A-Level Edexcel Maths Pure Graphs of Functions: 7. (a) Prove that $$\frac{\sin

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) On the axes on page 20, sketch the graph of $$y = 2 \csc 2\theta$$ for $0^{\circ} < \theta < 360^{\circ}$ - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 5

Question 7

$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)-On-the-axes-on-page-20,-sketch-the-graph-of---$$y-=-2-\csc-2\theta$$---for-$0^{\circ}-<-\theta-<-360^{\circ}$-Edexcel-A-Level Maths Pure-Question 7-2007-Paper 5.png$

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 $$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin \theta}{\sin \theta} = 2 \csc 2\theta, \quad \theta \neq 90^{\circ}.$$ (b) On the axes on page 20, sketch the graph of $$y = 2 \csc 2\theta$$ for $0^{\circ} < \theta < 360^{\circ}$ - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 5

Step 1

Prove that $$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin \theta}{\sin \theta} = 2 \csc 2\theta, \quad \theta \neq 90^{\circ}.$$

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Answer

To prove the given equation, we start by simplifying the left-hand side:

Combine the terms:
$\frac{\sin \theta \cdot \cos \theta + \cos^2 \theta}{\cos^2 \theta}$ .
Now, replace $\sin 2\theta$ using the double angle identity:
$\sin 2\theta = 2\sin \theta \cos \theta.$
The left-hand side then becomes: $\frac{2\sin \theta \cos \theta}{\cos^2 \theta} = \frac{\sin 2\theta}{\cos^2 \theta}.$
Recognizing that $\csc 2\theta = \frac{1}{\sin 2\theta}$ , we rewrite the equation to reach the conclusion:
$= 2\csc 2\theta.$
Thus, we have proved the identity.

Step 2

On the axes on page 20, sketch the graph of $$y = 2 \csc 2\theta$$ for $0^{\circ} < \theta < 360^{\circ}$.

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Answer

When sketching the graph of $y = 2\csc(2\theta)$ :

Identify asymptotes where $\sin(2\theta) = 0$ , which occurs at multiples of $90^{\circ}$ , specifically at $90^{\circ}$ , $270^{\circ}$ , etc.
The function has maximum and minimum values of $2$ and $-2$ , respectively, at the points in between the asymptotes.
Ensure the sketch properly represents the periodic nature of the cosecant function, reflecting the values at calculated points.

Step 3

Solve, for $0^{\circ} < \theta < 360^{\circ}$, the equation $$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta \cdot \sin \theta} = 3,$$

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Answer

To solve the equation:

Simplify it to: $\sin \theta = 3 \cos^2 \theta.$
Substitute $\sin \theta$ with $\sqrt{1 - \cos^2 \theta}$ and rearrange.
This leads to: $\cos^2 \theta = \frac{1}{3}.$
Hence, calculate: $\theta = 180^{\circ} - \cos^{-1}(\sqrt{\frac{1}{3}})$ and
$\theta = 180^{\circ} + \cos^{-1}(\sqrt{\frac{1}{3}}).$
Solve these equations to find the two angles and round to one decimal place.

Prove that $$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin \theta}{\sin \theta} = 2 \csc 2\theta, \quad \theta \neq 90^{\circ}.$$

On the axes on page 20, sketch the graph of $$y = 2 \csc 2\theta$$ for $0^{\circ} < \theta < 360^{\circ}$.

Solve, for $0^{\circ} < \theta < 360^{\circ}$, the equation $$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta \cdot \sin \theta} = 3,$$

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