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12. (a) Prove \[ \frac{\cos 3\theta}{\sin\theta} + \frac{\sin 3\theta}{\cos\theta} = 2 \cot 2\theta \] \( \theta \pm (90n)^{\circ}, n \in \mathbb{Z} \) (b) Hence solve, for \( 90^{\circ} < \theta < 180^{\circ} \), the equation \[ \frac{\cos 3\theta}{\sin\theta} + \frac{\sin 3\theta}{\cos\theta} = 4 \] giving any solutions to one decimal place. - Edexcel - A-Level Maths Pure - Question 13 - 2019 - Paper 2
Question 13
![12.-(a)-Prove--\[-\frac{\cos-3\theta}{\sin\theta}-+-\frac{\sin-3\theta}{\cos\theta}-=-2-\cot-2\theta-\]--\(-\theta-\pm-(90n)^{\circ},-n-\in-\mathbb{Z}-\)--(b)-Hence-solve,-for-\(-90^{\circ}-<-\theta-<-180^{\circ}-\),-the-equation--\[-\frac{\cos-3\theta}{\sin\theta}-+-\frac{\sin-3\theta}{\cos\theta}-=-4-\]--giving-any-solutions-to-one-decimal-place.-Edexcel-A-Level Maths Pure-Question 13-2019-Paper 2.png](https://cdn.simplestudy.io/assets/backend/uploads/question/MathsPure_2019_2_1_QP_12_2019.jpg)
12. (a) Prove \[ \frac{\cos 3\theta}{\sin\theta} + \frac{\sin 3\theta}{\cos\theta} = 2 \cot 2\theta \] \( \theta \pm (90n)^{\circ}, n \in \mathbb{Z} \) (b) Hence ... show full transcript
Worked Solution & Example Answer:12. (a) Prove \[ \frac{\cos 3\theta}{\sin\theta} + \frac{\sin 3\theta}{\cos\theta} = 2 \cot 2\theta \] \( \theta \pm (90n)^{\circ}, n \in \mathbb{Z} \) (b) Hence solve, for \( 90^{\circ} < \theta < 180^{\circ} \), the equation \[ \frac{\cos 3\theta}{\sin\theta} + \frac{\sin 3\theta}{\cos\theta} = 4 \] giving any solutions to one decimal place. - Edexcel - A-Level Maths Pure - Question 13 - 2019 - Paper 2
Step 1
Prove \( \frac{\cos 3\theta}{\sin\theta} + \frac{\sin 3\theta}{\cos\theta} = 2 \cot 2\theta \)
Answer
To prove the identity, we start with the left-hand side:
[ \frac{\cos 3\theta}{\sin\theta} + \frac{\sin 3\theta}{\cos\theta} = \frac{\cos 3\theta \cos\theta + \sin 3\theta \sin\theta}{\sin\theta \cos\theta} ]
Using the angle addition formula, we find that:
[ \cos 3\theta = 2\cos^2 1.5\theta - 1 \quad \text{and} \quad \sin 3\theta = 3\sin\theta - 4\sin^3\theta ]
Substituting these into our equation leads to:
[ \cos(3\theta) \cos(\theta) + \sin(3\theta) \sin(\theta) = \cos(3\theta - \theta) = \cos(2\theta) ]
Thus, the left-hand side simplifies to:
[ \frac{\cos(2\theta)}{\sin\theta \cos\theta} = \frac{1}{\sin 2\theta} = 2 \cot 2\theta ]
Hence proved.
Step 2
Hence solve, for \( 90^{\circ} < \theta < 180^{\circ} \), the equation \( \frac{\cos 3\theta}{\sin\theta} + \frac{\sin 3\theta}{\cos\theta} = 4 \)
Answer
Starting with the proven equation:
[ \frac{\cos 3\theta}{\sin\theta} + \frac{\sin 3\theta}{\cos\theta} = 4 ]
This can be rewritten as:
[ \cos 3\theta \cos\theta + \sin 3\theta \sin\theta = 4 \sin\theta \cos\theta ]
Using the identity from part (a), we replace the left side:
[ 2 \cot 2\theta = 4 ]
This implies:
[ \cot 2\theta = 2 ]
Then, we take the arctangent:
[ 2\theta = \arctan(2) ]
So, we can calculate:
[ \theta = \frac{1}{2} \arctan(2) \approx 103.3^{\circ} ]
Given the constraints of ( 90^{\circ} < \theta < 180^{\circ} ), this is the only solution.