Inverse Trig Function Unit Circle

Show Mobile Detect Prove All NotesHide All Notes

Mobile Discover

You appear to exist on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in mural way. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to curlicue to see them) and some of the bill of fare items will be cut off due to the narrow screen width.

Department 3-vii : Derivatives of Changed Trig Functions

In this section we are going to look at the derivatives of the inverse trig functions. In order to derive the derivatives of inverse trig functions we'll need the formula from the last section relating the derivatives of changed functions. If $f\left( x \right)$ and $1000\left( x \correct)$ are inverse functions and then,

\[g'\left( x \right) = \frac{1}{{f'\left( {g\left( x \right)} \right)}}\]

Think likewise that two functions are inverses if $f\left( {1000\left( x \right)} \correct) = x$ and $g\left( {f\left( 10 \correct)} \correct) = x$.

We'll get through inverse sine, changed cosine and inverse tangent in detail here and leave the other three to you to derive if you lot'd like to.

Inverse Sine

Let'southward start with changed sine. Here is the definition of the changed sine.

\[y = {\sin ^{ - i}}x\hspace{0.5in} \Leftrightarrow \hspace{0.5in}\sin y = 10\hspace{0.25in}{\mbox{for}}\,\,\,\,\,\,\,\,\, - \frac{\pi }{2} \le y \le \frac{\pi }{ii}\]

Then, evaluating an changed trig office is the same every bit asking what angle (i.e. $y$) did nosotros plug into the sine function to get $ten$. The restrictions on $y$ given above are there to brand certain that we become a consistent answer out of the changed sine. We know that there are in fact an infinite number of angles that will piece of work and we want a consistent value when nosotros work with inverse sine. Using the range of angles above gives all possible values of the sine role exactly once. If you're not sure of that sketch out a unit circumvolve and you lot'll run into that that range of angles (the $y$'s) will cover all possible values of sine.

Note as well that since $ - 1 \le \sin \left( y \right) \le 1$ nosotros also have $ - 1 \le x \le 1$.

Allow'due south work a quick example.

Example 1 Evaluate $\displaystyle {\sin ^{ - one}}\left( {\frac{1}{2}} \right)$

Show Solution

Then, we are really asking what bending $y$ solves the following equation.

\[\sin \left( y \correct) = \frac{one}{2}\]

and we are restricted to the values of $y$ above.

From a unit of measurement circle nosotros can rapidly come across that $y = \frac{\pi }{vi}$.

Nosotros take the post-obit human relationship betwixt the inverse sine function and the sine part.

\[\sin \left( {{{\sin }^{ - 1}}x} \correct) = x\hspace{0.5in}{\sin ^{ - 1}}\left( {\sin x} \right) = x\]

In other words they are inverses of each other. This means that we tin use the fact higher up to discover the derivative of inverse sine. Let'south start with,

\[f\left( x \right) = \sin 10\hspace{0.5in}yard\left( x \right) = {\sin ^{ - 1}}x\]

Then,

\[g'\left( x \right) = \frac{ane}{{f'\left( {g\left( x \correct)} \correct)}} = \frac{1}{{\cos \left( {{{\sin }^{ - one}}x} \right)}}\]

This is non a very useful formula. Let'due south see if nosotros can become a better formula. Permit's start by recalling the definition of the changed sine role.

\[y = {\sin ^{ - i}}\left( x \correct)\hspace{0.5in} \Rightarrow \hspace{0.5in}x = \sin \left( y \right)\]

Using the first office of this definition the denominator in the derivative becomes,

\[\cos \left( {{{\sin }^{ - one}}x} \correct) = \cos \left( y \right)\]

At present, recall that

\[{\cos ^2}y + {\sin ^ii}y = ane\hspace{0.5in} \Rightarrow \hspace{0.5in}\cos y = \sqrt {i - {{\sin }^2}y} \]

Using this, the denominator is now,

\[\cos \left( {{{\sin }^{ - ane}}ten} \correct) = \cos \left( y \right) = \sqrt {i - {{\sin }^2}y} \]

Now, use the second part of the definition of the changed sine part. The denominator is and then,

\[\cos \left( {{{\sin }^{ - 1}}x} \right) = \sqrt {1 - {{\sin }^2}y} = \sqrt {1 - {x^2}} \]

Putting all of this together gives the following derivative.

\[\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \frac{i}{{\sqrt {1 - {x^2}} }}\]

Inverse Cosine

Now let'south take a look at the inverse cosine. Hither is the definition for the changed cosine.

\[y = {\cos ^{ - i}}x\hspace{0.5in} \Leftrightarrow \hspace{0.5in}\cos y = x\hspace{0.25in}{\mbox{for}}\,\,\,\,\,\,\,\,\,0 \le y \le \pi \]

As with the changed sine we've got a restriction on the angles, $y$, that we get out of the inverse cosine role. Again, if yous'd like to verify this a quick sketch of a unit circle should convince y'all that this range will embrace all possible values of cosine exactly one time. Also, we also have $ - ane \le x \le ane$ because $ - ane \le \cos \left( y \right) \le 1$.

Case ii Evaluate $\displaystyle {\cos ^{ - one}}\left( { - \frac{{\sqrt 2 }}{2}} \right)$.

Prove Solution

As with the inverse sine we are really just request the following.

\[\cos y = - \frac{{\sqrt 2 }}{2}\]

where $y$ must meet the requirements given higher up. From a unit circle we can come across that nosotros must have $y = \frac{{3\pi }}{4}$.

The inverse cosine and cosine functions are also inverses of each other and and so nosotros have,

\[\cos \left( {{{\cos }^{ - ane}}10} \right) = x\hspace{0.5in}{\cos ^{ - 1}}\left( {\cos x} \right) = ten\]

To detect the derivative we'll practise the aforementioned kind of work that nosotros did with the inverse sine above. If we start with

\[f\left( x \correct) = \cos x\hspace{0.5in}thou\left( x \correct) = {\cos ^{ - 1}}10\]

then,

\[g'\left( x \correct) = \frac{1}{{f'\left( {grand\left( x \correct)} \right)}} = \frac{one}{{ - \sin \left( {{{\cos }^{ - 1}}x} \correct)}}\]

Simplifying the denominator here is almost identical to the piece of work we did for the inverse sine and so isn't shown here. Upon simplifying we get the following derivative.

\[\frac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right) = - \frac{1}{{\sqrt {1 - {x^2}} }}\]

So, the derivative of the inverse cosine is nigh identical to the derivative of the inverse sine. The merely difference is the negative sign.

Changed Tangent

Hither is the definition of the inverse tangent.

\[y = {\tan ^{ - 1}}x\hspace{0.5in} \Leftrightarrow \hspace{0.5in}\tan y = x\hspace{0.25in}{\mbox{for}}\,\,\,\,\,\,\,\,\, - \frac{\pi }{2} < y < \frac{\pi }{2}\]

Again, nosotros have a restriction on $y$, but detect that we can't let $y$ be either of the ii endpoints in the restriction in a higher place since tangent isn't fifty-fifty divers at those two points. To convince yourself that this range volition embrace all possible values of tangent do a quick sketch of the tangent function and we can come across that in this range we practice indeed cover all possible values of tangent. Besides, in this case at that place are no restrictions on $x$ because tangent can have on all possible values.

Example 3 Evaluate ${\tan ^{ - 1}}1$.

Show Solution

Here we are asking,

\[\tan y = 1\]

where $y$ satisfies the restrictions given in a higher place. From a unit circle nosotros can see that $y = \frac{\pi }{iv}$.

Because there is no restriction on $x$ nosotros tin ask for the limits of the changed tangent function as $ten$ goes to plus or minus infinity. To do this we'll demand the graph of the changed tangent function. This is shown below.

$Graph of $y={{\tan }^{-1}}\left( x \right)$. The graph starts on the left at the horizontal asymptote at $y=-\frac{\pi }{2}$ and increases going through (0,0) and then approaching another horizontal asymptote at $y=\frac{\pi }{2}$.$

From this graph we tin come across that

\[\mathop {\lim }\limits_{x \to \infty } {\tan ^{ - ane}}x = \frac{\pi }{2}\hspace{0.5in}\hspace{0.25in}\mathop {\lim }\limits_{x \to - \infty } {\tan ^{ - ane}}x = - \frac{\pi }{two}\]

The tangent and changed tangent functions are inverse functions and so,

\[\tan \left( {{{\tan }^{ - 1}}ten} \right) = x\hspace{0.5in}{\tan ^{ - one}}\left( {\tan x} \right) = 10\]

Therefore, to discover the derivative of the inverse tangent office we tin can start with

\[f\left( x \correct) = \tan x\hspace{0.5in}k\left( 10 \right) = {\tan ^{ - one}}x\]

Nosotros so take,

\[chiliad'\left( x \correct) = \frac{1}{{f'\left( {g\left( 10 \right)} \right)}} = \frac{1}{{{{\sec }^ii}\left( {{{\tan }^{ - 1}}x} \correct)}}\]

Simplifying the denominator is similar to the inverse sine, but unlike enough to warrant showing the details. We'll start with the definition of the inverse tangent.

\[y = {\tan ^{ - 1}}x\hspace{0.5in} \Rightarrow \hspace{0.5in}\tan y = x\]

The denominator is then,

\[{\sec ^two}\left( {{{\tan }^{ - 1}}x} \right) = {\sec ^2}y\]

At present, if we start with the fact that

\[{\cos ^2}y + {\sin ^two}y = i\]

and divide every term by cos² $y$ we will get,

\[1 + {\tan ^2}y = {\sec ^ii}y\]

The denominator is and then,

\[{\sec ^ii}\left( {{{\tan }^{ - 1}}10} \right) = {\sec ^2}y = 1 + {\tan ^2}y\]

Finally using the 2d portion of the definition of the inverse tangent function gives us,

\[{\sec ^ii}\left( {{{\tan }^{ - 1}}x} \correct) = i + {\tan ^ii}y = 1 + {x^two}\]

The derivative of the inverse tangent is then,

\[\frac{d}{{dx}}\left( {{{\tan }^{ - one}}x} \right) = \frac{1}{{1 + {x^2}}}\]

There are three more inverse trig functions but the iii shown here the most mutual ones. Formulas for the remaining three could be derived past a similar procedure as we did those above. Here are the derivatives of all six inverse trig functions.

\[\begin{array}{ll}\displaystyle \frac{d}{{dx}}\left( {{{\sin }^{ - ane}}x} \right) = \frac{1}{{\sqrt {ane - {ten^ii}} }} & \hspace{1.0in}\displaystyle \frac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right) = - \frac{i}{{\sqrt {1 - {ten^2}} }}\\ \displaystyle \frac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \frac{1}{{i + {10^2}}} & \hspace{one.0in}\displaystyle \frac{d}{{dx}}\left( {{{\cot }^{ - one}}x} \correct) = - \frac{ane}{{1 + {x^2}}}\\ \displaystyle \frac{d}{{dx}}\left( {{{\sec }^{ - 1}}x} \right) = \frac{1}{{\left| x \right|\sqrt {{x^2} - ane} }} & \hspace{ane.0in}\displaystyle \frac{d}{{dx}}\left( {{{\csc }^{ - 1}}ten} \right) = - \frac{i}{{\left| x \correct|\sqrt {{x^2} - 1} }}\end{array}\]

We should probably now do a couple of quick derivatives here earlier moving on to the next department.

Case four Differentiate the following functions.

$f\left( t \right) = 4{\cos ^{ - i}}\left( t \right) - ten{\tan ^{ - ane}}\left( t \correct)$
$y = \sqrt z \, {\sin ^{ - 1}}\left( z \right)$

Show All SolutionsHide All Solutions

a $f\left( t \correct) = iv{\cos ^{ - 1}}\left( t \right) - 10{\tan ^{ - 1}}\left( t \correct)$ Show Solution

Not much to exercise with this one other than differentiate each term.

\[f'\left( t \right) = - \frac{4}{{\sqrt {i - {t^ii}} }} - \frac{{10}}{{1 + {t^2}}}\]

b $y = \sqrt z \, {\sin ^{ - 1}}\left( z \right)$ Show Solution

Don't forget to convert the radical to fractional exponents before using the product rule.

\[y' = \frac{1}{2}{z^{ - \frac{1}{two}}}{\sin ^{ - 1}}\left( z \right) + \frac{{\sqrt z }}{{\sqrt {1 - {z^2}} }}\]

Alternating Notation

At that place is some alternate notation that is used on occasion to denote the inverse trig functions. This notation is,

\[\begin{array}{ll}{\sin ^{ - one}}x = \arcsin ten & \hspace{ane.0in}{\cos ^{ - i}}x = \arccos ten\\ {\tan ^{ - ane}}ten = \arctan x & \hspace{1.0in}{\cot ^{ - 1}}ten = {\mbox{arccot }}x\\ {\sec ^{ - 1}}ten = {\mathop{\rm arcsec}\nolimits} \,x & \hspace{1.0in}{\csc ^{ - 1}}x = {\mathop{\rm arccsc}\nolimits} \,x\end{array}\]