7.2: Derivatives of trigonometric functions

 

We will show that the derivative of the sine function is the cosine function, or

\[[\sin t]^{\prime}=\cos t \label{7.9}\]

From this formula and the Combination Derivative formulas 6.1 and 6.2 shown at the beginning of Chapter 6, the derivatives of the other five trigonometric functions are easily computed.

We first show that

\[[\sin t]_{t=0}^{\prime}=\sin ^{\prime}(0)=1\]

Assume \(h\) is a positive number less than \(\pi /2\). We know from Inequalities 7.1.4 and 7.1.5 that

\[\sin h<h<\tan h\]

We write

\[\left.\begin{array}{rl}\
\sin h & < h < \tan h &\\
\sin h & < h < \frac{\sin h}{\cos h} &\\
1 & < \frac{h}{\sin h}<\frac{1}{\cos h} &(i)\\
1 & > \frac{\sin h}{h}>\cos h &(ii)\\
\end{array} \right\} \label{7.10}\]

The inequalities

\[1>\frac{\sin h}{h}>\cos h\]

present an opportunity to reason in a rather clever way. We wish to know what \(\frac{\sin h}{h}\) approaches as the positive number \(h\) approaches 0 (as \(h\) approaches \(0^+\)). Because \(\cos x\) is continuous (Exercise 7.1.5) and \(\cos 0 = 1\), \(\cos h\) approaches 1 as \(h\) approaches \(0^+\). Now we have \(\frac{\sin h}{h}\) 'sandwiched' between two quantities, 1 and a quantity that approaches 1 as h approaches \(0^+\). We conclude that \(\frac{\sin h}{h}\) also approaches 1 as \(h\) approaches \(0^+\), and illustrate the result in the array:

\[\text { As } h \rightarrow 0^{+}\left\{\begin{array}{l}
1 ~< & \frac{\sin h}{h}  ~< & \cos h \\
\downarrow & \downarrow & \downarrow \\
1 ~\leq & 1 ~\leq & 1
\end{array}\right.\]

The argument can be formalized with the \(\epsilon, \delta\) definition of limit, but we leave it on an intuitive basis.

We have assumed \(h > 0\) in the previous steps. A similar argument can be made for \(h < 0\)

We now know that the slope of the graph of the sine function at (0,0) is 1. It is this result that makes radian measure so useful in calculus. For any other angular measure, the slope of the sine function at 0 is not 1. For example, the sine graph plotted in degrees has slope of \(\pi /180\) at (0,0).

Because of the continuity of the composition of two functions, Equation 4.1.8, the equation

\[\lim _{h \rightarrow 0} \frac{\sin h}{h}=1 \label{7.11}\]

may take a variety of forms:

\[\lim _{h \rightarrow 0} \frac{\sin 2 h}{2 h}=1 \quad \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}=1 \quad \lim _{h \rightarrow 0} \frac{\sin h^{2}}{h^{2}}=1\]

We write a general form:

\[\text { If } \theta(h) \neq 0 \quad \text { for } h \neq 0 \quad \text { and } \quad \lim _{h \rightarrow 0} \theta(h)=0 \quad \text { then } \quad \lim _{h \rightarrow 0} \frac{\sin \theta(h)}{\theta(h)}=1 \label{7.12}\]

We use

\[\lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}=1 .\]

in the next paragraph. We also use the fact that the cosine function is continuous, Exercise 7.1.5.

Now we compute \([ \sin t]^{\prime}\) for any \(t\). By Definition 3.3.3,

\[[\sin t]^{\prime}=\lim _{h \rightarrow 0} \frac{\sin (t+h)-\sin t}{h}\]

With the trigonometric identity

\[\sin x-\sin y=2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)\]

we get

\[\begin{array}\
{[\sin t]^{\prime} } &=\lim _{h \rightarrow 0} \frac{\sin (t+h)-\sin t}{h} & (i)\\
&=\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{t+h+t}{2}\right) \sin \left(\frac{t+h-t}{2}\right)}{h} & (ii)\\
&=\lim _{h \rightarrow 0} \frac{\cos \left(t+\frac{h}{2}\right) \sin \left(\frac{h}{2}\right)}{\frac{h}{2}} &\\
&=\lim _{h \rightarrow 0} \cos \left(t+\frac{h}{2}\right) \lim _{h \rightarrow 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} & (iii)\\
&=\cos t \lim _{h \rightarrow 0} \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}} & (iv)\\
&=\cos t \times 1=\cos t & (v)\\
\end{array} \label{7.13}\]

We have now shown that \([ \sin t] ^{\prime} = \cos t\). The derivatives of the other five trigonometric functions are easily computed using the derivative formulas 6.1 and 6.2 shown at the beginning of Chapter 6.

Derivative of the cosine. We will show that

\[[\cos t]^{\prime}=-\sin t \label{7.14}\]

Observe that

\[\sin \left(t+\frac{\pi}{2}\right)=\sin t \cos \frac{\pi}{2}+\cos t \sin \frac{\pi}{2}=\sin t \times 0+\cos t \times 1=\cos t\]

Therefore

\[\cos t=\sin \left(t+\frac{\pi}{2}\right) \quad \text { and } \quad[\cos t]^{\prime}=\left[\sin \left(t+\frac{\pi}{2}\right)\right]^{\prime}\]

We use the Chain Rule Equation 6.2.1

\[[G(u(t))]^{\prime}=G^{\prime}(u(t)) u^{\prime}(t) \quad \text { with } \quad G(u)=\sin u \quad \text { and } \quad u(t)=t+\frac{\pi}{2} .\]

Note that \(G ^{\prime} (u) = \cos u\) and \(u ^{\prime} (t) = 1\).

\[\begin{aligned}
{[\cos t]^{\prime} } &=\left[\sin \left(t+\frac{\pi}{2}\right)\right]^{\prime} \\
&=\cos \left(t+\frac{\pi}{2}\right)\left[t+\frac{\pi}{2}\right]^{\prime} \\
&=\cos \left(t+\frac{\pi}{2}\right) \times 1 \\
&=\cos t \cos \frac{\pi}{2}-\sin t \sin \frac{\pi}{2} \\
&=-\sin t
\end{aligned}\]

We have established Equation \ref{7.14}, \([\cos t]^{\prime}=-\sin t\).

Derivative of the tangent function. We will show that

\[[\tan t]^{\prime}=\sec ^{2} t \label{7.15}\]

using

\[\tan t=\frac{\sin t}{\cos t} \quad \text { and } \quad[\sin t]^{\prime}=\cos t \quad \text { and } \quad[\cos t]^{\prime}=-\sin t\]

and the quotient rule for derivatives from Equations 6.1 and 6.2,

\[\left[\frac{u}{v}\right]^{\prime}=\frac{v u^{\prime}-u v^{\prime}}{v^{2}} .\]

\[\begin{aligned}
{[\tan t]^{\prime} } &=\left[\frac{\sin t}{\cos t}\right]^{\prime} \\
&=\frac{\cos t[\sin t]^{\prime}-\sin t[\cos t]^{\prime}}{(\cos t)^{2}} \\
&=\frac{\cos t \cos t-\sin t(-\sin t)}{\cos ^{2} t} \\
&=\frac{\cos ^{2} t+\sin ^{2} t}{\cos ^{2} t} \\
&=\sec ^{2} t
\end{aligned}\]

In summary we write the formulas

\[\begin{aligned}
&[\sin t]^{\prime}=\cos t & \ref{7.9} & \quad [\cos t]^{\prime}=-\sin t & \ref{7.14}\\
&[\tan t]^{\prime}=\sec ^{2} t & \ref{7.15} & \quad [\cot t]^{\prime}=-\csc ^{2} t & \ref{7.16}\\
&[\sec t]^{\prime}=\sec t \tan t & \ref{7.17} & \quad [\csc t]^{\prime}=-\csc t \cot t & \ref{7.18}
\end{aligned}\]

You are asked to compute \([\cot t] ^{\prime}\), \([\sec t] ^{\prime} \), and \([\csc t] ^{\prime}\) in Exercise 7.2.5.

Example 7.2.1 We illustrate the use of the derivative formulas for sine, cosine, and tangent by computing the derivatives of

1. \(y=3 \sin t \cos t\)
2. \(y=\sin ^{4} t\)
3. \(y=\ln (\tan t)\)
4. \(y=e^{\sin t}\)

 

1. \(\begin{aligned}
   {[3 \sin t \cos t]^{\prime} } &=3\left(\sin t[\cos t]^{\prime}+[\sin t]^{\prime} \cos t\right) \\
   &=3(\sin t(-\sin t)+\cos t \cos t) \\
   &=-3 \sin ^{2} t+3 \cos ^{2} t
   \end{aligned}\)
2. \(\left[\sin ^{4} t\right]^{\prime}=4\left(\sin ^{3} t\right)[\sin t]^{\prime}=4\left(\sin ^{3} t\right) \cos t\)
3. \(\begin{array}\
   [\ln (\tan t)]^{\prime} & =[(\ln (\sin t))-(\ln (\cos t))]^{\prime}&=[\ln (\sin t)]^{\prime}-[\ln (\cos t)]^{\prime}\\
   &=\frac{1}{\sin t}[\sin t]^{\prime}-\frac{1}{\cos t}[\cos t]^{\prime}&=\frac{1}{\sin t} \cos t-\frac{1}{\cos t}(-\sin t)\\
   &=\frac{\cos ^{2} t+\sin ^{2} t}{\cos t \sin t}&=\sec t \csc t
   \end{array}\)
4. \(\left[e^{\sin t}\right]^{\prime}=e^{\sin t}[\sin t]^{\prime}=e^{\sin t} \cos t\)

Example 7.2.2 The function \(F(t)=e^{-t / 10} \sin t\) is an example of 'damped oscillation,' an important type of vibration. Its graph is shown in Figure \(\PageIndex{1}\). The peaks and valleys of the oscillation are marked by values of \(t\) for which \(F ^{\prime} (t) = 0\). We find them by

\[\begin{aligned}
{\left[e^{-t / 10} \sin t\right]^{\prime} } &=e^{-t / 10}[\sin t]^{\prime}+\left[e^{-t / 10}\right]^{\prime} \sin t \\
&=e^{-t / 10} \cos t-\frac{1}{10} e^{-t / 10} \sin t=e^{-t / 10}\left(\cos t-\frac{1}{10} \sin t\right)
\end{aligned}\]

Now \(e ^{-t/10} > 0\) for all \(t\); \(F ^{\prime} (t) = 0\) implies that

\[\cos t-\frac{1}{10} \sin t=0, \quad \tan t=10, \quad t=(\arctan 10)+n \pi \quad \text { for } n \text { an integer }\]

Figure \(\PageIndex{1}\): Graph of \(y=e^{-t / 10} \sin t\). The relative high and low points are marked by horizontal tangents and occur at \(t=\arctan 10+n \pi\) for \(n\) an integer.

 

Exercises for Section 7.2 Derivatives of trigonometric functions

Exercise 7.2.1 The difference quotient

\[\frac{F(t+h)-F(t)}{h} \quad \text { approximates } \quad F^{\prime}(t) \quad \text { when } h \text { is 'small.' }\]

Make a plot of

\[y=\cos t \quad \text { and of } \quad \frac{\sin (t+0.2)-\sin t}{0.2} \quad-\frac{\pi}{2} \leq t \leq 2 \pi .\]

Repeat, using \(h = 0.05\) instead of \(h = 0.2\).

Exercise 7.2.2 Compute the derivative of

\[y(t)=1+2 t-5 t^{7}+2 e^{3 t}-\ln 6 t+2 \sin t-3 \cos t\]

Exercise 7.2.3 Compute the derivatives of

1. \(y=2 \sin t \cos t\)
2. \(y=\sin ^{2} t+\cos ^{2} t\)
3. \(y=\sec t=\frac{1}{\cos t}\)
4. \(y=\cot t\)
5. \(y=\ln \cos t\)
6. \(y=\sin ^{2} t-\cos ^{2} t\)
7. \(y=\csc t=\frac{1}{\sin t}\)
8. \(y=\sec ^{2} t\)
9. \(y=e^{\cos t}\)
10. \(y=\ln (\sec t)\)
11. \(y=e^{-t} \sin t\)
12. \(y=\tan ^{2} t\)
13. \(y=\frac{e^{2 t}}{400}\)
14. \(y=\ln \left(\cos ^{20} t\right)\)
15. \(y=\frac{\ln t^{2}}{30}\)

Exercise 7.2.4 Compute \(y ^{\prime}\) and solve for \(t\) in \(y ^{\prime} (t) = 0\). Sketch the graphs and find the highest and the lowest points of the graphs of:

1. \(y=\sin t+\cos t \quad 0 \leq t \leq \pi\)
2. \(y=e^{-t} \sin t \quad 0 \leq t \leq \pi\)
3. \(y=\sqrt{3} \sin t+\cos t \quad 0 \leq t \leq \pi\)
4. \(y=\sin t \cos t \quad 0 \leq t \leq \pi\)
5. \(y=e^{-t} \cos t \quad 0 \leq t \leq \pi\)
6. \(y=e^{-\sqrt{3} t} \cos t \quad 0 \leq t \leq \pi\)

Note: \(\cos ^{2} t-\sin ^{2} t=\cos (2 t)\).

Exercise 7.2.5

1. Use \(\cot x=\frac{\cos x}{\sin x}\) and the quotient rule to show that \[[\cot x]^{\prime}=-\csc ^{2} x \label{7.16}\]
2. Use \(\sec x=\frac{1}{\cos x}=(\cos x)^{-1}\) and the power chain rule to show that \[[\sec x]^{\prime}=\sec x \tan x \label{7.17}\]
3. Show that \[[\csc x]^{\prime}=-\csc x \cot x \label{7.18}\]

Exercise 7.2.6 The graphs of \(y_{1} = \cos h\) and \(y_{2} = 1\) are shown in Figure Ex. 7.2.6. The inequality

\[1<\frac{\sin h}{h}<\cos h\]

implies that the graph of \(y=\frac{\sin h}{h}\), \(0<h<\pi / 2\) is 'sandwiched' between \(y_1\) and \(y_2\). Let \(F\) be any function defined on \(0 < z \leq 1\) whose graph lies above the graph of \(y_1\) and below the graph of \(y_2\).

1. Draw the graph of one such function, \(F\).
2. What number does \(F(h)\) approach as \(h\) approaches 0?

Figure for Exercise 7.2.6 Graphs of \(y_{1} = 1\) and \(y_{2} = \cos h\). See Exercise 7.2.6.

Exercise 7.2.7 Draw a tangent to the graph of the sine function at the point (0,0) in Figure Ex. 7.2.7. Choose two points of the tangent, measure the coordinates of the two points, and use those coordinates to compute the slope of the tangent and \(\left.[\sin t]^{\prime}\right|_{t=0}\).

Figure for Exercise 7.2.7 Graph of \(y = \sin x\). See Exercise 7.2.7.

Exercise 7.2.8 Give reasons for steps (i) and (ii) in Equation \ref{7.10} leading to the inequalities

\[1>\frac{\sin h}{h}>\cos h\]

It is important that \(h > 0\); why?

Exercise 7.2.9 The derivative of \(y = \cos x\) is defined by

\[[\cos x]^{\prime}=\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h}\]

Make a plot of

\[y=-\sin t \quad \text { and of } \quad \frac{\cos (t+0.2)-\cos t}{0.2} \quad-\frac{\pi}{2} \leq t \leq 2 \pi .\]

Repeat, using \(h = 0.05\) instead of \(h = 0.2\).

Exercise 7.2.10 Show that \([\cos t]^{\prime}=-\sin t\). To do so, use

\[[\cos t]^{\prime}=\lim _{h \rightarrow 0} \frac{\cos (t+h)-\cos t}{h}\]

and the trigonometric identity

\[\cos x-\cos y=-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)\]

The steps will be similar to those of Equation \ref{7.13}.