# Why is cot(t) not= tan^-1(t)?

#### B9766

I'm having trouble graphing the "Witch of Agnesi" on various calculators and on the Desmond graphing web site.

The correct parametric equations are: $x = 2\cot (t), y = 2\sin^2(t)$

I originally called Texas Instruments for help because I was getting the wrong graph on my TI-84 but could get the right graph on the Desmond website.

The TI-calculator doesn't have a $\cot$ key but does have a $tan^{-1}$ key, while the Desmond site does accept $\cot$ as an argument.

So, on the TI-84 I entered the equations as: $x = 2\tan^{-1}(t), y = 2\sin(t)^2$ (note that $\sin^2$ cannot be entered, so the exponent is applied to the variable)

But that gives a curve quite different than the actual "Witch of Agnesi".

TI told me I had to enter the equations on the calculator as: $x = 2(1/tan(t)), y = 2\sin(t)^2$ And it worked! But they couldn't explain why.

Does anyone have any idea why this should be the case? Because, it appears the $\tan^{-1}(t)$ can never be used for $\cot(t)$. It has to be $\dfrac{1}{tan(t)}$. And, apparently, they aren't the same.

#### HallsofIvy

MHF Helper
When working with functions, $$\displaystyle f^{-1}$$ means the inverse function to f, not the reciprocal. That is the standard convention.

• 2 people

#### MarkFL

• 1 person

#### Plato

MHF Helper
I'm having trouble graphing the "Witch of Agnesi" on various calculators and on the Desmond graphing web site.

The correct parametric equations are: $x = 2\cot (t), y = 2\sin^2(t)$

I originally called Texas Instruments for help because I was getting the wrong graph on my TI-84 but could get the right graph on the Desmond website.

The TI-calculator doesn't have a $\cot$ key but does have a $tan^{-1}$ key, while the Desmond site does accept $\cot$ as an argument.

So, on the TI-84 I entered the equations as: $x = 2\tan^{-1}(t), y = 2\sin(t)^2$ (note that $\sin^2$ cannot be entered, so the exponent is applied to the variable)

But that gives a curve quite different than the actual "Witch of Agnesi".

TI told me I had to enter the equations on the calculator as: $x = 2(1/tan(t)), y = 2\sin(t)^2$ And it worked! But they couldn't explain why.

Does anyone have any idea why this should be the case? Because, it appears the $\tan^{-1}(t)$ can never be used for $\cot(t)$. It has to be $\dfrac{1}{tan(t)}$. And, apparently, they aren't the same.
The answer to your question has a complicated history. The notation $\tan^{-1}(x)$ was commonly used for the $\arctan(x)$ function in pre-1970 text books. I for one fought to have that changed. I went so far as to veto textbook adoptions that used that notation if they were for general education courses. But unlike you, I feared that students would see $\tan^{-1}(x)$ and think it was $\dfrac{1}{\tan(x)}$ and not realize it was really meant to be $\arctan(x)$ Thus I can only guess that Ti made the decision to drop the notation altogether.

BTW LOOK HERE

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#### B9766

Of course! I totally forgot about the inverse functions and their notation. What a duh moment.

So then, it appears that calculators (at least TI anyway) only provides the sin, cos and tan functions (and their inverses) but the user has to input sec, csc and cot functions (and their inverses) as the reciprocals.

I wish the TI people had just said that on the phone.

#### B9766

Just in case you're interested, here's a thread I started on this plane curve on another site:

Thanks for the link. It's a great explanation. Except, my text book gives the parametric equations as: $(2cot(t), 2sin^2(t))$ as opposed to your $(2tan(t), 2cos^2(t))$

Yours makes more sense to me. If I'm calculating correctly, the one in my text book produces $x=div/0!, y=0\ \ at\ \ t=0$ which isn't anything close to the curve yours produces. Can you suggest what I may be missing?

#### MarkFL

Thanks for the link. It's a great explanation. Except, my text book gives the parametric equations as: $(2cot(t), 2sin^2(t))$ as opposed to your $(2tan(t), 2cos^2(t))$

Yours makes more sense to me. If I'm calculating correctly, the one in my text book produces $x=div/0!, y=0\ \ at\ \ t=0$ which isn't anything close to the curve yours produces. Can you suggest what I may be missing?
It would appear to me that in the implementation I derived, I am using for the parameter:

$$\displaystyle -\frac{\pi}{2}<t<\frac{\pi}{2}$$

Whereas you are using a complementary range:

$$\displaystyle -\pi<t<0$$

#### Monoxdifly

Wait, so... $$\displaystyle sin^2x∙sin^{–1}x\neq sinx$$?

#### Plato

MHF Helper
Wait, so... $$\displaystyle sin^2x∙sin^{–1}x\neq sinx$$?
Well that depends upon what the notation $sin^{-1}(x)$ means.
Mathematicians, such as myself, have spent many years working on issues in undergraduate studies,( note I did not say mathematics education).
Before the 1980 and the reform movement the notation $sin^{-1}(x)=\arcsin(x)$. But calculators changed all. There was a need for a context free notation.
If we need the reciprocal of the $\sin(x)$ it became $[\sin(x)]^{-1}$ as oppose to $\sin^{-1}(x)$ because the latter is $\arcsin(x)$

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#### B9766

Wait, so... $$\displaystyle sin^2x∙sin^{–1}x\neq sinx$$?
In the course I'm taking, the professor did warn us (I had forgotten) not to confuse $sin^{-1}(x)$ with $\dfrac{1}{sin(x)}$.

As Plato points out, the -1 exponent is easily confused because it usually means "the power of -1". But in this case it means "the inverse function of".

For the $sin$ function, $sin^{-1}(x) = arcsin(x)$, where $arcsin(x)$ is the preferable notation for the obvious reason that's it's less confusing to people who don't use it all the time.

So the question I would ask is: What is $sin^{-2}(x)$?

Is it $\dfrac{1}{sin^2(x)}$ or is it $arcsin^2(x)$?

Plato, what's your call on this one?