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8. (a) Express 3 cos θ + 4 sin θ in the form R cos(θ − a), where R and a are constants, R > 0 and 0 < a < 90° - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 5

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8.-(a)-Express-3-cos-θ-+-4-sin-θ-in-the-form-R-cos(θ-−-a),-where-R-and-a-are-constants,-R->-0-and-0-<-a-<-90°-Edexcel-A-Level Maths Pure-Question 2-2008-Paper 5.png

8. (a) Express 3 cos θ + 4 sin θ in the form R cos(θ − a), where R and a are constants, R > 0 and 0 < a < 90°. (b) Hence find the maximum value of 3 cos θ + 4 sin θ... show full transcript

Worked Solution & Example Answer:8. (a) Express 3 cos θ + 4 sin θ in the form R cos(θ − a), where R and a are constants, R > 0 and 0 < a < 90° - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 5

Step 1

Express 3 cos θ + 4 sin θ in the form R cos(θ − a)

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Answer

To express the given equation in the form R cos(θ − a), we first need to find R and a.

  1. Calculate R: R=sqrt32+42=sqrt9+16=sqrt25=5R = \\sqrt{3^2 + 4^2} = \\sqrt{9 + 16} = \\sqrt{25} = 5

  2. Calculate tan(a): tan(a)=43\tan(a) = \frac{4}{3} Thus, using inverse tangent: a=tan1(43)53°a = \\tan^{-1}(\frac{4}{3}) ≈ 53°

Now we can write: 3cosθ+4sinθ=5cos(θ53°)3 \cos θ + 4 \sin θ = 5 \cos(θ − 53°)

Step 2

Hence find the maximum value of 3 cos θ + 4 sin θ and the smallest positive value of θ for which this maximum occurs

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Answer

The maximum value of the expression occurs when: cos(θa)=1\cos(θ - a) = 1

This gives us: θa=0Rightarrowθ=a=53°θ - a = 0 \\Rightarrow θ = a = 53°

Thus, the maximum value of 3 cos θ + 4 sin θ is: 55

And the smallest positive value of θ for which this maximum occurs is: 53°53°

Step 3

Calculate the minimum temperature of the warehouse as given by this model

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Answer

We need to find the minimum temperature using the given function:

f(t)=10+3cos(15t)+4sin(15t)f(t) = 10 + 3 \cos(15t) + 4 \sin(15t)

To find the minimum, we consider the terms involving cosine and sine. The minimum occurs when: cos(15tα)=1\cos(15t - α) = -1 where α is determined by: tan(α)=43tan(α) = \frac{4}{3} Thus, the minimum temperature is: f(t)=10+3(1)=103=7°f(t) = 10 + 3(-1) = 10 - 3 = 7°

Step 4

Find the value of t when this minimum temperature occurs

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Answer

From the previous calculation, we know: 15tα=180°15t - α = 180° Therefore: 15t=180°+α15t = 180° + α Solving for t: t=180°+53°15=233°15=15.53t = \frac{180° + 53°}{15} = \frac{233°}{15} = 15.53

Thus, the minimum temperature occurs at: t15.53t ≈ 15.53 hours (from midday).

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