Question

The average daily high temperature in $^{\circ}C$ in Karachi, Pakistan, on the $t^{\text{th}}$ day of the year is approximately

$\qquad T(t) = 29 - 5\cos\left(\dfrac{2\pi (t- 12)}{365}\right)$

What are the highest and lowest average daily highs in Karachi? Give exact answers.

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Please explain it "thoroughly" in 'sentences' and with mathematical expression how exactly you'd figure out the answer. I know the answer. I need an elaborate explanation of the method. It's an imperative part of the question to explain your answer otherwise your answer would be considered incomplete

Giving 25 points, thanks (both in your answer and profile)!5 stars and of course brainliest!!!

Answer 1

Given expression

$\\ \rm\Rrightarrow T(t)=29-5cos\left(\cfrac{2\pi (t-12)}{365}\right)$

Now

We know according to property of cosine function

• It has range of [-1,1] for all intervals .

This property is enough for our calculations

Let's check

As t represents no of days hence it can't be negative

• t must be a whole number set member

So

for positive interval

• We shall use the cosine interval [0,π]

We know that

• cosπ is -1
• cos0=1

Hence it's enough for our range [-1,1]

Now

In the expression

• 29 is the vertical shift

Let

$\sf cos\left(\cfrac{2\pi(t-12)}{365}\right)=n$

Substitute n for lowest and highest value of range

When n=-1

$\\ \rm\Rrightarrow T(\pi)=29-5(-1)=29+5=34°C$

When n=1

$\\ \rm\Rrightarrow T(0)=29-5(1)=29-5=24°C$

• Highest average daily high=34°C
• lowest average daily high=24°C

Answer 2

Answer:

Highest temperature: 34°C

Lowest temperature: 24°C

Explanation:

In a Normal Trigonometry Function: y = A + B sin(x) or y = A + B cos(x)

Maximum value: A + |B|

Minimum value: A - |B|

In this trigonometric function, A = 29, B = -5

Hence find minimum/maximum:

Maximum value: A + |B| = 29 + |-5| = 34

Minimum value: A - |B| = 29 - |-5| = 24

Hence, the maximum/highest temperature is 34°C and minimum/lowest temperature is 24°C in Karachi, Pakistan.

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