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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) On the axes on page 20, sketch the graph of $$y = 2 \csc 2\theta$$ for $$0^\circ < \theta < 360^\circ.$$ (c) Solve, for $$0^\circ < \theta < 360^\circ$$, the equation $$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} \cdot \frac{\sin \theta}{\sin \theta} = 3,$$ giving your answers to 1 decimal place. - Edexcel - A-Level Maths Pure - Question 1 - 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)-On-the-axes-on-page-20,-sketch-the-graph-of---$$y-=-2-\csc-2\theta$$-for-$$0^\circ-<-\theta-<-360^\circ.$$----(c)-Solve,-for-$$0^\circ-<-\theta-<-360^\circ$$,-the-equation---$$\frac{\sin-\theta-\cdot-\cos-\theta}{\cos^2-\theta}-\cdot-\frac{\sin-\theta}{\sin-\theta}-=-3,$$---giving-your-answers-to-1-decimal-place.-Edexcel-A-Level Maths Pure-Question 1-2007-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) On... 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 $$0^\circ < \theta < 360^\circ.$$ (c) Solve, for $$0^\circ < \theta < 360^\circ$$, the equation $$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} \cdot \frac{\sin \theta}{\sin \theta} = 3,$$ giving your answers to 1 decimal place. - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 5

Step 1

Prove that \( \frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin \theta}{\sin \theta} = 2 \csc 2\theta \)

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Answer

To prove the identity, we'll start by simplifying the left-hand side:

  1. Combine the fractions onto a common denominator: sinθcosθ+cos2θcos2θ\frac{\sin \theta \cdot \cos \theta + \cos^2 \theta}{\cos^2 \theta}

  2. Notice that we can express the numerator as: sinθcosθ+cos2θ=sinθcosθ+122cos2θ\sin \theta \cdot \cos \theta + \cos^2 \theta = \sin \theta \cdot \cos \theta + \frac{1}{2} \cdot 2 \cos^2 \theta

  3. Use the double angle sine identity: sin2θ=2sinθcosθsin2θ2=sinθcosθ\sin 2\theta = 2 \sin \theta \cos \theta \\ \Rightarrow \frac{\sin 2\theta}{2} = \sin \theta \cos \theta

Thus, our expression becomes: 1cos2θ(sinθcosθ+12(2)cos2θ)=2csc2θ\frac{1}{\cos^2 \theta} (\sin \theta \cdot \cos \theta + \frac{1}{2}(2)\cos^2 \theta) = 2 \csc 2\theta.

This proves the identity correctly.

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

To sketch the graph of ( y = 2 \csc 2\theta ):

  1. Identify vertical asymptotes: The function ( \csc , \theta ) is undefined where ( \sin 2\theta = 0 ). This occurs at: 2θ=nπ,nZθ=nπ22\theta = n\pi, n \in \mathbb{Z} \Rightarrow \theta = \frac{n\pi}{2} Hence, the vertical asymptotes are at ( \theta = 0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ ).

  2. Determine values of the function: The maximum and minimum points of the function occur when ( \sin 2\theta = 1 ) and ( \sin 2\theta = -1 ), giving maximum and minimum values of 2 and -2 respectively.

  3. Sketch the graph: Plot the points, asymptotes, and minimum/maximum values to create a repeated wave pattern with the specified range.

Step 3

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

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Answer

To solve the equation:

  1. The given equation simplifies to: sin2θ=3cos2θ\sin^2 \theta = 3 \cos^2 \theta

  2. Use the identity ( \sin^2 \theta + \cos^2 \theta = 1 ): Substitute ( \sin^2 \theta = 3(1 - \sin^2 \theta) ) leading to: 4sin2θ=34\sin^2 \theta = 3 sin2θ=34sinθ=±32\sin^2 \theta = \frac{3}{4} \Rightarrow \sin \theta = \pm\frac{\sqrt{3}}{2}

  3. Find angles: Using the principal values for ( \sin \theta = \frac{\sqrt{3}}{2} ) gives:

    • ( \theta = 60^\circ )
    • ( \theta = 120^\circ ) and for ( \sin \theta = -\frac{\sqrt{3}}{2} ):
    • ( \theta = 240^\circ )
    • ( \theta = 300^\circ )

Thus, the solutions in the range are approximately:

  • ( \theta = 60.0^\circ, 120.0^\circ, 240.0^\circ, 300.0^\circ )

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