A-Level Edexcel Maths Pure Differentiation: 12. (a) Prove \[ \frac{\cos 3\t

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° < θ < 180°, the equation \[ \frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 4 \] - 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°-<-θ-<-180°,-the-equation--\[-\frac{\cos-3\theta}{\sin-\theta}-+-\frac{\sin-3\theta}{\cos-\theta}-=-4-\]-Edexcel-A-Level Maths Pure-Question 13-2019-Paper 2.png$

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° < θ < 180°, the equation \[ \frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 4 \] - Edexcel - A-Level Maths Pure - Question 13 - 2019 - Paper 2

Step 1

Prove \[ \frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 2 \cot 2\theta \]

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Answer

To prove the given equation, start by rewriting the left-hand side (LHS):

[ \frac{\cos 3\theta \cos \theta + \sin 3\theta \sin \theta}{\sin \theta \cos \theta} = \frac{\cos(3\theta - \theta)}{\sin \theta \cos \theta} ]

Using the identity for the cotangent:

[ \cot 2\theta = \frac{\cos 2\theta}{\sin 2\theta} \text{ and thus} \sin 2\theta = 2 \sin \theta \cos \theta ]

This means we can express the LHS as:

[ \frac{\cos(3\theta - \theta)}{\sin \theta \cos \theta} = 2 \cot 2\theta ]

This confirms the given equation is true.

Step 2

Hence solve, for 90° < θ < 180°, the equation \[ \frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 4 \]

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Answer

Starting with the equation:

[ \frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 4 ]

We can combine the fractions:

[ \frac{\cos 3\theta \cos \theta + \sin 3\theta \sin \theta}{\sin \theta \cos \theta} = 4 ]

This simplifies to:

[ \frac{\cos(3\theta - \theta)}{\sin \theta \cos \theta} = 4 ]

Solving for ( \theta ):

Cross-multiply to obtain: [ \cos 3\theta = 4 \sin \theta \cos \theta ]
Use ( \sin 2\theta = 2\sin \theta \cos \theta ) to yield: [ \cos 3\theta = 2 \sin 2\theta ]
Use the arctan function: [ \tan 2\theta = \frac{1}{2} ]
Therefore, (2\theta = an^{-1}(0.5) \Rightarrow 2\theta \approx 26.6°)
Finally, we have ( \theta \approx 13.3°), but we check the range, which only allows for: [ \theta = 103.3°\text{ (to one decimal place)} ]

Prove \[ \frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 2 \cot 2\theta \]

Hence solve, for 90° < θ < 180°, the equation \[ \frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 4 \]

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