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6. (a) Prove that \[ \frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta, \ \theta \neq 90^\circ, \ n \in \mathbb{Z} \] (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 6 - 2011 - Paper 3
Question 6
![6.-(a)-Prove-that--\[-\frac{1}{\sin-2\theta}-\cdot-\frac{\cos-2\theta}{\sin-2\theta}-=-\tan-\theta,-\-\theta-\neq-90^\circ,-\-n-\in-\mathbb{Z}-\]--(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 6-2011-Paper 3.png](https://cdn.simplestudy.cloud/assets/backend/uploads/question/MathsPure_2011_3_1_QP_6_2011.jpg)
6. (a) Prove that \[ \frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta, \ \theta \neq 90^\circ, \ n \in \mathbb{Z} \] (b) Hence, or othe... show full transcript
Worked Solution & Example Answer:6. (a) Prove that \[ \frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta, \ \theta \neq 90^\circ, \ n \in \mathbb{Z} \] (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 6 - 2011 - Paper 3
Step 1
Prove that \[ \frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta \]
Answer
To prove that
[ \frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta, ] we start with the left side:
[ \frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \frac{\cos 2\theta}{\sin^2 2\theta}. ] Now, using the double angle formula for sine, we have:
[ \sin 2\theta = 2 \sin \theta \cos \theta. ] Thus, substituting this into our expression:
[ \frac{\cos 2\theta}{\sin^2 2\theta} = \frac{\cos 2\theta}{(2 \sin \theta \cos \theta)^2} = \frac{\cos 2\theta}{4 \sin^2 \theta \cos^2 \theta}. ] Utilizing the identity ( \tan \theta = \frac{\sin \theta}{\cos \theta} ) and rearranging:
[ = \frac{\sin \theta}{\cos \theta} = \tan \theta. ] Thus, we conclude
[ \frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta. ]
Step 2
show that tan 15° = 2 - √3
Answer
To show that ( \tan 15^\circ = 2 - \sqrt{3} ), we can use the tangent subtraction formula:
[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} ] Here, we take ( a = 45^\circ ) and ( b = 30^\circ ):
[ \tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}}. ] Simplifying this expression:
[ = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = (\sqrt{3} - 1)(\frac{\sqrt{3} - 1}{\sqrt{3} - 1}) = \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}. ] Thus, we have shown that ( \tan 15^\circ = 2 - \sqrt{3}. )
Step 3
solve, for 0 < x < 360°, cosec 4x - cot 4x = 1
Answer
To solve ( \csc 4x - \cot 4x = 1 ), we can express them in terms of sine and cosine:
[ \frac{1}{\sin 4x} - \frac{\cos 4x}{\sin 4x} = 1. ] This leads to:
[ \frac{1 - \cos 4x}{\sin 4x} = 1. ] Cross multiplying gives:
[ 1 - \cos 4x = \sin 4x. ] Utilizing ( \sin^2 4x + \cos^2 4x = 1 ), we can substitute ( \cos 4x = 1 - \sin^2 4x ):
[ 1 - \cos 4x = 1 - (1 - \sin^2 4x) \Rightarrow \sin^2 4x = 1 - \sin 4x. ] This results in a quadratic equation:
[ \sin^2 4x + \sin 4x - 1 = 0. ] Using the quadratic formula ( \sin 4x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ):
Where ( a = 1, b = 1, c = -1 ):
[ \sin 4x = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}. ] Thus:
- If ( \sin 4x = \frac{-1 + \sqrt{5}}{2} ) (valid solution in the range),
- Setting ( 4x = n\pi + (-1)^n \arcsin \left( \frac{-1 + \sqrt{5}}{2} \right) )
- For solutions in ( 0 < x < 360^\circ ), we must evaluate:
[x = \frac{1}{4} \left( n\pi + (-1)^n \arcsin \left( \frac{-1 + \sqrt{5}}{2} \right) \right). ] Calculate for specific integers to find valid x-values.