Question: Tide depth at a harbour is $h = 3 - 2\cos\!\left(\dfrac{\pi t}{6}\right)$ metres, with $t$ in hours after midnight. (a) State the maximum and minimum tide depths and the times they first occur. (b) Find the times between which $h \geq 4$.

Step 1. Range of $\cos\!\left(\dfrac{\pi t}{6}\right)$ is $[-1, 1]$:

• Max $h = 3 - 2(-1) = 5$ when $\cos = -1$, i.e. $\dfrac{\pi t}{6} = \pi$, so $t = 6$ (6 a.m.).
• Min $h = 3 - 2(1) = 1$ when $\cos = 1$, i.e. $t = 0$ (midnight).

Step 2. $h \geq 4$:

$$3 - 2\cos\!\left(\dfrac{\pi t}{6}\right) \geq 4 \;\Rightarrow\; \cos\!\left(\dfrac{\pi t}{6}\right) \leq -\dfrac{1}{2}$$

Step 3. $\cos\theta \leq -\dfrac{1}{2}$ when $\theta \in \left[\dfrac{2\pi}{3}, \dfrac{4\pi}{3}\right]$.

$\dfrac{\pi t}{6} \in \left[\dfrac{2\pi}{3}, \dfrac{4\pi}{3}\right] \Rightarrow t \in [4, 8]$.

Answer: $h \geq 4$ from $t = 4$ (4 a.m.) to $t = 8$ (8 a.m.) — a 4-hour window.

Tip: Tidal models always have period 12 hours (half-day). Set up cosine with the appropriate phase so that high/low tide aligns with the given conditions.

Question: A Ferris wheel has carriages whose height is $h = 95 - 75\cos\!\left(\dfrac{\pi t}{10}\right)$ metres, with $t$ in minutes after boarding. (a) State the wheel's centre height and radius. (b) Find $t$ at which a passenger first reaches height $130$ m.

Step 1. Compare with $h = c - a\cos(bt)$ where:

• $c = 95$ is the centre height
• $a = 75$ is the radius (amplitude)
• Period $= \dfrac{2\pi}{\pi/10} = 20$ minutes per revolution

Step 2. Solve $h = 130$:

$$95 - 75\cos\!\left(\dfrac{\pi t}{10}\right) = 130$$ $$\cos\!\left(\dfrac{\pi t}{10}\right) = -\dfrac{35}{75} = -\dfrac{7}{15}$$

Step 3. First solution (smallest positive $t$):

$$\dfrac{\pi t}{10} = \cos^{-1}\!\left(-\dfrac{7}{15}\right) \approx 2.082 \;\text{rad}$$ $$t = \dfrac{10 \times 2.082}{\pi} \approx 6.63 \;\text{minutes}$$

Answer: Centre $95$ m, radius $75$ m, first reaches $130$ m at $t \approx 6.63$ min.

Tip: The "$95 - 75\cos$" form ensures the carriage starts at the BOTTOM ($\cos 0 = 1 \Rightarrow h = 20$) and rises to the top. If you see "$+\cos$", the carriage starts at the top.

Question: A side of a hinged frame is given by $FD = 14\cos\theta + 3\sin\theta + 6$, where $\theta$ is the hinge angle in degrees, $0° \leq \theta \leq 90°$. Find the maximum value of $FD$ and the angle $\theta$ at which it occurs.

Step 1. Apply R-formula on $14\cos\theta + 3\sin\theta$:

$$R = \sqrt{14^2 + 3^2} = \sqrt{205} \approx 14.32$$ $$\tan\alpha = \dfrac{3}{14}, \;\alpha = \tan^{-1}\!\dfrac{3}{14} \approx 12.1°$$

So $14\cos\theta + 3\sin\theta = \sqrt{205}\cos(\theta - 12.1°)$.

Step 2. $FD = \sqrt{205}\cos(\theta - 12.1°) + 6$.

Maximum when $\cos(\theta - 12.1°) = 1$, i.e. $\theta = 12.1°$.

$$FD_{\max} = \sqrt{205} + 6 \approx 20.32 \;\text{units}$$

Answer: Max $FD \approx 20.32$ units at $\theta \approx 12.1°$.

Tip: Geometric R-formula problems ALWAYS state a domain like $0° \leq \theta \leq 90°$ — verify your $\theta_{\max}$ falls in range. If it doesn't, the maximum occurs at the boundary.

Question: A sector of a circle has area $A = 20\sin 2\theta - 16\cos 2\theta + 16$. Find the maximum value of $A$ and the value of $\theta$ at which it occurs.

Step 1. Apply R-formula on $20\sin 2\theta - 16\cos 2\theta$:

$$R = \sqrt{20^2 + 16^2} = \sqrt{400 + 256} = \sqrt{656} \approx 25.61$$ $$\tan\alpha = \dfrac{16}{20} = \dfrac{4}{5}, \;\alpha \approx 38.7°$$

So $20\sin 2\theta - 16\cos 2\theta = \sqrt{656}\sin(2\theta - 38.7°)$.

Step 2. $A = \sqrt{656}\sin(2\theta - 38.7°) + 16$.

Maximum when $\sin(2\theta - 38.7°) = 1$, i.e. $2\theta = 90° + 38.7° = 128.7°$, so $\theta = 64.35°$.

$$A_{\max} = \sqrt{656} + 16 \approx 41.61 \;\text{(area units)}$$

Why argument is $2\theta$: When you see $\sin 2\theta$ or $\cos 2\theta$, the R-formula's argument matches — the resulting sinusoid is in $2\theta$ as well. Don't accidentally divide by 2 partway through.

Question: The area of an isosceles triangle is $A = \dfrac{1}{4}x^2(\sin 2\theta + 2\cos\theta)$ where $x$ is fixed and $\theta$ is a variable angle in $(0, \pi/2)$. Find the value of $\theta$ that maximises $A$.

Step 1. Differentiate $A$ with respect to $\theta$ (treating $x$ as constant):

$$\dfrac{dA}{d\theta} = \dfrac{x^2}{4}\left(2\cos 2\theta - 2\sin\theta\right) = \dfrac{x^2}{2}(\cos 2\theta - \sin\theta)$$

Step 2. Set $\dfrac{dA}{d\theta} = 0$:

$$\cos 2\theta = \sin\theta$$

Use $\cos 2\theta = 1 - 2\sin^2\theta$:

$$1 - 2\sin^2\theta = \sin\theta$$ $$2\sin^2\theta + \sin\theta - 1 = 0$$ $$(2\sin\theta - 1)(\sin\theta + 1) = 0$$ $$\sin\theta = \dfrac{1}{2} \quad\text{or}\quad \sin\theta = -1$$

Step 3. In $(0, \pi/2)$: $\sin\theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi}{6}$.

(Verify it's a max via second derivative test or sign change.)

Answer: $\theta = \dfrac{\pi}{6}$ maximises $A$.

Tip: Trig optimisation that mixes different argument types (e.g. $2\theta$ and $\theta$) usually CAN'T be reduced cleanly via R-formula — calculus is the tool. Differentiate, set to zero, convert all terms to one variable using identities, solve the polynomial in $\sin\theta$ or $\cos\theta$.

Question: In $\triangle ABC$, $\angle A = 60°$, $b = 8$, $c = 5$. Find $a$ exactly, then find the area of the triangle.

Step 1. Cosine rule for $a$:

$$a^2 = b^2 + c^2 - 2bc\cos A = 64 + 25 - 2(8)(5)\cos 60°$$ $$= 89 - 80 \cdot \dfrac{1}{2} = 89 - 40 = 49$$ $$a = 7$$

Step 2. Area using the sine area formula:

$$\text{Area} = \dfrac{1}{2}bc\sin A = \dfrac{1}{2}(8)(5)\sin 60° = 20 \cdot \dfrac{\sqrt{3}}{2} = 10\sqrt{3}$$

Answer: $a = 7$, Area $= 10\sqrt{3}$ square units.

Three triangle tools:

• Sine rule $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ — use when you have an angle-side pair.
• Cosine rule $a^2 = b^2 + c^2 - 2bc\cos A$ — use for SAS or SSS.
• Area $= \dfrac{1}{2}ab\sin C$ — use when you have two sides and the included angle.

Question: Show that $\sin 105° = \dfrac{\sqrt{6} + \sqrt{2}}{4}$, and find the exact value of $\sin 105° \cdot \cos 105°$.

Step 1. Write $105° = 60° + 45°$. Compound formula:

$$\sin(A + B) = \sin A\cos B + \cos A\sin B$$ $$\sin 105° = \sin 60°\cos 45° + \cos 60°\sin 45°$$ $$= \dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{2}}{2} + \dfrac{1}{2} \cdot \dfrac{\sqrt{2}}{2}$$ $$= \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6} + \sqrt{2}}{4} \;\checkmark$$

Step 2. For $\sin 105° \cdot \cos 105°$, use $\sin 2x = 2\sin x \cos x$:

$$\sin 105° \cdot \cos 105° = \dfrac{1}{2}\sin 210° = \dfrac{1}{2} \cdot \left(-\dfrac{1}{2}\right) = -\dfrac{1}{4}$$

Answer: $\sin 105° \cdot \cos 105° = -\dfrac{1}{4}$.

Tip: $\sin x \cos x$ is ALWAYS $\dfrac{1}{2}\sin 2x$ — much faster than computing each ratio separately and multiplying.