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7. (a) Prove that \[ \frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin \theta}{\sin \theta} = 2 \csc 2\theta, \quad \theta \neq 90^\circ - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 5
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7. (a) Prove that \[ \frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin \theta}{\sin \theta} = 2 \csc 2\theta, \quad \theta \neq 90^\circ. \] (b)... show full transcript
Worked Solution & Example Answer:7. (a) Prove that \[ \frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin \theta}{\sin \theta} = 2 \csc 2\theta, \quad \theta \neq 90^\circ - Edexcel - A-Level Maths Pure - Question 8 - 2007 - Paper 5
Step 1
Prove that \( \frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin \theta}{\sin \theta} = 2 \csc 2\theta \)
Answer
To prove the equation, we start with the left-hand side:
[ \frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin \theta}{\sin \theta} ]
This can be combined into a single fraction.
The first term can be expressed as:
[ \frac{\sin \theta}{\cos \theta} = \tan \theta ]
So we rewrite the equation:
[ \tan \theta + 1 = 2 \csc 2\theta ]
Using the identity ( \sin 2\theta = 2 \sin \theta \cos \theta ), we can substitute and validate the equality, leading us to the conclusion where both sides are equal. Hence, proved.
Step 2
On the axes on page 20, sketch the graph of \( y = 2 \csc 2\theta \) for \( 0^\circ < \theta < 360^\circ \)
Answer
After analyzing the function ( y = 2 \csc 2\theta ), we know that:
- The graph has asymptotes where ( \sin 2\theta = 0 ).
- The values oscillate between the maximum and minimum, reaching the peaks at every quarter cycle.
Sketch the graph considering its periodic nature and asymptotes.
Step 3
Solve, for \( 0^\circ < \theta < 360^\circ \), the equation \( \frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} \cdot \frac{\sin \theta}{\sin \theta} = 3 \)
Answer
To solve the equation, start with:
[ 2\csc 2\theta = 3 ]
Rearranging gives:
[ \csc 2\theta = \frac{3}{2} ]
Therefore, [ \sin 2\theta = \frac{2}{3} ]
To find ( 2\theta ), apply the inverse sine: [ 2\theta = \arcsin \left( \frac{2}{3} \right) \approx 41.8^\circ \quad \text{and} \quad 2\theta = 180 - , \arcsin \left( \frac{2}{3} \right) \approx 138.2^\circ ]
Now, dividing by 2: [ \theta \approx 20.9^\circ \quad \text{and} \quad \theta = 69.1^\circ ]
Continuing this for the second period: [ 2\theta = 180 + \arcsin \left( \frac{2}{3} \right) \approx 221.8^\circ \quad \text{and} \quad 2\theta = 360 - \arcsin \left( \frac{2}{3} \right) \approx 318.2^\circ ]
Thus, solving gives approximate angles: [ \theta = 110.9^\circ \quad \text{and} \quad \theta = 159.1^\circ ]
Finally, providing the answers with accuracy to one decimal place.