2.4 Derivatives of Sine and Cosine

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In this section we compute derivatives involving $\sin x$ and $\cos x$ .

We begin by computing the derivative of the trigonometric function $f(x) = \sin x$ . Two key trigonometric identities will be needed:

(a): The Pythagorean Identity: $\sin ^2 \theta + \cos ^2 \theta = 1,$

and
(b): The Sum Identity: $\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .$

We will also need the following limit, previously discussed in the Numerical Limits section and proved in the Squeeze Theorem section: $\lim _{\theta \to 0} \frac {\sin \theta }{\theta } = 1.$

We begin by using the Pythagorean Identity and the above limit to compute a second important limit involving the cosine function:

$\begin{align*} \lim_{\theta \to 0} \frac{1 - \cos \theta }{\theta} &= \lim_{\theta \to 0} \frac{1 - \cos \theta }{\theta} \cdot \frac{1 + \cos \theta }{1 + \cos \theta }\\ &= \lim_{\theta \to 0} \frac{1 - \cos^2 \theta }{\theta[1 + \cos \theta ]}\\ &= \lim_{\theta \to 0} \frac{\sin^2 \theta }{\theta[1 + \cos \theta ]}\\ &= \lim_{\theta \to 0} \frac{\sin \theta }{\theta} \cdot \frac{\sin \theta }{1 + \cos \theta }\\ &= 1 \cdot \frac{0}{2} = 0. \end{align*}$

If $f(x) = \sin (x)$ then

$\begin {aligned} f'(x) &= \lim _{h \to 0} \frac {f(x+h)-f(x)}{h} \\[5pt] &= \lim _{h \to 0} \frac {\sin (x+h) - \sin x }{h}\\[5pt] &= \lim _{h \to 0} \frac {\sin x \cos h + \cos x \sin h - \sin x }{h}\\[5pt] &= \lim _{h \to 0} \frac {\sin x \cos h - \sin x + \cos x \sin h }{h}\\[5pt] &= \lim _{h \to 0} \frac {\sin x [\cos h -1] + \cos x \sin h }{h}\\[5pt] &= \lim _{h \to 0} \frac {\sin x [\cos h -1]}{h} +\lim _{h \to 0} \frac {\cos x \sin h }{h}\\[5pt] &= \sin x\lim _{h \to 0} \frac { \cos h -1}{h} +\cos x\lim _{h \to 0} \frac { \sin h }{h}\\[5pt] &= (\sin x) \cdot 0 + (\cos x) \cdot 1 \\[5pt] &= \cos x. \end {aligned}$

Thus the derivative of sine is cosine: $\frac {d}{dx}\sin x = \cos x.$ In terms of the variables $\theta$ and $u$ : $\frac {d}{d\theta }\sin \theta = \cos \theta \text { and } \frac {d}{du}\sin u = \cos u .$

(problem 1)

Find the equation of the tangent line to the graph of $f(x) = \sin x$ at $x=0.$

The point of tangency is $(0, f(0))$

Use the derivative to find the slope, $m$

Point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {x}$

(problem 2)

Compute $\frac {d}{dx} \left [5\sin x\right ]$

The derivative of $\sin x$ is $\cos x$

Don’t forget to multiply by 5

The derivative of $5\sin x$ with respect to $x$ is $\answer {5\cos x}$

(problem 3)

Find the equation of the tangent line to the graph of $f(x) = \sin x$ at $x=\pi /2.$

The point of tangency is $(\pi /2, f(\pi /2))$

Use the derivative to find the slope, $m$

Point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {1}$

(problem 4)

Find $x$ -values in the interval $[0, 2\pi ]$ for which the tangent line to the graph of $f(x) = \sin x$ is horizontal.

Solve the equation $f'(x) = 0$ for $x$

The $x$ -values are:

$0, \pi , 2\pi$ $\pi /2, 3\pi /2$ $-\pi /2, \pi /2$

We now determine the derivative of the cosine function using the definition of the derivative.

If $f(x) = \cos x$ then

$\begin {aligned} f'(x) &= \lim _{h \to 0} \frac {f(x+h)-f(x)}{h} \\[5pt] &= \lim _{h \to 0} \frac {\cos (x+h) - \cos x}{h}\\[5pt] &= \lim _{h \to 0} \frac {\cos x\cos h - \sin x\sin (h) - \cos x}{h}\\[5pt] &= \lim _{h \to 0} \frac {\cos x\cos h - \cos x - \sin x\sin h}{h}\\[5pt] &= \lim _{h \to 0} \frac {\cos x[\cos h -1] - \sin x\sin h}{h}\\[5pt] &= \lim _{h \to 0} \frac {\cos x[\cos h -1]}{h} - \lim _{h \to 0}\frac {\sin x\sin h}{h}\\[5pt] &= \cos x\lim _{h \to 0} \frac {\cos h -1}{h} - \sin x\lim _{h \to 0}\frac {\sin h}{h}\\[5pt] &= \cos x \cdot 0 - \sin x\cdot 1\\[5pt] &= -\sin x.\\[5pt] \end {aligned}$

Using other variables, this can be written as $\frac {d}{dt} \cos t = -\sin t \\ \text {or}\\ \frac {d}{du} \cos u = -\sin u.$

(problem 5) Find the equation of the tangent line to the graph of $f(x) = \cos x$ at $x=\pi /2.$

The point of tangency is $(\pi /2, f(\pi /2))$

Use the derivative to find the slope, $m$

Point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {-x + \pi /2}$

example 6 If $f(x) = -3\cos x$ then we use the constant multiple rule with $c = -3$ and we get $f'(x) = -3(-\sin x) = 3\sin x.$

(problem 6a) Compute $\frac {d}{dx} \left [\pi \cos x\right ]$

The derivative of $\cos x$ is $-\sin x$

Don’t forget to multiply by $\pi$

The derivative of $\pi \cos x$ with respect to $x$ is $\answer {-\pi \sin x}$

(problem 6b)

Find $x$ -values in the interval $[0, 2\pi ]$ for which the tangent line to the graph of $f(x) = \cos x$ is horizontal.

Solve the equation $f'(x) = 0$ for $x$

The $x$ -values are:

$\pi /2, 3\pi /2$ $-\pi ,0, \pi$ $0, \pi , 2\pi$

Below is a graph of $f(x) = \cos x$ (in blue) and its derivative, $f'(x) = -\sin x$ (in purple). Notice that the slope of the tangent line to $f(x)$ (in red) is the height of the corresponding point on $f'(x)$ . Use it to see a graphical representation of the answers to the problem above.

example 7 Find $f'(x)$ if $f(x) = 4\sin x + 5\cos x.$
Using the sum and constant multiple rules, we get: $f'(x) = 4\cos x - 5\sin x$

(problem 7) Compute $\frac {d}{dx} \left [3\sin x + 7\cos x\right ]$

Use the Constant Multiple Rule on each term

The Constant Multiple Rule says: $\frac {d}{dx} cf(x) = cf'(x)$

The derivative of $3\sin x + 7\cos x$ with respect to $x$ is $\answer {3\cos x - 7\sin x}$

example 8 Find $f'(x)$ if $f(x) = 2\sin x - 3\cos x.$
Using the Difference and Constant Multiple Rules, we get: $f'(x) = 2\cos x - (-3\sin x) = 2\cos x + 3\sin x$

(problem 8a) Compute $\frac {d}{dx} \left [3\sin x - 7\cos x\right ]$

Use the difference and constant multiple rules

Constant Multiple Rule: $(cu)' = cu'$

The derivative of $3\sin x - 7\cos x$ with respect to $x$ is $\answer {3\cos x + 7\sin x}$

(problem 8b) Compute $y'$ if $y= \left [6\cos x - 8\sin x\right ]$

Use the difference and constant multiple Rules

Difference Rule: $(u-v)' = u' - v'$

$y' = \answer {-6\sin x - 8\cos x}$

We close this section with an example involving rectilinear motion.

example 9, Rectilinear Motion A point on a vibrating guitar string is moving vertically. The position, $s$ , of the point at time $t$ , is given by $s = 2\cos t$ . Find the velocity and acceleration of the point.
The velocity of the point is given by $v = s' = -2\sin t,$ and its acceleration is given by $a = s'' = -2\cos t$

(problem 9) A weight is suspended from a spring. The height, $s$ , of the weight at time $t$ , is given by $s = 10-\sin t$ . Find the velocity and acceleration of the weight.
The velocity of the point is given by
$v = \answer { -\cos t},$
and its acceleration is given by
$a = \answer {\sin t}.$