A-Level Edexcel Maths Pure Differentiation: 3. (a) Express $2 \, \cos \theta

3. (a) Express $2 \, \cos \theta - \sin \theta$ in the form $R \cos(\theta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90^\circ$ - Edexcel - A-Level Maths Pure - Question 5 - 2016 - Paper 3

Question 5

$3.-(a)-Express-$2-\,-\cos-\theta---\sin-\theta$-in-the-form-$R-\cos(\theta-+-\alpha)$,-where-$R$-and-$\alpha$-are-constants,-$R->-0$-and-$0-<-\alpha-<-90^\circ$-Edexcel-A-Level Maths Pure-Question 5-2016-Paper 3.png$

3. (a) Express $2 \, \cos \theta - \sin \theta$ in the form $R \cos(\theta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90^\circ$. Giv... show full transcript

Worked Solution & Example Answer:3. (a) Express $2 \, \cos \theta - \sin \theta$ in the form $R \cos(\theta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90^\circ$ - Edexcel - A-Level Maths Pure - Question 5 - 2016 - Paper 3

Step 1

Express $2 \cos \theta - \sin \theta$ in the form $R \cos(\theta + \alpha)$

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Answer

To express $2 \cos \theta - \sin \theta$ in the required form, we recognize that we can rewrite it using the cosine addition formula.

Let:

$R = \sqrt{(2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$
To find $\alpha$ , use: $\tan \alpha = \frac{-1}{2}$ Thus, we find: $\alpha = \tan^{-1}\left(-\frac{1}{2}\right)$ which gives approximately $\alpha = -26.57^\circ$ . Since we need $0 < \alpha < 90^\circ$ , we take: $\alpha = 26.57^\circ$ .

Step 2

Hence solve, for $0 \leq \theta < 360^\circ$, $\frac{2}{2 \cos \theta - \sin \theta - 1} = 15$

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Answer

Starting from the equation: $\frac{2}{2 \cos \theta - \sin \theta - 1} = 15$ Rearranging gives: $2 = 15(2 \cos \theta - \sin \theta - 1)$ Expanding yields: $15(2 \cos \theta - \sin \theta) - 15 = 2$ $30 \cos \theta - 15 \sin \theta = 17$ Substituting from part (a): $30 \cos \theta - \sqrt{5} \cos(\theta + 26.57^\circ) = 17$ This requires numerical methods or graphing techniques to solve within the range $0 \leq \theta < 360^\circ$ .

Step 3

Use your solutions to parts (a) and (b) to deduce the smallest positive value of $\theta$ for which $\frac{2}{2 \cos \theta + \sin \theta - 1} = 15$

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Answer

To solve the equation: $\frac{2}{2 \cos \theta + \sin \theta - 1} = 15$ Following similar steps as in part (b) results in: $2 = 15(2 \cos \theta + \sin \theta - 1)$ After rearranging and simplifying, we solve for $\theta$ using the previously found values from parts (a) and (b) to determine the smallest positive solution for $\theta$ .

3. (a) Express $2 \, \cos \theta - \sin \theta$ in the form $R \cos(\theta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90^\circ$ - Edexcel - A-Level Maths Pure - Question 5 - 2016 - Paper 3

Worked Solution & Example Answer:3. (a) Express $2 \, \cos \theta - \sin \theta$ in the form $R \cos(\theta + \alpha)$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha < 90^\circ$ - Edexcel - A-Level Maths Pure - Question 5 - 2016 - Paper 3

Express $2 \cos \theta - \sin \theta$ in the form $R \cos(\theta + \alpha)$

Hence solve, for $0 \leq \theta < 360^\circ$, $\frac{2}{2 \cos \theta - \sin \theta - 1} = 15$

Use your solutions to parts (a) and (b) to deduce the smallest positive value of $\theta$ for which $\frac{2}{2 \cos \theta + \sin \theta - 1} = 15$

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