1. (a) Show that \[\frac{\sin 2\theta}{1 + \cos 2\theta} = \tan \theta\] (b) Hence find, for \(-180^\circ \leq \theta < 180^\circ\), all the solutions of \[\frac{2\sin 2\theta}{1 + \cos 2\theta} = 1\] Give your answers to 1 decimal place. - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 5

To prove this identity, we start with the left-hand side:

[\frac{\sin 2\theta}{1 + \cos 2\theta}]

Using the double angle identity for sine, (\sin 2\theta = 2\sin \theta \cos \theta) and for cosine, (\cos 2\theta = 2\cos^2 \theta - 1), we substitute:

[\frac{2\sin \theta \cos \theta}{1 + (2\cos^2 \theta - 1)} = \frac{2\sin \theta \cos \theta}{2\cos^2 \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta]

Thus, the equation is confirmed.

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(b) Hence find, for (-180^\circ \leq \theta < 180^\circ), all the solutions of (\frac{2\sin 2\theta}{1 + \cos 2\theta} = 1)

Starting from the equation:

[\frac{2\sin 2\theta}{1 + \cos 2\theta} = 1]

Multiplying both sides by (1 + \cos 2\theta) gives:

[2\sin 2\theta = 1 + \cos 2\theta]

Rearranging yields:

[2\sin 2\theta - \cos 2\theta - 1 = 0]

Now, we utilize the double angle identity to express (2\sin 2\theta) as (\sin(2 \theta + 90^\circ)) or any applicable method. This could allow us to solve it algebraically or graphically. Solving this would involve finding the angles by using the sine or cosine inverses as needed within the specified range.

The critical points usually occur at interval checks: (\theta = 0, \pm 90^\circ, \pm 180^\circ). The final solutions should be calculated and rounded to 1 decimal place as per the instructions.