Trigonometry：練習問題

Find the sum of all solutions to the equation \(4\sin^2\theta = 5 - 8\cos\theta\) in the interval \(0^\circ \le \theta < 360^\circ\).

The expression \(6 \sin \theta - 8 \cos \theta\) is written in the form \(R \sin(\theta - \alpha)\), where \(R > 0\) and \(0^\circ < \alpha < 90^\circ\). Determine the maximum value of the expression and the smallest positive value of \(\theta\) (to one decimal place) at which this maximum occurs.

Find the sum of all solutions for the equation \(4\sin x - 3\cos x = 2\) in the interval \(0 \le x \le 2\pi\). Give your answer correct to 4 significant figures.

Solve the equation \(2 \sec^2 x = 5 \tan x\) for \(0 \le x < \pi\), giving your answers correct to 3 significant figures.

The function $f(\theta)$ is defined by $f(\theta) = 4\cos \theta - 3\sin \theta$.

Part (a)

Express $f(\theta)$ in the form $R\cos(\theta + \alpha)$, where $R > 0$ and $0^\circ < \alpha < 90^\circ$. Give the value of $\alpha$ correct to one decimal place.

Part (b)

Write down the minimum value of $f(\theta)$ and the smallest positive value of $\theta$ (in degrees, to one decimal place) for which this minimum occurs.

Part (c)

Solve the equation $4\cos \theta - 3\sin \theta = 2$ for $0^\circ \leq \theta < 360^\circ$. Give your answers to one decimal place.

The temperature \(T\) in a room, measured in degrees Celsius, is modelled by the equation \(T = 15 + 3\cos(2t) - 4\sin(2t)\), where \(t\) is the time in hours after midnight, and \(0 \leq t < 12\).

Part (a)

Express \(3\cos(2t) - 4\sin(2t)\) in the form \(R\cos(2t + \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{\pi}{2}\).

Part (b)

Hence, find the maximum temperature predicted by this model and the time \(t\) at which it first occurs in the interval \(0 \leq t < 12\).

Part (c)

Determine the total length of time, during the interval \(0 \leq t < 12\), for which the temperature \(T\) is above \(17^\circ \mathrm{C}\).