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

In summary, the conversation discusses finding the absolute maximum and minimum values of a function on a given interval through solving a calculus 1 problem. The approach involves using identities and taking reciprocals to find candidate values for the variable. There is also a discussion on the use of inverse trigonometric functions in solving the problem.
  • #1
TitoSmooth
158
6
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
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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
 
  • #4
Sorry, I didn't read the whole of your first post earlier. Arccot or arcsin won't work, the square messes them up.
 

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

1. What is the purpose of finding critical values of a trig function?

The critical values of a trigonometric function are the values that produce undefined or infinite results, such as division by zero. These values are important to identify in order to understand the behavior and limitations of the function.

2. How do you determine critical values of a trig function?

To find the critical values of a trig function, you must set the function equal to zero and solve for the variable. This will give you the values where the function is undefined or infinite. You can also graph the function and identify the points where the graph has vertical asymptotes or crosses the x-axis.

3. Can critical values of a trig function be negative?

Yes, critical values of a trig function can be negative. The critical values depend on the specific trigonometric function and can be positive, negative, or zero.

4. Why is it important to consider critical values when using trigonometric functions?

Critical values are important because they determine the domain and range of a trigonometric function. They also help identify any potential errors or limitations in calculations involving trigonometric functions.

5. Is there a formula for finding critical values of a trig function?

There is not a specific formula for finding critical values of a trig function, as it depends on the specific function and its properties. However, setting the function equal to zero and solving for the variable is a common method for identifying critical values.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
389
  • Calculus and Beyond Homework Help
Replies
27
Views
3K
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
113
  • Calculus and Beyond Homework Help
Replies
3
Views
608
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
468
  • Calculus and Beyond Homework Help
Replies
1
Views
738
Back
Top