Question: Triangles MOP and MPQ with $\angle MOP = \angle MPQ = 90°$, $\angle OMP = \theta$. Line QR is parallel to PO. $MP = 5$, $PQ = 2$.

(i) Show $QR = 2\cos\theta + 5\sin\theta$.

(ii) R-form:

$$QR = \sqrt{2^2 + 5^2}\sin\!\left(\theta + \tan^{-1}\frac{2}{5}\right) = \sqrt{29}\sin(\theta + 21.8°)$$

(iii) Maximum $QR = \sqrt{29}$ when $\theta + 21.8° = 90° \Rightarrow \theta = 68.2°$.

Question: A length is modelled by $L = 5\sin\theta + 12\cos\theta$. Express $L$ in the form $R\sin(\theta + \alpha)$ and find the maximum value of $L$ and the value of $\theta$ at which it occurs.

Step 1. $R = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$.

Step 2. $\tan\alpha = \dfrac{12}{5}$ (coefficient of $\cos$ over coefficient of $\sin$), so $\alpha = \tan^{-1}\!\dfrac{12}{5} \approx 67.4°$.

So $L = 13\sin(\theta + 67.4°)$.

Step 3. Maximum: $\sin(\theta + 67.4°) = 1$ at $\theta + 67.4° = 90°$, i.e. $\theta = 22.6°$.

$$L_{\max} = 13 \;\text{when}\; \theta \approx 22.6°$$

Tip: For $a\sin\theta + b\cos\theta = R\sin(\theta + \alpha)$:

• $R = \sqrt{a^2 + b^2}$
• $\tan\alpha = \dfrac{b}{a}$ (cos coefficient ÷ sin coefficient)

Question: OPQR is two triangles OPR and QPR with $OR = 4$, $RP = 5$, $RQ = 8$, $\angle ORQ = 90°$, $\angle PRQ = \theta$ (acute). Area is $L$ cm².

(i) Show $L = 20\sin\theta + 10\cos\theta$.

(Set up using $\frac{1}{2}ab\sin C$ for each triangle and combine.)

(ii) R-form:

$$L = \sqrt{20^2 + 10^2}\sin\!\left(\theta + \tan^{-1}\frac{10}{20}\right) = \sqrt{500}\sin(\theta + 26.6°)$$

(iii) Maximum: Max $L = \sqrt{500}$ cm² is achieved when $\sin(\theta + 26.6°) = 1$, i.e. $\theta + 26.6° = 90° \Rightarrow \theta = 63.4°$.

Question: Enclosure with $OR = 5$, $QR = 3$, $\angle ORQ = 90°$. Side $OR$ makes acute angle $\theta$ with $OP$. $OP \perp PQ$. Total length is $T$ m.

(i) Show $T = 8 + 2\cos\theta + 8\sin\theta$.

(ii) R-form:

$$T = 8 + \sqrt{2^2 + 8^2}\cos\!\left(\theta - \tan^{-1}\frac{8}{2}\right) = 8 + 2\sqrt{17}\cos(\theta - 76.0°)$$

(iii) Given $T = 11.8$:

$$2\sqrt{17}\cos(\theta - 76.0°) = 3.8 \Rightarrow \cos(\theta - 76.0°) = \frac{3.8}{2\sqrt{17}}$$

Basic angle $\approx 62.6°$, so $\theta - 76.0° = \pm 62.6°$, giving $\theta = 13.4°$ (acute, valid).

Question: A rectangular table $1 \text{ m} \times 2.5$ m is turned until it wedges in a 1.5 m wide corridor. Show that $\cos\theta + 2.5\sin\theta = 1.5$, and find $\theta$ in degrees.

Solution:

The geometric setup gives the relation $\cos\theta + 2.5\sin\theta = 1.5$.

Apply R-formula:

$$\cos\theta + 2.5\sin\theta = \sqrt{1^2 + 2.5^2}\cos\!\left(\theta - \tan^{-1}\frac{2.5}{1}\right) = \sqrt{7.25}\cos(\theta - 68.2°)$$

Thus:

$$\sqrt{7.25}\cos(\theta - 68.2°) = 1.5 \Rightarrow \cos(\theta - 68.2°) = \frac{1.5}{\sqrt{7.25}}$$

Basic angle $= 56.1°$, giving $\theta - 68.2° = \pm 56.1°$. The acute solution is $\theta \approx 12.1°$.

Question: Pentagon OPQRW with $OM = MW$ on a straight line, $PM = QM = 4$, $QR = 5$, $OW = 8$, QRWM rectangle, $\angle PMQ = \theta$, area $A$ m².

(i) $A = 6\cos\theta + 8\sin\theta + 20$.

(ii) R-form: $A = 20 + \sqrt{6^2 + 8^2}\cos(\theta - \tan^{-1}(8/6)) = 20 + 10\cos(\theta - 53.1°)$.

(iii) Max $A$ $= 20 + 10 = 30$ m², when $\theta = 53.1°$.

(iv) Can $A = 20$?

$20 = 20 + 10\cos(\theta - 53.1°) \Rightarrow \cos(\theta - 53.1°) = 0 \Rightarrow \theta - 53.1° = 90° \Rightarrow \theta = 143.1°$.

Since $\theta$ must be acute (here $0° < \theta < 90°$ from geometry), $\theta = 143.1°$ is rejected. No, $A$ cannot equal 20 within the valid range.

Question: A playground OPQR is a quadrilateral with $\angle POR = \angle PQR = 90°$, $OP = 10$ m, $OR = 24$ m, $\angle ORQ = \theta$ where $0° < \theta < 90°$. $W$ is on $RQ$ with $OW \parallel PQ$.

(i) Show that perimeter $L = 34 + 14\cos\theta + 34\sin\theta$.

From geometry: $OW = 24\sin\theta$, $RW = 24\cos\theta$, $PW = 10\sin\theta$, $OW' = 10\cos\theta$. Then $PQ = OW - OW' = 24\sin\theta - 10\cos\theta$ and $WQ = 10\sin\theta$.

$$L = OR + RW + WQ + QP + PO = 24 + 24\cos\theta + 10\sin\theta + 24\sin\theta - 10\cos\theta + 10 = 34 + 14\cos\theta + 34\sin\theta$$

(ii) Express in R-form:

$$L = 34 + \sqrt{14^2 + 34^2}\cos\!\left(\theta - \tan^{-1}\frac{34}{14}\right) = 34 + 26\sqrt{2}\cos(\theta - 67.6°)$$

(iii) Given $L = 50$:

$$\cos(\theta - 67.62°) = \frac{16}{26\sqrt{2}} \Rightarrow \text{basic angle} = 64.21°$$ $$\theta - 67.62° = 64.21°,\, -64.21° \Rightarrow \theta = 131.8° \text{ (rej)},\, 3.4°$$

(iv) Maximum $L = 34 + 26\sqrt{2} = 70.8$ m, when $\theta - 67.62° = 0° \Rightarrow \theta \approx 67.6°$.

Question: Window OPQR is a trapezium OPWR ($OR \parallel PW$) plus a right triangle PQW. $OP = 17.5$, $PQ = 42.5$, $\angle WPQ = 2\theta$.

(i) Show $L = 60 + 60\cos 2\theta + 25\sin 2\theta$ (perimeter).

(ii) R-form:

$$L = 60 + \sqrt{60^2 + 25^2}\sin(2\theta + \alpha), \quad \tan\alpha = \frac{60}{25}$$ $$L = 60 + 65\sin(2\theta + 67.4°)$$

(iii) $L = 100$:

$$\sin(2\theta + 67.4°) = \frac{40}{65} \Rightarrow 2\theta + 67.4° = 142.0° \Rightarrow \theta = 37.3°$$

(iv) Maximum $L = 60 + 65 = 125$ cm, when $2\theta + 67.4° = 90° \Rightarrow \theta = 11.3°$.

Question: The area of a triangle is $A = 16\sin^2\theta + 4\sin\theta\cos\theta$. Express $A$ in the form $R\sin(2\theta - \alpha) + c$ and find $\theta$ when $A$ is maximum.

Step 1. Apply double-angle identities:

• $\sin\theta\cos\theta = \dfrac{1}{2}\sin 2\theta$, so $4\sin\theta\cos\theta = 2\sin 2\theta$.
• $\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}$, so $16\sin^2\theta = 8(1 - \cos 2\theta) = 8 - 8\cos 2\theta$.

Step 2. Combine:

$$A = 8 - 8\cos 2\theta + 2\sin 2\theta = 2\sin 2\theta - 8\cos 2\theta + 8$$

Step 3. Apply R-formula on $2\sin 2\theta - 8\cos 2\theta$: $R = \sqrt{2^2 + 8^2} = \sqrt{68} = 2\sqrt{17}$ $\tan\alpha = \dfrac{8}{2} = 4$, so $\alpha = \tan^{-1} 4 \approx 76.0°$.

$$A = 2\sqrt{17}\sin(2\theta - 76.0°) + 8$$

Step 4. Maximum: $\sin(2\theta - 76.0°) = 1$ at $2\theta - 76.0° = 90°$ → $\theta = 83.0°$.

$$A_{\max} = 8 + 2\sqrt{17} \approx 16.25$$

Tip: Whenever you see $\sin^2$ or $\sin\theta\cos\theta$, your first move is double-angle reduction — that brings you to a pure $a\sin 2\theta + b\cos 2\theta$ form where R-formula applies on argument $2\theta$.

Question: Using $R\cos(\theta - \alpha)$, solve $5\cos\theta + 8\sin\theta = 6$ for $0° \leq \theta \leq 180°$.

Step 1. For $a\cos\theta + b\sin\theta = R\cos(\theta - \alpha)$: $R = \sqrt{5^2 + 8^2} = \sqrt{89} \approx 9.43$ $\tan\alpha = \dfrac{b}{a} = \dfrac{8}{5}$, so $\alpha = \tan^{-1}\!\dfrac{8}{5} \approx 58.0°$.

So $5\cos\theta + 8\sin\theta = \sqrt{89}\cos(\theta - 58.0°)$.

Step 2. Solve $\sqrt{89}\cos(\theta - 58.0°) = 6$:

$$\cos(\theta - 58.0°) = \dfrac{6}{\sqrt{89}} \approx 0.6361$$ $$\theta - 58.0° = \pm\cos^{-1}(0.6361) \approx \pm 50.5°$$

Step 3. Solutions:

• $\theta = 58.0° + 50.5° = 108.5°$ ✓ (in range)
• $\theta = 58.0° - 50.5° = 7.5°$ ✓ (in range)

Answer: $\theta \approx 7.5°, 108.5°$.

Tip: For "$R\cos(\theta - \alpha)$" form, $\tan\alpha$ uses $\dfrac{\sin\text{ coeff}}{\cos\text{ coeff}}$ (note: the OPPOSITE of the $R\sin$ form).

Question: Find the minimum value of $\dfrac{1}{(3\cos\theta + 5\sin\theta)^2 + 1}$.

Step 1. Let $f(\theta) = 3\cos\theta + 5\sin\theta$. R-formula: $R = \sqrt{9 + 25} = \sqrt{34}$.

So $f(\theta) = \sqrt{34}\sin(\theta + \alpha)$ (or equivalent form), and $f(\theta)$ ranges over $[-\sqrt{34}, \sqrt{34}]$.

Step 2. $f(\theta)^2$ ranges over $[0, 34]$. So $(3\cos\theta + 5\sin\theta)^2 + 1$ ranges over $[1, 35]$.

Step 3. The reciprocal $\dfrac{1}{(3\cos\theta + 5\sin\theta)^2 + 1}$ ranges over $\left[\dfrac{1}{35}, 1\right]$.

Minimum $= \dfrac{1}{35}$, occurring when $(3\cos\theta + 5\sin\theta)^2 = 34$, i.e. when $f(\theta) = \pm\sqrt{34}$.

Tip: For composite extrema, find the range of the INSIDE first (using R-formula), then apply the outer transformation (squared, reciprocal, etc.) to those bounds. Track which input value gives which extremum.

Question: A perimeter is given by $P = 3\sin\theta + 4\cos\theta + 5$ for $0° < \theta < 90°$. Explain whether $P$ can equal $11$.

Step 1. R-formula on the trig part: $R = \sqrt{3^2 + 4^2} = 5$. So $3\sin\theta + 4\cos\theta = 5\sin(\theta + \alpha)$ where $\alpha = \tan^{-1}\!\dfrac{4}{3} \approx 53.1°$.

Step 2. $P = 5\sin(\theta + \alpha) + 5$ has maximum $5 + 5 = 10$ (when $\sin(\theta + \alpha) = 1$).

Step 3. Since $P_{\max} = 10 < 11$, the perimeter CANNOT reach $11$.

Answer: No, $P$ cannot equal $11$ because the maximum value of $3\sin\theta + 4\cos\theta$ is $\sqrt{3^2 + 4^2} = 5$, so $P \leq 10$.

Tip: "Can $X$ equal $k$?" → find max/min of $X$ via R-formula. If $k$ lies outside $[X_{\min}, X_{\max}]$, the answer is no.

Question: Express $h = 5\sin\theta + 3\sin(\theta - 60°)$ in the form $R\sin(\theta - \alpha)$.

Step 1. Expand $\sin(\theta - 60°) = \sin\theta\cos 60° - \cos\theta\sin 60° = \dfrac{1}{2}\sin\theta - \dfrac{\sqrt{3}}{2}\cos\theta$.

Step 2. Substitute:

$$h = 5\sin\theta + 3\left(\dfrac{1}{2}\sin\theta - \dfrac{\sqrt{3}}{2}\cos\theta\right) = \left(5 + \dfrac{3}{2}\right)\sin\theta - \dfrac{3\sqrt{3}}{2}\cos\theta$$ $$= \dfrac{13}{2}\sin\theta - \dfrac{3\sqrt{3}}{2}\cos\theta$$

Step 3. Apply R-formula:

$$R = \sqrt{\left(\tfrac{13}{2}\right)^2 + \left(\tfrac{3\sqrt{3}}{2}\right)^2} = \sqrt{\tfrac{169}{4} + \tfrac{27}{4}} = \sqrt{\tfrac{196}{4}} = 7$$ $$\tan\alpha = \dfrac{3\sqrt{3}/2}{13/2} = \dfrac{3\sqrt{3}}{13}$$ $$\alpha = \tan^{-1}\!\dfrac{3\sqrt{3}}{13} \approx 22.4°$$

Answer: $h = 7\sin(\theta - 22.4°)$.

Tip: When the question has $\sin(\theta + \text{fixed angle})$, ALWAYS expand first using compound-angle. Then collect like terms in $\sin\theta$ and $\cos\theta$, then apply R-formula.

Question: The sound-wave amplitude at a microphone is modelled by $A(t) = 8\sin(\omega t) + 6\cos(\omega t)$, in arbitrary units. The microphone's safe upper limit is $9$. Decide whether the signal ever exceeds the safe limit.

Step 1. Apply R-formula:

$$R = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$$

So $A(t) = 10\sin(\omega t + \alpha)$ for some phase $\alpha$.

Step 2. Maximum value of $A(t)$ is $R = 10$, which is greater than the safe limit of $9$.

Answer: Yes, the signal exceeds the safe limit. At its peak it reaches $10$, which is $1$ unit above the threshold.

Practical implication: The microphone is at risk of clipping or damage; an attenuator (e.g. $-1$ dB pad) would be needed to keep peaks under $9$ units.

Tip: Decision-context questions hinge on comparing the threshold to $R$ (the amplitude). The phase $\alpha$ doesn't affect whether the threshold is crossed — only WHEN it's crossed.