Trigonometry|Precalculus

Question

D(t) models the distance (in thousands of km) from the earth to the Moon t days after the moon's perigee (when it's closest to the earth). Here, t is entered in radians.

D(t) = −21cos (2π​/29.5 t) + 384

How many days after its perigee does the moon first reach 

380 thousands of km from the Earth?

Round your final answer to the nearest whole day.

Answer 1

D(t) models the distance (in thousands of km) from the earth to the Moon t days after the moon's perigee (when it's closest to the earth). Here, t is entered in radians.

D(t) = −21cos (2π​/29.5 t) + 384

I set the computer into rad.

3.8*10^5(t)=-21cos(.213t)+384

" " =-21*

How do you put time into radians? A rad=180/π=57.2957...........deg.

I'm still trying to figure this.

3.8*10^5=(-21*cos(.213t)+384)

" " =(-21*cos(.213t/57.2957)+384)/t

I can't get any further. Please check your labels.

Answer 2

D(t) = −21cos (2π​/29.5 t) + 384, and D(t) = 380

=>

380 = −21cos (2π​/29.5 t) + 384

=>

380 - 384 = −21cos (2π​/29.5 t) + 384 - 384

=>

-4 = −21cos (2π​/29.5 t)

=>

-4/-21 = −21cos (2π​/29.5 t)/-21

=>

4/21 = cos (2π​/29.5 t)

9.5 t)/-21

=>

arccos(4/21) = arccos(cos (2π​/29.5 t))

=>

arccos(4/21) = 2π​/29.5 t

=>

29.5 * t * arccos(4/21) = 29.5 t * 2π​/29.5 t

=>

29.5 * t * arccos(4/21) = 2π​

=>

29.5 * t * arccos(4/21)/29.5 arccos(4/21) = 2π​/29.5 arccos(4/21)

=>

t = 2π​/(29.5*arccos(4/21))

Answer 3

You plug into the equation 380 for D(t) since they tell you that this equation gives D in 1000s of km. Now, solve for t which will be in days.

Here is the the solution for t done algebraically:

29.5/(2pi) * cos^-1((D-384)/(-21)) = t