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6. (a) Express sin x + 2 cos x in the form R sin (x + a) where R and a are constants, R > 0 and 0 < a < \frac{\pi}{2} Give the exact value of R and give the value of a in radians to 3 decimal places - Edexcel - A-Level Maths Pure - Question 8 - 2020 - Paper 1

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6.-(a)-Express-sin-x-+-2-cos-x-in-the-form-R-sin-(x-+-a)-where-R-and-a-are-constants,-R->-0-and-0-<-a-<-\frac{\pi}{2}--Give-the-exact-value-of-R-and-give-the-value-of-a-in-radians-to-3-decimal-places-Edexcel-A-Level Maths Pure-Question 8-2020-Paper 1.png

6. (a) Express sin x + 2 cos x in the form R sin (x + a) where R and a are constants, R > 0 and 0 < a < \frac{\pi}{2} Give the exact value of R and give the value o... show full transcript

Worked Solution & Example Answer:6. (a) Express sin x + 2 cos x in the form R sin (x + a) where R and a are constants, R > 0 and 0 < a < \frac{\pi}{2} Give the exact value of R and give the value of a in radians to 3 decimal places - Edexcel - A-Level Maths Pure - Question 8 - 2020 - Paper 1

Step 1

Express sin x + 2 cos x in the form R sin (x + a)

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Answer

To express the function in the form R sin (x + a), we need to find R and a. We begin with the equation:

R=(1)2+(2)2=5R = \sqrt{(1)^2 + (2)^2} = \sqrt{5}

For the angle a, we use the relationship:

tan(a)=21    a=tan1(2)1.107 radians\tan(a) = \frac{2}{1} \implies a = \tan^{-1}(2) \approx 1.107 \text{ radians}

Thus, the values are: R=5,a1.107extradiansR = \sqrt{5}, \quad a \approx 1.107 ext{ radians}

Step 2

Deduce the maximum temperature of the room during this day

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Answer

Using the equation given for the temperature:

θ=5+5sin(π12+1.1073)\theta = 5 + \sqrt{5} \sin\left(\frac{\pi}{12} + 1.107 - 3\right)

The maximum temperature occurs at the maximum value of the sine function, which is 1. Therefore:

θmax=5+5(1)=5+57.236 degrees Celsius\theta_{max} = 5 + \sqrt{5}(1) = 5 + \sqrt{5} \approx 7.236 \text{ degrees Celsius}

Step 3

Find the time of day when the maximum temperature occurs

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Answer

To find the time 'i' when the maximum occurs, we set:

π12+1.1073=π2\frac{\pi}{12} + 1.107 - 3 = \frac{\pi}{2}

This gives:

π123=π21.107    i=π12+1.1073\frac{\pi}{12} - 3 = \frac{\pi}{2} - 1.107 \implies i = \frac{\pi}{12} + 1.107 - 3

After solving,

i13.2i \approx 13.2

Converting this to standard time gives:

  • Either 13:14 or 13 hours 14 minutes after midnight.

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