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
Question 2
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.
Calculate R:
R=sqrt32+42=sqrt9+16=sqrt25=5
Calculate tan(a):
tan(a)=34
Thus, using inverse tangent:
a=tan−1(34)≈53°
Now we can write:
3cosθ+4sinθ=5cos(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(θ−a)=1
This gives us:
θ−a=0Rightarrowθ=a=53°
Thus, the maximum value of 3 cos θ + 4 sin θ is:
5
And the smallest positive value of θ for which this maximum occurs is:
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)
To find the minimum, we consider the terms involving cosine and sine. The minimum occurs when:
cos(15t−α)=−1
where α is determined by:
tan(α)=34
Thus, the minimum temperature is:
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:
15t−α=180°
Therefore:
15t=180°+α
Solving for t:
t=15180°+53°=15233°=15.53
Thus, the minimum temperature occurs at:
t≈15.53 hours (from midday).