Trig derivatives: find the extrema of 2sec(theta)+tan(theta)

[kinerry]

Here's the problem:

• Find the indicated extrema of \(\displaystyle f(\theta)\, =\, 2sec\theta\, +\, tan\theta\)
  
  A) Relative extrema on the interval \(\displaystyle (0,\, 2\pi)\)
  
  B) Absolute extrema on the interval \(\displaystyle \left(\frac{3\pi}{4},\, \frac{4\pi}{3}\right)\)

I never actually took trig (not a requirement at my college), so I am at a loss as to how to handle the theta.
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Edited by stapel -- Reason for edit: formatting clarity

 

[royhaas]

Start by taking the derivative and set it equal to zero.

 

[kinerry]

\(\displaystyle \L \frac{\cos^2{(\theta)}\,+\,\sin{(\theta)}\left[\sin{(\theta)}\,+\,2 \right]}{\cos^2{(\theta)}}\,=\,0\)

Correct so far?
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Edited by stapel -- Reason for edit: formatting clarity

 

[stapel]

Looks good so far. Now multiply through by cos²(theta), and multiply out the numerator.

. . . . .cos²(theta) + sin²(theta) + 2sin(theta) = 0

. . . . .1 + 2sin(theta) = 0

. . . . .sin(theta) = -1/2

Then what?

Eliz.

 

[kinerry]

Hmmm, I can't seem to get anywhere on paper and my calculator is giving me strange answers

theta=2*@n5pi+(7*pi) or theta = 2*@n5pi-(pi/6)

 

[stapel]

It looks like maybe you needed to take that trig course, because "-1/2" is one of the "basic reference angle values" that you'd have memorized, and is something that this course expects you to know.

Your book should have a trig appendix or a table printed on the endpapers. Look for a table of values corresponding to 0°, 30°, 45°, 90°, and maybe some more angles (and maybe listed in radians instead of degrees). You'll want to memorize that table.

Eliz.

 

[kinerry]

I'll figure that part out and come back.

 

[kinerry]

Is the answer going to be in the format
theta = whatever

 

[kinerry]

7pi/6?

 

[stapel]

7pi/6?

For which part is the above your answer?

Eliz.