A-Level Edexcel Maths Pure Trigonometric Equations: 12. (a) Prove $$\frac{\cos 3\th

12. (a) Prove $$\frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 2\cot 2\theta$$ $\theta \pm (90n)^{\circ}, n \in \mathbb{Z}$ (4) (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 12 - 2019 - Paper 2

Question 12

$12.-(a)-Prove--$$\frac{\cos-3\theta}{\sin-\theta}-+-\frac{\sin-3\theta}{\cos-\theta}-=-2\cot-2\theta$$--$\theta-\pm-(90n)^{\circ},-n-\in-\mathbb{Z}$-(4)--(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 12-2019-Paper 2.png$

12. (a) Prove $$\frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 2\cot 2\theta$$ $\theta \pm (90n)^{\circ}, n \in \mathbb{Z}$ (4) (b) Hence... show full transcript

Worked Solution & Example Answer:12. (a) Prove $$\frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 2\cot 2\theta$$ $\theta \pm (90n)^{\circ}, n \in \mathbb{Z}$ (4) (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 12 - 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 identity, we start with the left-hand side:

$\frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta}$

We can find a common denominator:

$= \frac{\cos 3\theta \cos \theta + \sin 3\theta \sin \theta}{\sin \theta \cos \theta}$

Using the cosine addition formula, this simplifies to:

$= \frac{\cos(3\theta - \theta)}{\sin \theta \cos \theta}$

Then, using the double angle identity for cotangent, we have:

$= \frac{\cos 2\theta}{\sin \theta \cos \theta}$

Identifying this with the right-hand side, we recognize that:

$= 2\cot 2\theta$

Thus, proving the identity is complete.

Step 2

Hence solve, for $90^{\circ} < \theta < 180^{\circ}$, the equation $$\frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 4$$

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Answer

Starting from the earlier rearrangement:

$\frac{\cos 3\theta \sin 3\theta}{\sin \theta \cos \theta} = 4$

This simplifies to:

$\sin 3\theta = 4 \frac{\sin \theta \cos \theta}{\cos 3\theta}$

Next, we rearrange this equation to isolate (\tan 3\theta):

$\tan 3\theta = 4\tan \theta$

Using the arctangent function:

$3\theta = \arctan(4 \tan \theta)$

This gives us:

\arctan(4) + 180n}{3}, n \in \mathbb{Z}$$ Calculating the principal solution leads us to:\n \[ \theta \approx 103.3^{\circ} \] \nGiving only this solution for \(90^{\circ} < \theta < 180^{\circ}\).

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

Hence solve, for \(90^{\circ} < \theta < 180^{\circ}\), the equation $$\frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 4$$

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