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Parents Pricing Home A-Level Edexcel Maths Pure Basic Trigonometry 7. (a) Prove that
$$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\sin \theta} = 2 \csc 2\theta, \quad \theta \neq 90^\circ.$$
(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
$$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} \cdot \frac{\cos \theta}{\sin \theta} = 3,$$
giving your answers to 1 decimal place.
7. (a) Prove that
$$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\sin \theta} = 2 \csc 2\theta, \quad \theta \neq 90^\circ.$$
(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
$$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} \cdot \frac{\cos \theta}{\sin \theta} = 3,$$
giving your answers to 1 decimal place. - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 5 Question 1
View full question 7. (a) Prove that
$$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\sin \theta} = 2 \csc 2\theta, \quad \theta \neq 90^\circ.$$
(b) ... show full transcript
View marking scheme Worked Solution & Example Answer:7. (a) Prove that
$$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\sin \theta} = 2 \csc 2\theta, \quad \theta \neq 90^\circ.$$
(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
$$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} \cdot \frac{\cos \theta}{\sin \theta} = 3,$$
giving your answers to 1 decimal place. - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 5
Prove that $$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\sin \theta} = 2 \csc 2\theta$$ Only available for registered users.
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To prove the equation, start by rewriting the left-hand side:
Combine terms by finding a common denominator:
sin θ ⋅ cos θ + sin 2 θ cos 2 θ \frac{\sin \theta \cdot \cos \theta + \sin^2 \theta}{\cos^2 \theta} c o s 2 θ s i n θ ⋅ c o s θ + s i n 2 θ
Factor out sin θ \sin \theta sin θ from the numerator:
sin θ ( cos θ + sin θ ) cos 2 θ \frac{\sin \theta (\cos \theta + \sin \theta)}{\cos^2 \theta} c o s 2 θ s i n θ ( c o s θ + s i n θ )
Using the double angle identity, recall that sin 2 θ = 2 sin θ cos θ \sin 2\theta = 2\sin \theta \cos \theta sin 2 θ = 2 sin θ cos θ . Thus, we can express this as:
1 sin 2 θ \frac{1}{\sin 2\theta} s i n 2 θ 1
Conclude that this equals the right-hand side: 2 csc 2 θ 2 \csc 2\theta 2 csc 2 θ , thereby proving the equation.
On the axes on page 20, sketch the graph of $$y = 2 \csc 2\theta$$ for $$0^\circ < \theta < 360^\circ$$ Only available for registered users.
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To sketch the graph:
Identify the asymptotes where 2 θ = n π 2\theta = n\pi 2 θ = nπ for n ∈ Z n \in \mathbb{Z} n ∈ Z , which gives asymptotes at:
θ = 0 ∘ , 9 0 ∘ , 18 0 ∘ , 27 0 ∘ , 36 0 ∘ \theta = 0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ θ = 0 ∘ , 9 0 ∘ , 18 0 ∘ , 27 0 ∘ , 36 0 ∘ .
Mark the points where the function attains its maximum and minimum, which occur at θ = 4 5 ∘ , 13 5 ∘ , 22 5 ∘ , 31 5 ∘ \theta = 45^\circ, 135^\circ, 225^\circ, 315^\circ θ = 4 5 ∘ , 13 5 ∘ , 22 5 ∘ , 31 5 ∘ resulting in values of 2 and -2.
Plot the curve, noting the general shape of the cosecant function - it will have branches extending infinitely in both positive and negative directions.
Solve, for $$0^\circ < \theta < 360^\circ$$, the equation $$\frac{\sin \theta \cdot \cos \theta}{\cos^2 \theta} \cdot \frac{\cos \theta}{\sin \theta} = 3$$ Only available for registered users.
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To solve the equation:
Rearrange and simplify the equation:
2 csc 2 θ = 3 2\csc 2\theta = 3 2 csc 2 θ = 3 .
Therefore, csc 2 θ = 3 2 \csc 2\theta = \frac{3}{2} csc 2 θ = 2 3 implies:
sin 2 θ = 2 3 \sin 2\theta = \frac{2}{3} sin 2 θ = 3 2 .
Find the general solution for 2 θ 2\theta 2 θ :
2 θ = sin − 1 ( 2 3 ) , 18 0 ∘ − sin − 1 ( 2 3 ) 2\theta = \sin^{-1}\left(\frac{2}{3}\right), 180^\circ - \sin^{-1}\left(\frac{2}{3}\right) 2 θ = sin − 1 ( 3 2 ) , 18 0 ∘ − sin − 1 ( 3 2 )
which gives:
2 θ ≈ 41.8 1 ∘ , 138.1 9 ∘ , 401.8 1 ∘ , 498.1 9 ∘ 2\theta \approx 41.81^\circ, 138.19^\circ, 401.81^\circ, 498.19^\circ 2 θ ≈ 41.8 1 ∘ , 138.1 9 ∘ , 401.8 1 ∘ , 498.1 9 ∘ .
Hence, for θ \theta θ , divide each result by 2:
θ ≈ 20.9 0 ∘ , 69.1 0 ∘ , 200.9 0 ∘ , 249.1 0 ∘ . \theta \approx 20.90^\circ, 69.10^\circ, 200.90^\circ, 249.10^\circ. θ ≈ 20.9 0 ∘ , 69.1 0 ∘ , 200.9 0 ∘ , 249.1 0 ∘ .
Format the answers to 1 decimal place as:
20. 9 ∘ 20.9^\circ 20. 9 ∘ , 69. 1 ∘ 69.1^\circ 69. 1 ∘ , 200. 9 ∘ 200.9^\circ 200. 9 ∘ , 249. 1 ∘ 249.1^\circ 249. 1 ∘ .
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