Finding stationary point of y = (sqrt(3))cos 2x - sin 2x

[Monkeyseat]

Hi,

Question:



Working:

a)

I'll just skip straight to the answer, because this is not the part that I need help on.

Answer: 2cos(2theta + 0.524). This is correct, I have checked.

b)

Rewriting y, you get y = 2cos(2x + 0.524).

The maximum value of 2cos(2x + 0.524) is 2 and occurs when:

cos(2x + 0.524) = 1
2x + 0.524 = cos^-1 (1)
2x + 0.524 = 0
2x = -0.524
x = -0.262

Therefore the maximum point is (-0.262, 2).

The minimum value of 2cos(2x + 0.524) is -2 and occurs when:

cos(2x + 0.524) = -1
2x + 0.524 = cos^-1 (-1)
2x + 0.524 = pi
2x = pi - 0.524
x = 1.31

Therefore the minimum point is (1.31, -2).

So those are the two stationary points as I have figured them out. My answer for the minimum point is correct in the textbook, but the textbook gives a different answer for the maximum point. It says that the maximum point is (2.88, 2). Is this correct?

To me, I can't see how they got that answer. To get 2.88 for x you must do:

cos(2x + 0.524) = 1
2x + 0.524 = cos^-1 (1)
2x + 0.524 = 2pi

But 2pi is outside the domain stated in the question...

Could someone please clarify which answer is correct? If mine is incorrect, where have I gone wrong?

Many thanks.

 

[Deleted member 4993]

Hi,

Question:



...but the textbook gives a different answer for the maximum point. It says that the maximum point is (2.88, 2). Is this correct? <<< Yes

To me, I can't see how they got that answer. To get 2.88 for x you must do:

cos(2x + 0.524) = 1
2x + 0.524 = cos^-1 (1)
2x + 0.524 = 2pi

But 2pi is outside the domain stated in the question...

No x = 2? is outside the domain - however x = ? - 0.264 is not (when 2x + 0.524 = 2?)

However, your answer x <0, is out of domain


Could someone please clarify which answer is correct? If mine is incorrect, where have I gone wrong?

Many thanks.