Question

Help with a question for a medal?
(I have part of it solved. I just need help with the rest)

Answer 1

I have the derivative. I just need help finding the zeroes

Answer 2

and what did you get for your derivative?

Answer 3

\[2\sec \theta(\tan \theta)+\sec^2 \theta\]

Answer 4

so, 2sec(tan+sec) is equal to zero, or undefined, give us critical points right?

Answer 5

Yes

Answer 6

recall that sec = 1/cos and tan = sin\cos

so when are these undefinied?

Answer 7

When cos=0?

Answer 8

correct

and to determine the zero .. well 1/cos is never equal to zero since the top of the fraction is always 1

tan + sec = 0

or rather, sin + 1 = 0 is what we have to look for right?

Answer 9

Sorry. My page just randomly timed out and reloaded.

Answer 10

and yes.
(though I have to admit that I am terrible at recalling the trig identities)

Answer 11

trig is like 98% memorization

Answer 12

Yep, that's why I suck at it

Answer 13

so our critical points of interest are;when cos=0, when sin=-1, and at out endpoints

Answer 14

Okay

Answer 15

How would we solve that?

Answer 16

then we are asked to determine when f'=pi

Answer 17

those are basic trig stuff. use a unit circle if you must.

Answer 18

Okay. Just a second

Answer 19

so cos=0 at pi/2
and sin=-1 at 3pi/2

Answer 20

yes, and cos=0 at 3pi/2 but thats a bit redundant eh :)

Answer 21

Oh yeah that too

Answer 22

2sec tan + sec^2 = pi

this one might not be so simple to do ...

2sin + 1
--------= pi
cos^2

hmm

Answer 23

might have a bad zero ... forgot about the 2 i think

2sin+1 = 0, sin = -1/2

Answer 24

so sin would actually be 2pi/3 or 4pi/3?

Answer 25

60 degrees ... yeah

Answer 26

4pi/3 and 5pi/3 .... is what im getting in my mind

sin is negative on the bottom part of the circle

Answer 27

I believe I misread the circle...so 210 and 330 degrees?

Answer 28

180+60, and +120

240 and 300

Answer 29

I'm looking on the 3rd page of here: http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf
by the way, if that helps.

Answer 30

yep, 240 and 300 should give us zeros, and 270 should be undefined, and then our endpoint of 0 and/or 2pi all give critical values

Answer 31

240 makes sense but why 300?
I though we were looking for -1/2, or maybe I'm missing something?

Answer 32

ugh .. nah, its me. im trying to remember trig and confusing the issue :)

sin = -1/2 that is when the y coord is -1/2 on the unit circle you posted

210 and 330

Answer 33

our reference angle is 30 degrees, not the 60 i mistook it for ... sigh

Answer 34

cos=0 at pi/2 and 3pi/2 for the undefinied

Answer 35

That makes sense

Answer 36

So sin would be 210 and 330?

Answer 37

yes

and as for the next part of it, the mean value thrm works on continuous functions ... we have an undefined part of the interval at pi/2 which would cause us some issue.

Answer 38

So if it's not continuous, we can't do the mean theorem correct?

Answer 39

it does not apply, correct

Answer 40

our graph is something like this, notice that slope between 0 and pi .... due to the discontinuity there is no x=c such that f'= pi