Is this a valid approach to finding critical values of a trig function

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Homework Help Overview

The discussion revolves around finding the absolute maximum and minimum values of the function f(t) = t + cot(t/2) over the interval [π/4, 7π/4]. Participants are exploring the differentiation of the function and the implications of critical values derived from the first derivative.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the differentiation of the function and the resulting equation 2 = csc²(t/2). They explore different methods to proceed from this equation, including using trigonometric identities and taking reciprocals. Questions arise about the validity of applying inverse functions and the implications of squaring in the context of trigonometric equations.

Discussion Status

There are multiple approaches being considered regarding how to handle the equation involving csc²(t/2). Some participants express uncertainty about the application of inverse functions, while others provide insights into potential methods for finding candidate values for t/2. The discussion is ongoing, with no clear consensus yet.

Contextual Notes

Participants note the importance of considering the restricted interval for solutions and the potential for multiple candidate values arising from the equations discussed. There is also mention of the need to be cautious with signs when taking square roots.

TitoSmooth
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I was stuck for an hour trying to do this calculus 1 problem. Think I figured it out but it's a even problm.

Find the absolute maximum and absolute minimum values of f on the given interval.

f(t)=t+cot (t/2), [pie/4,7pie/4]

f'=1-(1/2) csc^2 (t/2)

So 1=1/2*csc^2 (t/2)

2=csc^2 (t/2)

For some reason I didn't know what to to do with cc squared so I applied an identity.

+/_ 1=cot (t/2)

Took arc cot on both side.

Arccot (+-1) =t/2

The arc cot gives me pie/4 and 3pie/4

So now I multiply both by 2.

So I get pie/2=t or 3pie/2=t.


Then I just plug the values in f and solve.

To see which ones are global max n min including the endpoints.

Also my question was.


Are we allowed to take the inverse cot on both sides of

Arccot (1)=arccot (cot^2 (t/2)) ?

If so how would we work with it?

I was thinking arccot1=1 and on the right side. Arccot will cancel one cot so I'm left with cot(t/2)

Now 1=cot (t/2)


Then I use pie/4 +pie (k)=t/2 and solve for solutions in the restricted interval? Is this correct
 
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At this step: 2=csc^2 (t/2), I see two ways to proceed.

1) Use the identity for csc^2, remember that it is very similar to the sec^2 formula: csc^2 = 1 + cot^2. This gives 1 = cot^2(t/2).
OR
2) Take reciprocals. This gives 1/2 = sin^2(t/2).

Then if you are careful to include the minus sign when taking square roots, you should get 4 candidate values for t/2.
 
verty said:
At this step: 2=csc^2 (t/2), I see two ways to proceed.

1) Use the identity for csc^2, remember that it is very similar to the sec^2 formula: csc^2 = 1 + cot^2. This gives 1 = cot^2(t/2).
OR
2) Take reciprocals. This gives 1/2 = sin^2(t/2).

Then if you are careful to include the minus sign when taking square roots, you should get 4 candidate values for t/2.

Aha. Thank you I over thought the problem.

I forgot that say sin^2 (x)=(sin x)^2.

Thanks verty.


Not will say arcsin (sin ^2 (x)) that's the same as sin x? I think I need to relearn trig equations lol
 
Sorry, I didn't read the whole of your first post earlier. Arccot or arcsin won't work, the square messes them up.
 

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