Calculus of Trigonometric Functions

Lesson

We understand the tangent function in right-angled triangles to be the ratio of sides $\frac{\text{opposite}}{\text{adjacent}}$oppositeadjacent with respect to one of the non right-angles in the triangle. More generally, we use the unit circle definitions of the trigonometric functions in order to give meaning to functions of any angle. The tangent function is defined in terms of the sine and cosine functions.

$\tan x\equiv\frac{\sin x}{\cos x}$`t``a``n``x`≡`s``i``n``x``c``o``s``x`

The graph below shows the tangent function between $-\pi$−π and $\pi$π.

Observe that the function has asymptotes at $\frac{\pi}{2}\pm n\pi$π2±`n`π, the points where the cosine function in the denominator in the definition is zero.

From the graph, it can be seen that the gradient of the tangent function is always positive and it has a minimum value of $1$1 at $x=0\pm n\pi$`x`=0±`n`π. The gradient has no maximum.

We deduce the tangent function explicitly by differentiating $f(x)=\tan x=\frac{\sin x}{\cos x}$`f`(`x`)=`t``a``n``x`=`s``i``n``x``c``o``s``x`. It is convenient to use the quotient rule for this.

We find, $f'\left(x\right)=\frac{\cos x\times\cos x-\left(-\sin x\times\sin x\right)}{\cos^2x}=\frac{\cos^2x+\sin^2x}{\cos^2x}=\frac{1}{\cos^2x}=\sec^2x$`f`′(`x`)=`c``o``s``x`×`c``o``s``x`−(−`s``i``n``x`×`s``i``n``x`)`c``o``s`2`x`=`c``o``s`2`x`+`s``i``n`2`x``c``o``s`2`x`=1`c``o``s`2`x`=`s``e``c`2`x` .

In Leibniz's notation, if $y=\tan x$`y`=`t``a``n``x`, then $\frac{\mathrm{d}y}{\mathrm{d}x}=\sec^2x$`d``y``d``x`=`s``e``c`2`x`.

In the diagram below, the graph of $\sec x^2$`s``e``c``x`2 (in red) has been superimposed on the tangent graph. It confirms the observations made above about the gradient of the tangent function.

Expressions of the type $a\tan\left(bx+c\right)+d$`a``t``a``n`(`b``x`+`c`)+`d` are differentiated by the usual rules, so that $\frac{\mathrm{d}}{\mathrm{d}x}\left(a\tan\left(bx+c\right)+d\right)=ab\sec^2\left(bx+c\right)$`d``d``x`(`a``t``a``n`(`b``x`+`c`)+`d`)=`a``b``s``e``c`2(`b``x`+`c`).

It is essential that the argument of the tangent function be expressed in radian measure. This requirement comes from the way in which the derivative of the sine function was obtained.

Differentiate $f\left(x\right)=4\tan x+4$`f`(`x`)=4`t``a``n``x`+4.

Differentiate $y=\tan3x$`y`=`t``a``n`3`x`.

At what values of $x$`x` does $\tan2x$`t``a``n`2`x` have a gradient of $2$2?

We need to find out where the derivative is equal to $2$2. That is, we must solve $2\sec^22x=2$2`s``e``c`22`x`=2. So, $\sec^22x=1$`s``e``c`22`x`=1. From the graph above, it is clear that $\sec^22x=1$`s``e``c`22`x`=1 at $2x=0,\pm\pi,\pm2\pi,...$2`x`=0,±π,±2π,.... Hence, the required solutions are $x=0,\pm\frac{\pi}{2},\pm\pi,...$`x`=0,±π2,±π,....

Consider the graph of $y=\tan x$`y`=`t``a``n``x`.

Loading Graph...

Which of the following best describes the graph of $y=\tan x$

`y`=`t``a``n``x`?The graph increases and decreases periodically.

AIt is constantly decreasing.

BIt is constantly increasing.

CThe graph increases and decreases periodically.

AIt is constantly decreasing.

BIt is constantly increasing.

CWhich of the following best describes the nature of the gradient of the curve?

Select all the correct options.

The gradient is always negative.

AThe gradient to the curve is never $0$0.

BThe gradient function has the same period as the curve itself.

CThe gradient increases more and more rapidly as the curve approaches the asymptotes.

DThe gradient is always positive.

EThe gradient to the curve is $0$0 every $\pi$π radians.

FThe gradient is always negative.

AThe gradient to the curve is never $0$0.

BThe gradient function has the same period as the curve itself.

CThe gradient increases more and more rapidly as the curve approaches the asymptotes.

DThe gradient is always positive.

EThe gradient to the curve is $0$0 every $\pi$π radians.

FThe tangent lines at the intercepts of the curve have been graphed as well.

Loading Graph...Using the graph, write down the gradient to the curve at $x=0$

`x`=0, $\pi$π, $2\pi$2π, $3\pi$3π, $\text{. . .}$. . .Gradient $=$= $\editable{}$.

The gradient function of $y=\tan x$

`y`=`t``a``n``x`is $y'$`y`′. Which of the following is the correct graph of $y'$`y`′ for each value of $x$`x`?Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DLoading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DWhich of the following is the equation of the gradient function $y'$

`y`′?$y'=\sec^2\left(x\right)$

`y`′=`s``e``c`2(`x`)A$y'=\sec x$

`y`′=`s``e``c``x`B$y'=\csc^2\left(x\right)$

`y`′=`c``s``c`2(`x`)C$y'=\csc x$

`y`′=`c``s``c``x`D$y'=\sec^2\left(x\right)$

`y`′=`s``e``c`2(`x`)A$y'=\sec x$

`y`′=`s``e``c``x`B$y'=\csc^2\left(x\right)$

`y`′=`c``s``c`2(`x`)C$y'=\csc x$

`y`′=`c``s``c``x`D

Differentiate $\tan\left(x^\circ\right)$`t``a``n`(`x`°).

Solution:

First, the argument $x^\circ$`x`° needs to be converted to circular measure.

To convert a degrees measure to radians, we can multiply it by $\frac{\pi}{180}$π180. This means $x^\circ$`x`° is equal to $\frac{x\pi}{180}$`x`π180 radians.

Now we can look to differentiate $\tan\left(\frac{x\pi}{180}\right)$`t``a``n`(`x`π180).

Using the chain rule, we get $\frac{d}{dx}\tan\left(\frac{x\pi}{180}\right)=\frac{\pi}{180}\sec^2\left(\frac{x\pi}{180}\right)$`d``d``x``t``a``n`(`x`π180)=π180`s``e``c`2(`x`π180).

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply differentiation methods in solving problems