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Figure 1 shows the curve C, with equation $y = 6 \, ext{cos} \, x + 2.5 \, ext{sin} \, x$ for $0 \leq x \leq 2\pi$ - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 7

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Figure-1-shows-the-curve-C,-with-equation-$y-=-6-\,--ext{cos}-\,-x-+-2.5-\,--ext{sin}-\,-x$-for-$0-\leq-x-\leq-2\pi$-Edexcel-A-Level Maths Pure-Question 2-2013-Paper 7.png

Figure 1 shows the curve C, with equation $y = 6 \, ext{cos} \, x + 2.5 \, ext{sin} \, x$ for $0 \leq x \leq 2\pi$. (a) Express $6 \, ext{cos} \, x + 2.5 \, ext... show full transcript

Worked Solution & Example Answer:Figure 1 shows the curve C, with equation $y = 6 \, ext{cos} \, x + 2.5 \, ext{sin} \, x$ for $0 \leq x \leq 2\pi$ - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 7

Step 1

Express $6 \, \text{cos} \, x + 2.5 \, \text{sin} \, x$ in the form $R \, \text{cos}(x - \alpha)$

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Answer

To express the equation in the required form, we first calculate the resultant amplitude RR using:

R=(62+2.52)=36+6.25=42.25≈6.50R = \sqrt{(6^2 + 2.5^2)} = \sqrt{36 + 6.25} = \sqrt{42.25} \approx 6.50

Next, we find the angle α\alpha using:

tan⁡(α)=2.56\tan(\alpha) = \frac{2.5}{6}

This yields:

α=tan⁡−1(2.56)≈0.396\alpha = \tan^{-1}\left(\frac{2.5}{6}\right) \approx 0.396

Thus, we have R≈6.500R \approx 6.500 and α≈0.395\alpha \approx 0.395.

Step 2

Find the coordinates of the points on the graph where the curve C crosses the coordinate axes

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Answer

  1. To find where the curve crosses the x-axis (where y=0y = 0):

    6cos⁡(x)+2.5sin⁡(x)=06 \cos(x) + 2.5 \sin(x) = 0

    Solving for xx gives two points approximately x≈1.97x \approx 1.97 and x≈5.11x \approx 5.11.

    So, the points are (1.97,0)(1.97, 0) and (5.11,0)(5.11, 0).

  2. To find where the curve crosses the y-axis (substituting x=0x = 0):

    y=6cos⁡(0)+2.5sin⁡(0)=6y = 6 \cos(0) + 2.5 \sin(0) = 6

    So, the point is (0,6)(0, 6).

Step 3

Use this function to find the maximum and minimum values of $H$ predicted by the model

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Answer

The function for hours of daylight is given by:

H=12+6cos⁡(2πt52)+2.5sin⁡(2πt52)H = 12 + 6 \cos\left(\frac{2\pi t}{52}\right) + 2.5 \sin\left(\frac{2\pi t}{52}\right)

To find the maximum HmaxH_{max} and minimum HminH_{min}, we need to find the maximum and minimum of the cosine and sine components, which are:

  • The maximum of 6cos⁡(.)6\cos(.) is 6 and the maximum of 2.5sin⁡(.)2.5\sin(.) is 2.5.

  • Thus:
    Hmax=12+6+2.5=20.5H_{max} = 12 + 6 + 2.5 = 20.5

  • The minimum of 6cos⁡(.)6\cos(.) is -6 and the minimum of 2.5sin⁡(.)2.5\sin(.) is -2.5.

  • Thus:
    Hmin=12−6−2.5=3.5H_{min} = 12 - 6 - 2.5 = 3.5

Step 4

Find the value of $t$ when $H = 16$

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Answer

To find when H=16H = 16, we substitute into the function:

16=12+6cos⁡(2πt52)+2.5sin⁡(2πt52)16 = 12 + 6 \cos\left(\frac{2\pi t}{52}\right) + 2.5 \sin\left(\frac{2\pi t}{52}\right)

Rearranging gives: 4=6cos⁡(2πt52)+2.5sin⁡(2πt52)4 = 6 \cos\left(\frac{2\pi t}{52}\right) + 2.5 \sin\left(\frac{2\pi t}{52}\right)

This equation can be solved either graphically or numerically. After solving, we find a value of tt approximately equal to 4 weeks, providing: t≈4t \approx 4

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