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f(x) = 7cos x + sin x Given that f(x) = Rcos(x - α), where R > 0 and 0 < α < 90°, a) find the exact value of R and the value of α to one decimal place - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 8

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f(x)-=-7cos-x-+-sin-x--Given-that-f(x)-=-Rcos(x---α),-where-R->-0-and-0-<-α-<-90°,--a)-find-the-exact-value-of-R-and-the-value-of-α-to-one-decimal-place-Edexcel-A-Level Maths Pure-Question 5-2013-Paper 8.png

f(x) = 7cos x + sin x Given that f(x) = Rcos(x - α), where R > 0 and 0 < α < 90°, a) find the exact value of R and the value of α to one decimal place. b) Hence s... show full transcript

Worked Solution & Example Answer:f(x) = 7cos x + sin x Given that f(x) = Rcos(x - α), where R > 0 and 0 < α < 90°, a) find the exact value of R and the value of α to one decimal place - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 8

Step 1

find the exact value of R and the value of α to one decimal place.

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Answer

To find the value of R, we use the formula:

R=(7)2+(1)2=49+1=50=52R = \sqrt{(7)^2 + (1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2}

Next, to determine α, we calculate:

tanα=17\tan α = \frac{1}{7}

This gives us:

α=arctan(17)8.1°α = \arctan\left( \frac{1}{7} \right) \approx 8.1°

Step 2

Hence solve the equation 7cos x + sin x = 5 for 0 ≤ x < 360°, giving your answers to one decimal place.

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Answer

Starting with the equation:

7cosx+sinx=57cos x + sin x = 5

We rearrange it as follows:

50cos(x8.1°)=5\sqrt{50}cos(x - 8.1°) = 5

This simplifies to:

cos(x8.1°)=550=552=12cos(x - 8.1°) = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}

Thus:

x8.1°=45° or 315°x - 8.1° = 45° \text{ or } 315°

Solving for x gives:

x=53.1° or 323.1°x = 53.1°\text{ or } 323.1°

Step 3

State the values of k for which the equation 7cos x + sin x = k has only one solution in the interval 0 ≤ x < 360°.

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Answer

For the equation to have only one solution, the value of k must equal the maximum or minimum value of the function:

k=±50k = \pm \sqrt{50}

Thus, the values of k are:

k=50 and k=50k = \sqrt{50} \text{ and } k = -\sqrt{50}

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