Express 6cosθ – 8sinθ in the form Rcos(θ + α). Hence or otherwise, solve the equation 6cosθ – 8sinθ = 2.5 for –180° ≤ θ ≤ 180°.

Here's the solution to Question 6:

Part (i): Express $6\cos\theta - 8\sin\theta$ in the form $R\cos(\theta + \alpha)$.

Step 1.1: Compare the given expression with the R-formula. We use the identity $R\cos(\theta + \alpha) = R(\cos\theta\cos\alpha - \sin\theta\sin\alpha) = (R\cos\alpha)\cos\theta - (R\sin\alpha)\sin\theta$. Comparing $6\cos\theta - 8\sin\theta$ with $(R\cos\alpha)\cos\theta - (R\sin\alpha)\sin\theta$, we get: $R\cos\alpha = 6 \quad (1)$ $R\sin\alpha = 8 \quad (2)$

Step 1.2: Calculate the value of $R$. Square equations (1) and (2) and add them: $(R\cos\alpha)^2 + (R\sin\alpha)^2 = 6^2 + 8^2$ $R^2(\cos^2\alpha + \sin^2\alpha) = 36 + 64$ $R^2(1) = 100$ $R = \sqrt{100} = 10 \quad (\text{since } R > 0)$

Step 1.3: Calculate the value of $\alpha$. Divide equation (2) by equation (1): $\frac{R\sin\alpha}{R\cos\alpha} = \frac{8}{6}$ $\tan\alpha = \frac{4}{3}$ $\alpha = \arctan\left(\frac{4}{3}\right) \approx 53.130^\circ$ Rounding to one decimal place, $\alpha \approx 53.1^\circ$.

Step 1.4: Write the expression in the required form. Therefore, $6\cos\theta - 8\sin\theta = \boxed{10\cos(\theta + 53.1^\circ)}$.

Part (ii): Hence or otherwise, solve the equation $6\cos\theta - 8\sin\theta = 2.5$ for $-180^\circ \le \theta \le 180^\circ$.

Step 2.1: Substitute the R-formula expression into the equation. Using the result from Part (i): $10\cos(\theta + 53.1^\circ) = 2.5$ $\cos(\theta + 53.1^\circ) = \frac{2.5}{10}$ $\cos(\theta + 53.1^\circ) = 0.25$

Step 2.2: Determine the range for the compound angle and find the principal value. Let $\phi = \theta + 53.1^\circ$. The given range for $\theta$ is $-180^\circ \le \theta \le 180^\circ$. The range for $\phi$ is $-180^\circ + 53.1^\circ \le \phi \le 180^\circ + 53.1^\circ$, which simplifies to $-126.9^\circ \le \phi \le 233.1^\circ$. The principal value for $\phi$ is $\arccos(0.25) \approx 75.522^\circ$. Rounding to one decimal place, $\phi_0 = 75.5^\circ$.

Step 2.3: Find all possible values of $\phi$ within the determined range. Since $\cos\phi$ is positive, $\phi$ lies in the first or fourth quadrant. The general solutions are $\phi = \pm \phi_0 + 360^\circ n$, where $n$ is an integer.

1. $\phi_1 = 75.5^\circ$ (This is within the range $[-126.9^\circ, 233.1^\circ]$).
2. $\phi_2 = -75.5^\circ$ (This is within the range $[-126.9^\circ, 233.1^\circ]$). Other values (e.g., $75.5^\circ + 360^\circ$ or $-75.5^\circ - 360^\circ$) fall outside the range.

Step 2.4: Solve for $\theta$. Case 1: $\theta + 53.1^\circ = 75.5^\circ$ $\theta = 75.5^\circ - 53.1^\circ$ $\theta = 22.4^\circ$ Case 2: $\theta + 53.1^\circ = -75.5^\circ$ $\theta = -75.5^\circ - 53.1^\circ$ $\theta = -128.6^\circ$ The solutions are $\boxed{\theta = 22.4^\circ, -128.6^\circ}$. (Note: The answers provided in the image for part (ii) correspond to the equation $6\cos\theta - 8\sin\theta = -2.5$, not $2.5$.)

Part (iii): State the minimum and the maximum values of $\frac{1}{6\cos\theta - 8\sin\theta + 13}$.

Step 3.1: Determine the range of the denominator's trigonometric part. From Part (i), we know that $6\cos\theta - 8\sin\theta = 10\cos(\theta + 53.1^\circ)$. The range of $\cos(\theta + 53.1^\circ)$ is $[-1, 1]$. Therefore, the range of $10\cos(\theta + 53.1^\circ)$ is $[-10, 10]$.

Step 3.2: Determine the range of the entire denominator. Let $X = 6\cos\theta - 8\sin\theta$. So, $-10 \le X \le 10$. The denominator of the expression is $X + 13$. The minimum value of the denominator is $-10 + 13 = 3$. The maximum value of the denominator is $10 + 13 = 23$. So, $3 \le X + 13 \le 23$.

Step 3.3: Calculate the minimum and maximum values of the fraction. To find the maximum value of the fraction $\frac{1}{X+13}$, the denominator must be at its minimum: $\text{Maximum value} = \frac{1}{3}$ To find the minimum value of the fraction $\frac{1}{X+13}$, the denominator must be at its maximum: $\text{Minimum value} = \frac{1}{23}$ The maximum value is $\boxed{\frac{1}{3}}$ and the minimum value is $\boxed{\frac{1}{23}}$.