Prove the identity $\cos(a + b) + \cos(a - b) = 2 \cos a \cos b$ - HSC - SSCE Mathematics Extension 2 - Question 6 - 2003 - Paper 1

Using the identity we just proved, we rewrite the integral:

[ \int \cos 3x \cos 2x , dx = \frac{1}{2} \int (\cos(3x - 2x) + \cos(3x + 2x)) , dx ]

This simplifies to:

[ \frac{1}{2} \int (\cos x + \cos 5x) , dx ]

Calculating this gives:

[ \frac{1}{2} \left( \sin x + \frac{1}{5} \sin 5x \right) + C ]

To find $s_3$, we use the recurrence relation:

[ s_3 = s_2 + (3-1)s_1 = 2 + 2 \times 1 = 4 ]

Now for $s_4$:

[ s_4 = s_3 + (4-1)s_2 = 4 + 3 \times 2 = 10 ]

We start by observing that:

[ \sqrt{x + x^2} = \sqrt{x(1 + x)} = \sqrt{x(x + 1)} ]

This holds for all $x \geq 0$ as both expressions are non-negative.

Base case: For $n=1$, $s_1 = 1 \geq \sqrt{1}.$

Inductive step: Assume for $k$, $s_k \geq \sqrt{k}$. Then for $n = k + 1$:

[ s_{k+1} = s_k + (k) s_{k-1} ]

By induction hypothesis, we know $s_k \geq \sqrt{k}$ and $s_{k-1} \geq \sqrt{k-1}$. Therefore:

[ s_{k+1} \geq \sqrt{k} + (k) \sqrt{k-1} \geq \sqrt{k + 1} ]

Thus, the statement holds for $n = k + 1$.

Using the Arithmetic Mean-Geometric Mean inequality:

[ \frac{x + y}{2} \geq \sqrt{xy} ]

This holds by the fundamental properties of non-negative numbers.

Using the AM-GM inequality again:

[ a^2 + b^2 \geq 2ab \quad ext{(and similar for others)} ]

Adding these gives:

[ a^2 + b^2 + c^2 \geq ab + bc + ca ]

From the Pythagorean results derived previously, we replace $d$:

[ a^2 + b^2 + c^2 \geq ab + bc + ca \geq abc ]

This follows by the symmetry of the situation.