Trigonometry: แบบฝึกหัด

A sector of a circle has a radius of \( 6 \) cm and a central angle of \( \frac{2\pi}{3} \) radians. Calculate the area of the sector in terms of \( \pi \).

Given that \( \tan \theta + \cot \theta = 4 \) and \( 0 < \theta < \frac{\pi}{4} \), find the exact value of \( \cos 2\theta \).

Find all values of \( x \) in the interval \( 0^\circ \le x < 360^\circ \) that satisfy the equation:
\( 2\cos^2 x - \sin x - 1 = 0 \)

In triangle \( ABC \), side \( AB = (x + 3) \) cm, side \( BC = (2x + 4) \) cm, and side \( AC = 7 \) cm. Given that angle \( \angle ABC = 60^\circ \), find the area of the triangle.

A sector of a circle has radius \( r \) cm and a central angle of \( \theta \) radians. The perimeter of the sector is fixed at \( 24 \) cm. Find the value of \( \theta \) that results in the maximum possible area for the sector.

In triangle \( ABC \), side \( AB = 8 \text{ cm} \), side \( BC = 11 \text{ cm} \) and angle \( \angle ABC = 150^\circ \). Calculate the area of the triangle, giving your answer in \( \text{cm}^2 \).

In triangle \( ABC \), \( AB = 5 \text{ cm} \), \( AC = 7 \text{ cm} \), and angle \( \angle ABC = 60^\circ \). Given that \( BC = x \text{ cm} \), find the value of \( x \) by forming and solving a quadratic equation.

In triangle \( ABC \), side \( a = 7 \text{ cm} \), side \( b = 5 \text{ cm} \) and the area of the triangle is \( 10\sqrt{3} \text{ cm}^2 \). Find the two possible exact lengths of side \( c \).
[6 marks]

(a) Express \(2\sin \theta + 5\cos \theta\) in the form \(R\sin(\theta + \alpha)\), where \(R > 0\) and \(0 < \alpha < 90^\circ\). Give the value of \(\alpha\) to 2 decimal places.
(b) Hence, solve the equation \(2\sin \theta + 5\cos \theta = 3\) for \(0 \le \theta < 360^\circ\), giving your answers to one decimal place.
(c) Write down the maximum value of \(2\sin \theta + 5\cos \theta\) and the smallest positive value of \(\theta\) at which this maximum occurs.

(a) Show that the equation \(3\cos 2\theta + \sin \theta = 2\) can be written in the form \(6\sin^2 \theta - \sin \theta - 1 = 0\).
(b) Hence solve, for \(0 \le \theta < 360^\circ\), the equation \(3\cos 2\theta + \sin \theta = 2\), giving your answers to one decimal place where appropriate.