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

1. Jul 6, 2014

### TitoSmooth

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

2. Jul 6, 2014

### verty

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. Jul 6, 2014

### TitoSmooth

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. Jul 6, 2014

### verty

Sorry, I didn't read the whole of your first post earlier. Arccot or arcsin won't work, the square messes them up.