A-Level Edexcel Maths Pure Differentiation: 6. (a) Prove that \[ \frac{1}{\

6. (a) Prove that \[ \frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta, \quad \theta \neq 90^\circ, n \in \mathbb{Z} \] (b) Hence, or otherwise, (i) show that $\tan 15^\circ = 2 - \sqrt{3}$ (ii) solve, for $0 < x < 360^\circ$, \[ \csc 4x - \cot 4x = 1 \] - Edexcel - A-Level Maths Pure - Question 7 - 2011 - Paper 3

Question 7

$6.-(a)-Prove-that--\[-\frac{1}{\sin-2\theta}-\cdot-\frac{\cos-2\theta}{\sin-2\theta}-=-\tan-\theta,-\quad-\theta-\neq-90^\circ,-n-\in-\mathbb{Z}-\]--(b)-Hence,-or-otherwise,--(i)-show-that-$\tan-15^\circ-=-2---\sqrt{3}$--(ii)-solve,-for-$0-<-x-<-360^\circ$,--\[-\csc-4x---\cot-4x-=-1-\]-Edexcel-A-Level Maths Pure-Question 7-2011-Paper 3.png$

6. (a) Prove that \[ \frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta, \quad \theta \neq 90^\circ, n \in \mathbb{Z} \] (b) Hence, or ot... show full transcript

Worked Solution & Example Answer:6. (a) Prove that \[ \frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta, \quad \theta \neq 90^\circ, n \in \mathbb{Z} \] (b) Hence, or otherwise, (i) show that $\tan 15^\circ = 2 - \sqrt{3}$ (ii) solve, for $0 < x < 360^\circ$, \[ \csc 4x - \cot 4x = 1 \] - Edexcel - A-Level Maths Pure - Question 7 - 2011 - Paper 3

Step 1

Prove that $\frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta$

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Answer

We start with the left-hand side:

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

Using the double angle identity for sine, we have:

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

Thus, substituting this into our expression gives:

[ \sin^2 2\theta = (2 \sin \theta \cos \theta)^2 = 4 \sin^2 \theta \cos^2 \theta ]

Now substituting this back in:

[ \frac{\cos 2\theta}{4 \sin^2 \theta \cos^2 \theta} = \frac{\cos 2\theta}{2 \sin^2 \theta \cos^2 \theta} \cdot \frac{1}{2} ]

Using the identity (\tan \theta = \frac{\sin \theta}{\cos \theta}), we rewrite the expression as:

[ \frac{\sin \theta}{\cos \theta} = \tan \theta ]

Hence, it is proven that: [ \frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta ]

Step 2

show that $\tan 15^\circ = 2 - \sqrt{3}$

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Answer

Using the tangent subtraction formula:

[ \tan(60^\circ - 45^\circ) = \frac{\tan 60^\circ - \tan 45^\circ}{1 + \tan 60^\circ \tan 45^\circ} ]

Substituting in the known values:

[ \tan 60^\circ = \sqrt{3}, \quad \tan 45^\circ = 1 ]

We get:

[ \tan 15^\circ = \frac{\sqrt{3} - 1}{1 + \sqrt{3} \cdot 1} = \frac{\sqrt{3} - 1}{1 + \sqrt{3}} ]

Rationalizing the denominator:

[ \tan 15^\circ = \frac{(\sqrt{3} - 1)(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{\sqrt{3} - 3 + 1}{1 - 3} = 2 - \sqrt{3} ]

Step 3

solve, for $0 < x < 360^\circ$, \csc 4x - \cot 4x = 1

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Answer

Start with the given equation:

[ \csc 4x - \cot 4x = 1 ]

Rearranging gives:

[ \csc 4x = 1 + \cot 4x ]

Substituting (\csc 4x = \frac{1}{\sin 4x}) and (\cot 4x = \frac{\cos 4x}{\sin 4x}) leads to:

[ \frac{1}{\sin 4x} = 1 + \frac{\cos 4x}{\sin 4x} \implies 1 = \sin 4x + \cos 4x ]

This implies:

[ \sin 4x + \cos 4x = 1 ]

Square both sides:

[\sin^2 4x + \cos^2 4x + 2\sin 4x \cos 4x = 1 \rightarrow 1 + \sin 8x = 1 \rightarrow \sin 8x = 0]

Thus, (4x = n\pi) gives possible solutions for:

[ 4x = 0, 180, 360, ...]

Solving for (x:)

[ x = 0, 45, 90, ...]

Limiting to (0 < x < 360) gives solutions: (x = 45^\circ, 90^\circ)

Prove that \(\frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta\)

show that \(\tan 15^\circ = 2 - \sqrt{3}\)

solve, for \(0 < x < 360^\circ\), \csc 4x - \cot 4x = 1

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