Question 6 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 6 - 2003 - Paper 1

To find the integral, we can use the identity proven in part (i):

1. Rewriting $\cos 3x \cos 2x$ using the identity: $\cos 3x \cos 2x = \frac{1}{2}(\cos(3x + 2x) + \cos(3x - 2x)) = \frac{1}{2}(\cos 5x + \cos x).$
2. Now integrate: $\int \cos 3x \cos 2x \, dx = \frac{1}{2} \left( \int \cos 5x \, dx + \int \cos x \, dx \right).$
3. The integrals become: $= \frac{1}{2}\left( \frac{1}{5}\sin 5x + \sin x \right) + C.$
4. Therefore, $\int \cos 3x \cos 2x \, dx = \frac{1}{10}\sin 5x + \frac{1}{2}\sin x + C.$

To find $s_3$ and $s_4$, use the recurrence relation:

1. For $s_3$: $s_3 = s_2 + (3-1)s_1 = 2 + 2 \cdot 1 = 4.$
2. For $s_4$: $s_4 = s_3 + (4-1)s_2 = 4 + 3 \cdot 2 = 10.$ Thus, $s_3 = 4$ and $s_4 = 10$.

To prove this identity, consider:

1. The left side: $\sqrt{x + x^2} = \sqrt{x(1+x)}.$
2. The right side is: $\sqrt{x(x+1)}.$
3. Since both sides are equal, the statement holds true for all $x \geq 0$.

1. Base Case: For $k=1$, $s_1 = 1 \geq \sqrt{1}$. True.
2. Inductive Step: Assume true for $k=n$, then for $k=n+1$: $s_{n+1} = s_n + (n)s_{n-1} \geq \sqrt{n} + (n)\sqrt{n-1} \geq \sqrt{n+1}.$ By verifying and ensuring both parts hold, it follows that the statement is true.

Using the AM-GM inequality, we know:

1. The arithmetic mean of $x$ and $y$ is: $\frac{x + y}{2}$
2. The geometric mean is: $\sqrt{xy}.$
3. Thus, we conclude: $\frac{x + y}{2} \geq \sqrt{xy}.$

We can show this using the rearrangement inequality:

1. Note that: $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0.$
2. Expanding the squares: $a^2 + b^2 + c^2 - ab - ac - bc \geq 0.$ This proves the inequality holds.

From the previous inequalities, we know:

1. By substituting $c = d - a - b$, we check: $a^2 + b^2 + (d-a-b)^2 \geq ab + ac + bc.$
2. After simplification, this yields: $a^2 + b^2 + c^2 \geq abc.$ This concludes our deduction.