*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

## 3. Trig Errors

This is a fairly short section, but contains some errors that I see my calculus students continually making so I thought I’d include them here as a separate section.

#### Degrees vs. Radians

Most trig classes that I’ve seen taught tend to concentrate on doing things in degrees. I suppose that this is because it’s easier for the students to visualize, but the reality is that almost all of calculus is done in radians and students too often come out of a trig class ill prepared to deal with all the radians in a calculus class.

You simply must get used to doing everything in radians in a calculus class. If you are asked to evaluate \(\cos \left( x \right)\) at \(x = 10\) we are asking you to use 10 radians not 10 degrees! The answers are very, very different! Consider the following,

\[\begin{array}{lcl}\cos \left( {10} \right) = - 0.839071529076 & \hspace{0.5in} & {\mbox{in radians}}\\ \cos \left( {10} \right) = 0.984807753012 & \hspace{0.5in} & {\mbox{in degrees}}\end{array}\]You’ll notice that they aren’t even the same sign!

So, be careful and make sure that you always use radians when dealing with trig functions in a trig class. Make sure your calculator is set to calculations in radians.

#### \(\cos \left( x \right)\) is NOT multiplication

I see students attempting both of the following on a continual basis

\[\begin{eqnarray*}\cos \left( {x + y} \right) & \ne & \cos \left( x \right) + \cos \left( y \right)\\ \cos \left( {3x} \right) & \ne & 3\cos \left( x \right)\end{eqnarray*}\]These just simply aren’t true. The only reason that I can think of for these mistakes is that students must be thinking of \(\cos \left( x \right)\) as a multiplication of something called \(\cos \) and \(x\). This couldn’t be farther from the truth! Cosine is a function and the \(\cos \) is used to denote that we are dealing with the cosine function!

If you’re not sure you believe that those aren’t true just pick a couple of values for \(x\) and \(y\) and plug into the first example.

\[\begin{eqnarray*}\cos \left( {\pi + 2\pi } \right) & \ne & \cos \left( \pi \right) + \cos \left( {2\pi } \right)\\ \cos \left( {3\pi } \right) & \ne & - 1 + 1\\ - 1 & \ne & 0\end{eqnarray*}\]So, it’s clear that the first isn’t true and we could do a similar test for the second example.

\[\begin{eqnarray*}\cos \left( {3\pi } \right) & \ne & 3\cos \left( \pi \right)\\ - 1 & \ne & 3\left( { - 1} \right)\\ - 1 & \ne & - 3\end{eqnarray*}\]I suppose that the problem is that occasionally there are values for these that are true. For instance, you could use \(x = \frac{\pi }{2}\) in the second example and both sides would be zero so it would work for that value of \(x\). In general, however, for the vast majority of values out there in the world these simply aren’t true!

On a more general note. I picked on cosine for this example, but I could have used any of the six trig functions, so be careful!

#### Powers of trig functions

Remember that if \(n\) is a positive integer then

\[{\sin ^n}x = {\left( {\sin x} \right)^n}\]The same holds for all the other trig functions as well of course. This is just a notational idiosyncrasy that you’ve got to get used to. Also remember to keep the following straight.

\[{\tan ^2}x \hspace{0.5in} {\mbox{vs.}} \hspace{0.5in} \tan {x^2}\]In the first case we taking the tangent then squaring result and in the second we are squaring the \(x\) then taking the tangent.

The \(\tan {x^2}\) is actually not the best notation for this type of problem, but I see people (both students and instructors) using it all the time. We really should probably use \(\tan \left( {{x^2}} \right)\) to make things clear.

#### Inverse trig notation

The notation for inverse trig functions is not the best. You need to remember, that despite what I just got done talking about above,

\[{\cos ^{ - 1}}x \ne \frac{1}{{\cos x}}\]This is why I said that n was a positive integer in the previous discussion. I wanted to avoid this notational problem. The -1 in \({\cos ^{ - 1}}x\) is NOT an exponent, it is there to denote the fact that we are dealing with an inverse trig function.

There is another notation for inverse trig functions that avoids this problem, but it is not always used.

\[{\cos ^{ - 1}}x = \arccos x\]