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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 8 - 2011 - Paper 3
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![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 8-2011-Paper 3.png](https://cdn.simplestudy.io/assets/backend/uploads/question/MathsPure_2011_3_1_QP_6_2011.jpg)
6. (a) Prove that \[ \frac{1}{\sin 2\theta} \cdot \frac{\cos 2\theta}{\sin 2\theta} = \tan \theta, \quad \theta \neq 90^\circ, \; n \in \mathbb{Z} \] (b) Hence, or... 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 8 - 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 the given equation, 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, we know that:
[ \sin 2\theta = 2 \sin \theta \cos \theta ]
Thus, substituting this into our equation:
[ \sin^2 2\theta = (2 \sin \theta \cos \theta)^2 = 4 \sin^2 \theta \cos^2 \theta ]
Now substituting this back into the left-hand side:
[ \frac{\cos 2\theta}{4 \sin^2 \theta \cos^2 \theta} = \frac{\cos 2\theta}{4} \cdot \frac{1}{\sin^2 \theta \cos^2 \theta} ]
Now we can simplify the right-hand side:
[ \tan \theta = \frac{\sin \theta}{\cos \theta} ]
This leads us to the required result of ( \tan \theta ).
Step 2
show that \( \tan 15^\circ = 2 - \sqrt{3} \)
Answer
To demonstrate that ( \tan 15^\circ = 2 - \sqrt{3} ), we can use the angle subtraction formula for tangent:
[ \tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} ]
Let ( A = 45^\circ ) and ( B = 30^\circ ):
[ \tan 45^\circ = 1 \quad \tan 30^\circ = \frac{1}{\sqrt{3}} ]
Substituting these values into the formula:
[ \tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} ]
Rationalizing the numerator:
[ \tan 15^\circ = \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{2} = 2 - \sqrt{3} ]
Step 3
solve, for \( 0 < x < 360^\circ \), \( \csc 4x - \cot 4x = 1 \)
Answer
To solve the equation ( \csc 4x - \cot 4x = 1 ), we start by writing it in sine and cosine:
[ \frac{1}{\sin 4x} - \frac{\cos 4x}{\sin 4x} = 1 ]
This simplifies to:
[ \frac{1 - \cos 4x}{\sin 4x} = 1 ]
Multiplying both sides by ( \sin 4x ):
[ 1 - \cos 4x = \sin 4x ]
Now, rearranging gives:
[ 1 = \sin 4x + \cos 4x ]
The maximum value of ( \sin 4x + \cos 4x ) occurs when both sine and cosine are equal, specifically:
[\sin 4x = \cos 4x]
Thus, setting ( 4x = 45^\circ + k \cdot 180^\circ ) for integer values of ( k ). Solving for ( x ):
[x = \frac{45 + k \cdot 180}{4}]
Finding values within ( 0 < x < 360 ):
- For ( k = 0 ) => ( x = 11.25^\circ )
- For ( k = 1 ) => ( x = 56.25^\circ )
- For ( k = 2 ) => ( x = 101.25^\circ )
- Increase until ( x < 360^\circ ) gives at least 2 valid solutions.