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In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 15 - 2020 - Paper 1

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In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. (a) Show that cosec θ − sin θ ≡ ... show full transcript

Worked Solution & Example Answer:In this question you must show all stages of your working - Edexcel - A-Level Maths Pure - Question 15 - 2020 - Paper 1

Step 1

Show that cosec θ − sin θ ≡ cos θ cot θ

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Answer

To show that cosec θ − sin θ ≡ cos θ cot θ, we start from the left-hand side:

  1. Rewrite cosec θ in terms of sin θ: cosecθ=1sinθcosec θ = \frac{1}{sin θ} Therefore, cosecθsinθ=1sinθsinθcosec θ - sin θ = \frac{1}{sin θ} - sin θ

  2. Combine the terms under a common denominator: cosecθsinθ=1sin2θsinθcosec θ - sin θ = \frac{1 - sin^{2} θ}{sin θ}

  3. Use the Pythagorean identity, where: 1sin2θ=cos2θ1 - sin^{2} θ = cos^{2} θ So, cosecθsinθ=cos2θsinθcosec θ - sin θ = \frac{cos^{2} θ}{sin θ}

  4. This can be factored as: cos2θsinθ=cosθcosθsinθ=cosθcotθ\frac{cos^{2} θ}{sin θ} = cos θ \cdot \frac{cos θ}{sin θ} = cos θ \, cot θ

Thus, we have shown that: cosecθsinθcosθcotθcosec θ - sin θ ≡ cos θ cot θ

  1. Finally, state that θ ≠ (180n)° for any integer n, as this condition is necessary to ensure that the expressions are defined.

Step 2

Hence, or otherwise, solve for 0 < x < 180°

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Answer

Starting from: cosecxsinx=cosxcot(3x50°)cosec x - sin x = cos x \cdot cot(3x - 50°)

  1. Rewrite cosec x in terms of sin x: 1sinxsinx=cosxcos(3x50°)sin(3x50°)\frac{1}{sin x} - sin x = cos x \cdot \frac{cos(3x - 50°)}{sin(3x - 50°)}

  2. Combine the left-hand side: 1sin2xsinx=cosxcot(3x50°)\frac{1 - sin^{2} x}{sin x} = cos x \cdot cot(3x - 50°)

  3. Using the identity, we have: cos2xsinx=cosxcot(3x50°)\frac{cos^{2} x}{sin x} = cos x \cdot cot(3x - 50°)

  4. Cancel cos x (assuming cos x ≠ 0): cosxsinx=cot(3x50°\frac{cos x}{sin x} = cot(3x - 50°

  5. Set equations equal: cotx=cot(3x50°)cot x = cot(3x - 50°) Implying: x=3x50°+kimes180°, kZx = 3x - 50° + k imes 180°,\ k \in Z

  6. Solve for x: a) First solution: x=3x50°    2x=50°    x=25°x = 3x - 50° \implies 2x = 50° \implies x = 25°

    b) Second solution: x=3x50°+180°    2x=230°    x=115°x = 3x - 50° + 180° \implies 2x = 230° \implies x = 115°

  7. Check that both solutions fall within the interval 0 < x < 180°:

    • Both 25° and 115° are valid solutions.

Thus, the solutions are: x=25°,x=115°x = 25°, x = 115°

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