1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1

Prove that $$\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos 2\theta.$$ [4]

1. Find, in the form \(y = f(x)\), the general solution of the equation $$\frac{dy}{dx} = 2y \tan x + \sin 2x, \quad 0 < x < \frac{\pi}{2}$$ [6]

Given that \(y = 2\) at \(x = \frac{\pi}{6}\),

1. find the value of \(y\) at \(x = \frac{\pi}{4}\), giving your answer in the form \(a + k \ln b\), where \(a\) and \(b\) are integers and \(k\) is rational. [4]

Find the general solution of the differential equation $$\frac{dy}{dx} + 2y \cot 2x = \sin x, \quad 0 < x < \frac{\pi}{2},$$ giving your answer in the form \(y = f(x)\). [7]

\includegraphics{figure_1} A particle has mass \(2\) kg. It is attached at \(B\) to the ends of two light inextensible strings \(AB\) and \(BC\). When the particle hangs in equilibrium, \(AB\) makes an angle of \(30°\) with the vertical, as shown in Fig. 1. The magnitude of the tension in \(BC\) is twice the magnitude of the tension in \(AB\).

1. Find, in degrees to one decimal place, the size of the angle that \(BC\) makes with the vertical. [4]
2. Hence find, to 3 significant figures, the magnitude of the tension in \(AB\). [4]

\includegraphics{figure_1} A tennis ball \(P\) is attached to one end of a light inextensible string, the other end of the string being attached to a the top of a fixed vertical pole. A girl applies a horizontal force of magnitude 50 N to \(P\), and \(P\) is in equilibrium under gravity with the string making an angle of \(40°\) with the pole, as shown in Fig. 1. By modelling the ball as a particle find, to 3 significant figures,

1. the tension in the string, [3]
2. the weight of \(P\). [4]

A particle is projected from a point with speed \(u\) at an angle of elevation \(\alpha\) above the horizontal and moves freely under gravity. When it has moved a horizontal distance \(x\), its height above the point of projection is \(y\).

1. Show that $$y = x \tan \alpha - \frac{gx^2}{2u^2}(1 + \tan^2 \alpha).$$ [5]

A shot-putter puts a shot from a point \(A\) at a height of 2 m above horizontal ground. The shot is projected at an angle of elevation of 45° with a speed of 14 m s\(^{-1}\). By modelling the shot as a particle moving freely under gravity,

1. find, to 3 significant figures, the horizontal distance of the shot from \(A\) when the shot hits the ground, [5]
2. find, to 2 significant figures, the time taken by the shot in moving from \(A\) to reach the ground. [2]

Find the values of \(\theta\), to 1 decimal place, in the interval \(-180 \leq \theta < 180\) for which $$2 \sin^2 \theta° - 2 \sin \theta° = \cos^2 \theta°.$$ [8]

Given that \(2 \sin 2\theta = \cos 2\theta\),

1. show that \(\tan 2\theta = 0.5\). [1]
2. Hence find the values of \(\theta\), to one decimal place, in the interval \(0 \leq \theta < 360\) for which \(2 \sin 2\theta° = \cos 2\theta°\). [5]

Find, in degrees, the value of \(\theta\) in the interval \(0 \leq \theta < 360°\) for which $$2\cos^2 \theta - \cos \theta - 1 = \sin^2 \theta$$ Give your answers to 1 decimal place where appropriate. [8]

Given that \(2 \sin 2\theta = \cos 2\theta\),

1. show that \(\tan 2\theta = 0.5\). [1]
2. Hence find the values of \(\theta\), to one decimal place, in the interval \(0 \leq \theta < 360\) for which \(2 \sin 2\theta° = \cos 2\theta°\). [5]

Find, in degrees, the value of \(\theta\) in the interval \(0 \leq \theta < 360°\) for which $$2\cos^2\theta - \cos\theta - 1 = \sin^2\theta.$$ Give your answers to \(1\) decimal place where appropriate. [8]

You are given that \(\sin \theta = \frac{\sqrt{2}}{3}\) and that \(\theta\) is an acute angle. Find the exact value of \(\tan \theta\). [3]

\(\theta\) is an acute angle and \(\sin \theta = \frac{1}{4}\). Find the exact value of \(\tan \theta\). [3]

Showing your method, solve the equation \(2\sin^2\theta = \cos\theta + 2\) for values of \(\theta\) between \(0°\) and \(360°\). [5]

Showing your method clearly, solve the equation \(4 \sin^2 \theta = 3 + \cos^2 \theta\), for values of \(\theta\) between \(0°\) and \(360°\). [5]

Simplify \(\frac{\sqrt{1 - \cos^2 \theta}}{\tan \theta}\), where \(\theta\) is an acute angle. [3]

Find all values of \(x\) in the interval \(0 \leq x < 360°\) for which $$2\sin^2 x - 2\cos x - \cos^2 x = 1.$$ [8]