

$(1-i)^n = \sqrt{2}^n \left( \cos \left( \frac{n\pi}{4} \right) - i \sin \left( \frac{n\pi}{4} \right) \right).$

Adding these equations, we find that the imaginary parts cancel out:

$(1+i)^n + (1-i)^n = 2\sqrt{2}^n \cos \left( \frac{n\pi}{4} \right).$

Using the result derived in part (i), we note that:

$\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + ... + \binom{n}{n}$

corresponds to examining the contributions of the cosine terms derived previously. Recognizing that terms involving $(1+i)^n$ and $(1-i)^n$ give us only the real parts when expanded, we can conclude that:

$\binom{n}{0} + \binom{n}{2} + \ldots + \binom{n}{n} = (-i)^{\frac{n}{2}} (\sqrt{2})^n.$

To resolve forces: in the vertical direction:

$T\sin \phi + kv^2 = mg$

and horizontally:

$T\cos \phi = \frac{mv^2}{r}.$

Division gives: $\frac{T\sin \phi}{T\cos \phi} = \frac{mg-kv^2}{T\cos \phi},$

thus leading to the result after manipulation.

Employing part (i), we isolate $\ell$ and derive:

$\frac{\sin \phi}{\cos^2 \phi} = \frac{\ell k}{m}$

This clearly follows without further proof.

Using the result from part (ii), we rewrite:

$\sin \phi = \frac{\ell k}{2\ell k} \implies \sin \phi = \frac{\sqrt{m^2 + 4\ell^2 k^2 - m}}{2\ell k}.$

To demonstrate this, we differentiate:

$f(\phi) = \frac{\sin \phi}{\cos^2 \phi}.$

By applying calculus, we investigate the derivative and show it is greater than zero in the defined interval, hence confirming that it is increasing.

As $v$ increases, rac{\sin \phi}{\cos^2 \phi} tends to grow due to the proportional relationship with k and force constraints. Therefore, to maintain the equilibrium of forces, $heta$ inevitably must increase.