A-Level Edexcel Maths Pure Differentiation: 1. (a) Show that $$\frac{\sin 2

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

Question 2

$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.png$

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\si... show full transcript

Worked Solution & Example Answer: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

Step 1

Show that $$\frac{\sin 2\theta}{1+\cos 2\theta} = \tan \theta$$

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Answer

To show that

$\frac{\sin 2\theta}{1+\cos 2\theta} = \tan \theta$ ,

we start by using the double angle identities:

Recall that $\sin 2\theta = 2\sin \theta \cos \theta$
Also, $\cos 2\theta = 2\cos^2 \theta - 1$ Therefore, $1 + \cos 2\theta = 1 + (2\cos^2 \theta - 1) = 2\cos^2 \theta$ .
Substitute these into the fraction:

$\frac{\sin 2\theta}{1+\cos 2\theta} = \frac{2\sin \theta \cos \theta}{2\cos^2 \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta$ .

Thus, we have shown the required identity.

Step 2

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

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Answer

Starting from the previous result, we replace the fraction with its equivalent:

$\tan \theta = 1$ .

To find the angles where this is true, we can write:

$\theta = 45^\circ + k\cdot180^\circ,$ for integer values of $k$ .
Considering the interval $-180^\circ \leq \theta < 180^\circ$ $- 18 0^{\circ} \leq θ < 18 0^{\circ}$ :
- For $k = -1$ : $\theta = 45^\circ - 180^\circ = -135^\circ$ .
- For $k = 0$ : $\theta = 45^\circ$ .
- For $k = 1$ : $\theta = 45^\circ + 180^\circ = 225^\circ,$ (which is outside the interval)

Thus, the solutions are:

$\theta = -135^\circ, 45^\circ$ .

To convert these into decimal form (noting that $45^\circ = 45$ and $-135^\circ = -135$ ) gives us:

$\theta = -135.0$ to 1 decimal place.
$\theta = 45.0$ to 1 decimal place.

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

Show that $$\frac{\sin 2\theta}{1+\cos 2\theta} = \tan \theta$$

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

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