Given that $y = ext{cosec } \theta$ (a)(i) Express $y$ in terms of $ ext{sin } \theta$. (a)(ii) Hence, prove that \[ \frac{dy}{d\theta} = -\text{cosec } \theta \cot \theta \] (a)(iii) Show that \[ \frac{\sqrt{y^2 - 1}}{y} = \cos \theta \quad \text{for } 0 2 \] can be written as $k \int \sin u \, du$ where $k$ is a constant to be found. (b)(ii) Hence, show \[ \int \frac{1}{\sqrt{2^2 - 4}} \, dx = \frac{\sqrt{x^2 - 4}}{4} + c \quad \text{for } x > 2 \] where $c$ is a constant.

A-Level AQA Maths Pure 8.1 Integration: Given that $y = ext{cosec } \t

Given that $y = ext{cosec } \theta$ (a)(i) Express $y$ in terms of $ ext{sin } \theta$ - AQA - A-Level Maths Pure - Question 15 - 2022 - Paper 1

Question 15

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Given that $y = ext{cosec } \theta$ (a)(i) Express $y$ in terms of $ ext{sin } \theta$. (a)(ii) Hence, prove that \[ \frac{dy}{d\theta} = -\text{cosec } \theta ... show full transcript

Worked Solution & Example Answer:Given that $y = ext{cosec } \theta$ (a)(i) Express $y$ in terms of $ ext{sin } \theta$ - AQA - A-Level Maths Pure - Question 15 - 2022 - Paper 1

Step 1

Express $y$ in terms of $ ext{sin } \theta$

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Answer

From the definition of cosecant:

$y = \text{cosec } \theta = \frac{1}{\sin \theta}$

Step 2

Hence, prove that $\frac{dy}{d\theta} = -\text{cosec } \theta \cot \theta$

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Answer

Using the derivative of $y$ :

Start with $y = \frac{1}{\sin \theta}$ .
Applying the quotient rule or chain rule for derivatives:

$\frac{dy}{d\theta} = -\frac{1}{\sin^2 \theta} \cdot \frac{d}{d\theta} \sin \theta$ $= -\frac{1}{\sin^2 \theta} \cdot \cos \theta$ $= -\text{cosec } \theta \cot \theta$

Step 3

Show that $\frac{\sqrt{y^2 - 1}}{y} = \cos \theta$ for $0 < \theta < \frac{\pi}{2}$

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Answer

Using the identity of cosecant:

From $y = \text{cosec } \theta$ , we know $y = \frac{1}{\sin \theta}$ , which gives: $y^2 = \frac{1}{\sin^2 \theta}$
Therefore: $y^2 - 1 = \frac{1}{\sin^2 \theta} - 1 = \frac{1 - \sin^2 \theta}{\sin^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta}$
Taking the square root: $\sqrt{y^2 - 1} = \frac{\cos \theta}{\sin \theta} = \cot \theta$
Substituting back gives: $\frac{\sqrt{y^2 - 1}}{y} = \frac{\cot \theta}{\text{cosec } \theta} = \cos \theta$

Step 4

Use the substitution $x = 2 \text{cosec } u$ to show that $\int \frac{1}{x^2 \sqrt{2^2 - 4}} \, dx$ can be written as $k \int \sin u \, du$

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Answer

Using the substitution:

Calculate: $dx = -2 \text{cosec } u \cot u \, du$
Substitute: $\int \frac{1}{(2 \text{cosec } u)^2 \sqrt{4 - 4}} (-2 \text{cosec } u \cot u) \, du$ This simplifies to: $\int \frac{-2 \text{cosec } u \cot u}{4 \text{cosec}^2 u} \, du = k \int \sin u \, du$ where $k$ must be determined.

Step 5

Hence, show that $\int \frac{1}{\sqrt{2^2 - 4}} \, dx = \frac{\sqrt{x^2 - 4}}{4} + c$

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Answer

From the previous results: $\int \sin u \, du = -\cos u$
Deducing $\cos u$ in terms of $x$ : $\cos u = \frac{\sqrt{x^2 - 4}}{2}$
Thus, integrating gives: $\int \frac{1}{\sqrt{2^2 - 4}} \, dx = \frac{\sqrt{x^2 - 4}}{4} + c$

Given that $y = ext{cosec } \theta$ (a)(i) Express $y$ in terms of $ ext{sin } \theta$ - AQA - A-Level Maths Pure - Question 15 - 2022 - Paper 1

Worked Solution & Example Answer:Given that $y = ext{cosec } \theta$ (a)(i) Express $y$ in terms of $ ext{sin } \theta$ - AQA - A-Level Maths Pure - Question 15 - 2022 - Paper 1

Express $y$ in terms of $ ext{sin } \theta$

Hence, prove that $\frac{dy}{d\theta} = -\text{cosec } \theta \cot \theta$

Show that $\frac{\sqrt{y^2 - 1}}{y} = \cos \theta$ for $0 < \theta < \frac{\pi}{2}$

Use the substitution $x = 2 \text{cosec } u$ to show that $\int \frac{1}{x^2 \sqrt{2^2 - 4}} \, dx$ can be written as $k \int \sin u \, du$

Hence, show that $\int \frac{1}{\sqrt{2^2 - 4}} \, dx = \frac{\sqrt{x^2 - 4}}{4} + c$

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