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

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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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