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<!DOCTYPE html>
<meta charset="utf-8">
<title>How to deal with Sines and Cosines | Calculus Made Easy</title>
<link rel="stylesheet" href="screen.css">
<style>
body{counter-reset:h1 15}
</style>
<script type="text/x-mathjax-config">
MathJax.Hub.Config({tex2jax: {inlineMath: [['$','$'], ['\\(','\\)']]}});
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</script>
<h1><br>How to deal with Sines and Cosines</h1>
<p class="a rotatedFloralHeartBullet">
<p>Greek letters being usual to denote angles, we will
take as the usual letter for any variable angle the
letter $\theta$ (“theta”).
<p>Let us consider the function
\[
y= \sin \theta.
\]
<a name="figure3">
<p><img src="33283-t/images/177a.pdf.png-1.png">
<p>What we have to investigate is the value of $\dfrac{d(\sin \theta)}{d \theta}$;
or, in other words, if the angle $\theta$ varies, we have to
find the relation between the increment of the sine
and the increment of the angle, both increments being
indefinitely small in themselves. Examine <a href="#figure43">Figure 43</a>,
wherein, if the radius of the circle is unity, the height
of $y$ is the sine, and $\theta$ is the angle. Now, if $\theta$ is
supposed to increase by the addition to it of the
small angle $d \theta$–an element of angle–the height
of $y$, the sine, will be increased by a small element $dy$.
The new height $y + dy$ will be the sine of the new
angle $\theta + d \theta$, or, stating it as an equation,
\[
y+dy = \sin(\theta + d \theta);
\]
and subtracting from this the first equation gives
\[
dy = \sin(\theta + d \theta)- \sin \theta.
\]
<p>The quantity on the right-hand side is the difference
between two sines, and books on trigonometry tell
us how to work this out. For they tell us that if
$M$ and $N$ are two different angles,
\[
\sin M - \sin N = 2 \cos\frac{M+N}{2}·\sin\frac{M-N}{2}.
\]
<p>If, then, we put $M= \theta + d \theta$ for one angle, and
$N= \theta$ for the other, we may write
\begin{align*}
dy &= 2 \cos\frac{\theta + d\theta + \theta}{2}
· \sin\frac{\theta + d\theta - \theta}{2},\\
\text{or,}\;
dy &= 2\cos(\theta + \tfrac{1}{2}d\theta)
· \sin\tfrac{1}{2} d\theta.
\end{align*}
<p>But if we regard $d \theta$ as indefinitely small, then in
the limit we may neglect $\frac{1}{2} d \theta$ by comparison with $\theta$,
and may also take $\sin\frac{1}{2} d \theta$ as being the same as $\frac{1}{2} d \theta$.
The equation then becomes:<a name="differsin"/>
\begin{align*}
dy &= 2 \cos \theta × \tfrac{1}{2} d \theta; \\
dy &= \cos \theta · d \theta, \\
\text{ and, finally,}\;
\dfrac{dy}{d \theta} &= \cos \theta.
\end{align*}
<p>The accompanying curves, <a href="#figure44">Fig. 44</a> and <a href="#figure45">Fig. 45</a> show,
plotted to scale, the values of $y=\sin \theta$, and $\dfrac{dy}{d\theta}=\cos\theta$,
for the corresponding values of $\theta$.
<a name="figure44">
<p><img src="33283-t/images/179a.pdf.png-1.png"><a name="erratum2"/>
<a name="figure45">
<p><img src="33283-t/images/179b.pdf.png-1.png">
<hr>
<p>Take next the cosine.<a name="differcos"/>
<p>Let $y=\cos \theta$.
<p>Now $\cos \theta=\sin\left(\dfrac{\pi}{2}-\theta\right)$.
<p>Therefore
\begin{align*}
&\begin{aligned}
dy = d\left(\sin\left(\frac{\pi}{2} - \theta\right)\right)
&= \cos\left(\frac{\pi}{2} - \theta\right) × d(-\theta), \\
&= \cos\left(\frac{\pi}{2} - \theta\right) × (-d\theta),
\end{aligned} \\
&\frac{dy}{d\theta} = -\cos\left(\frac{\pi}{2} - \theta\right).
\end{align*}
And it follows that
\begin{align*}
&\frac{dy}{d\theta} = -\sin \theta.
\end{align*}
<p><hr>
<p>Lastly, take the tangent.
Let
\begin{align*}
y &= \tan \theta, \\
dy &= \tan(\theta + d\theta) - \tan\theta. \\
\end{align*}
Expanding, as shown in books on trigonometry,
\begin{align*}
\tan(\theta + d\theta)
&= \frac{\tan\theta + \tan d\theta}
{1 - \tan\theta·\tan d\theta}; \\
\text{whence}\;
dy &= \frac{\tan\theta + \tan d\theta}
{1-\tan\theta·\tan d\theta} - \tan\theta \\
&= \frac{(1 + \tan^2\theta)\tan d\theta}
{1-\tan\theta·\tan d\theta}.
\end{align*}
<p>Now remember that if $d\theta$ is indefinitely diminished,
the value of $\tan d\theta$ becomes identical with $d\theta$, and
$\tan\theta · d\theta$ is negligibly small compared with $1$, so that
the expression reduces to
\begin{align*}
dy &= \frac{(1+\tan^2 \theta)\, d\theta}{1}, \\
\text{so that}\;
\frac{dy}{d\theta} &= 1 + \tan^2\theta, \\
\text{or}\;
\frac{dy}{d\theta} &= \sec^2 \theta.
\end{align*}
<p>Collecting these results, we have:
<table>
<tr><th>$y$</th><th>$\dfrac{dy}{d\theta}$</th></tr>
<tr><td>$\sin\theta$</td><td>$\cos\theta$</td></tr>
<tr><td>$\cos\theta$</td><td>$-\sin\theta$</td></tr>
<tr><td>$\tan\theta$</td><td>$\sec^2\theta$</td></tr>
</table>
<p>Sometimes, in mechanical and physical questions,
as, for example, in simple harmonic motion and in
wave-motions, we have to deal with angles that increase
in proportion to the time. Thus, if $T$ be the
time of one complete <em>period</em>, or movement round the
circle, then, since the angle all round the circle is $2\pi$ radians,
or $360°$, the amount of angle moved through
in time $t$, will be
\begin{align*}
\theta &= 2\pi\frac{t}{T},\quad \text{in radians,} \\
\text{or}\;
\theta &= 360\frac{t}{T},\quad \text{in degrees.}
\end{align*}
<p>If the <em>frequency</em>, or number of periods per second,
be denoted by $n$, then $n = \dfrac{1}{T}$, and we may then write:
\[
\theta=2\pi nt.
\]
Then we shall have
\[
y = \sin 2\pi nt.
\]
<p>If, now, we wish to know how the sine varies with
respect to time, we must differentiate with respect, not
to $\theta$, but to $t$. For this we must resort to the artifice
explained in <a href="9.html">Chapter IX</a>., and put
\[
\frac{dy}{dt} = \frac{dy}{d\theta} · \frac{d\theta}{dt}.
\]
<p>Now $\dfrac{d\theta}{dt}$ will obviously be $2\pi n$; so that
\begin{align*}
\frac{dy}{dt} &= \cos \theta × 2\pi n \\
&= 2\pi n · \cos 2\pi nt. \\
\end{align*}
Similarly, it follows that
\begin{align*}
\frac{d(\cos 2\pi nt)}{dt} &= -2\pi n · \sin 2\pi nt.
\end{align*}
<p>
<h2>Second Differential Coefficient of Sine or Cosine.</h2>
<p>We have seen that when $\sin \theta$ is differentiated with
respect to $\theta$ it becomes $\cos \theta$; and that when $\cos \theta$ is
differentiated with respect to $\theta$ it becomes $-\sin \theta$;
or, in symbols,
\[
\frac{d^2(\sin \theta)}{d\theta^2} = -\sin \theta.
\]
<p>So we have this curious result that we have found
a function such that if we differentiate it twice over,
we get the same thing from which we started, but
with the sign changed from $+$ to $-$.
<p>The same thing is true for the cosine; for differentiating
$\cos\theta$ gives us $-\sin\theta$, and differentiating
$-\sin\theta$ gives us $-\cos\theta$; or thus:
\[
\frac{d^2(\cos\theta)}{d\theta^2} = -\cos\theta.
\]
<p><em>Sines and cosines are the only functions of which
the second differential coefficient is equal (and of
opposite sign to) the original function.</em>
<p><hr>
<p><em>Examples</em><a name="intex3"/>
With what we have so far learned we can now
differentiate expressions of a more complex nature.
<p>(1) $y=\arcsin x$.
<p>If $y$ is the arc whose sine is $x$, then $x = \sin y$.
\[
\frac{dx}{dy}=\cos y.
\]
<p>Passing now from the inverse function to the original
one, we get
\begin{align*}
\frac{dy}{dx}
&= \frac{1}{\;\dfrac{dx}{dy}\;} = \frac{1}{\cos y}. \\
\text{Now}\;
\cos y
&= \sqrt{1-\sin^2 y}=\sqrt{1-x^2}; \\
\text{hence}\;
\frac{dy}{dx}
&= \frac{1}{\sqrt{1-x^2}},
\end{align*}
a rather unexpected result.
<p>(2) $y=\cos^3 \theta$.
<p>This is the same thing as $y=(\cos \theta)^3$.
<p>Let $\cos\theta=v$; then $y=v^3$; $\dfrac{dy}{dv}=3v^2$.
\begin{align*}
\frac{dv}{d\theta} &= -\sin\theta.\\
\frac{dy}{d\theta} &= \frac{dy}{dv} × \frac{dv}{d\theta}
= -3 \cos^2 \theta \sin\theta.
\end{align*}
<p>(3) $y=\sin(x+a)$.
<p>Let $x+a=v$; then $y=\sin v$.
\[
\frac{dy}{dv}=\cos v;\qquad
\frac{dv}{dx}=1 \quad\text{and}\quad
\frac{dy}{dx}=\cos(x+a).
\]
<p>(4) $y=\log_\epsilon \sin \theta$.
<p>Let $\sin\theta=v$; $y=\log_\epsilon v$.
\begin{align*}
\frac{dy}{dv} &= \frac{1}{v};\quad \frac{dv}{d\theta}=\cos\theta;\\
\frac{dy}{d\theta} &= \frac{1}{\sin\theta} × \cos\theta = \cot\theta.
\end{align*}
<p>(5) $y=\cot\theta=\dfrac{\cos\theta}{\sin\theta}$.
\begin{align*}
\frac{dy}{d\theta}
&= \frac{-\sin^2\theta - \cos^2 \theta}{\sin^2 \theta}\\
&= -(1+\cot^2 \theta) = -\text{cosec}^2 \theta.
\end{align*}
<p>(6) $y=\tan 3\theta$.
<p>Let $3\theta=v$; $y=\tan v$; $\dfrac{dy}{dv}=\sec^2 v$.
\[
\frac{dv}{d\theta}=3;\quad
\frac{dy}{d\theta}=3 \sec^2 3\theta.
\]
<p>(7) $y = \sqrt{1+3\tan^2\theta}$; $y=(1+3 \tan^2 \theta)^{\frac{1}{2}}$.
<p>Let $3\tan^2\theta=v$.
\begin{align*}
y &= (1+v)^{\frac{1}{2}};\quad
\frac{dy}{dv} = \frac{1}{2\sqrt{1+v}}
\end{align*}
(see <a href="9.html#ExNo1">here</a>);
\begin{align*}
\frac{dv}{d\theta}
&= 6\tan\theta \sec^2 \theta \\
\end{align*}
for, if $\tan \theta = u$,
\begin{align*}
v &= 3u^2;\quad \frac{dv}{du} = 6u;\quad \frac{du}{d\theta} = \sec^2 \theta; \\
hence
\frac{dv}{d\theta}
&= 6 (\tan \theta \sec^2 \theta) \\
hence
\frac{dy}{d\theta}
&= \frac{6\tan\theta \sec^2\theta}{2\sqrt{1 + 3\tan^2\theta}}.
\end{align*}
<p>(8) $y=\sin x \cos x$. <a name="example1"/>
\begin{align*}
\frac{dy}{dx}
&= \sin x(-\sin x) + \cos x × \cos x \\
&= \cos^2 x - \sin^2 x.
\end{align*}
<p>
<hr><h3>Exercises XIV</h3>
(1) Differentiate the following:
\begin{align*}
\text{(i)}\quad y &= A \sin\left(\theta - \frac{\pi}{2}\right).\\
\text{(ii)}\quad y &= \sin^2 \theta;\quad \text{and } y = \sin 2\theta.\\
\text{(iii)}\quad y &= \sin^3 \theta;\quad \text{and } y = \sin 3\theta.
\end{align*}
<p>(2) Find the value of $\theta$ for which $\sin\theta × \cos\theta$ is a
maximum.
<p>(3) Differentiate $y=\dfrac{1}{2\pi} \cos 2\pi nt$.
<p>(4) If $y = \sin a^x$, find $\dfrac{dy}{dx}$.
<p>(5) Differentiate $y=\log_\epsilon \cos x$.
<p>(6) Differentiate $y=18.2 \sin(x+26°)$.
<p>(7) Plot the curve $y=100 \sin(\theta-15°)$; and show
that the slope of the curve at $\theta = 75°$ is half the
maximum slope.
<p>(8) If $y=\sin \theta·\sin 2\theta$, find $\dfrac{dy}{d\theta}$.
<p>(9) If $y=a·\tan^m(\theta^n)$, find the differential coefficient
of $y$ with respect to $\theta$.
<p>(10) Differentiate $y=\epsilon^x \sin^2 x$.
<p>(11) Differentiate the three equations of Exercises XIII.
(<a href="14b.html">here</a>), No. 4, and compare their differential
coefficients, as to whether they are equal, or nearly
equal, for very small values of $x$, or for very large
values of $x$, or for values of $x$ in the neighbourhood
of $x=30$.
<p>(12) Differentiate the following:
\begin{align*}
\text{(i)}\quad y &= \sec x. \\
\text{(ii)}\quad y &= \arccos x. \\
\text{(iii)}\quad y &= \arctan x. \\
\text{(iv)}\quad y &= \text{arcsec} x. \\
\text{(v)}\quad y &= \tan x × \sqrt{3 \sec x}. &&
\end{align*}
<p>(13) Differentiate $y=\sin(2\theta +3)^{2.3}$.
<p>(14) Differentiate $y=\theta^3+3 \sin(\theta+3)-3^{\sin \theta} - 3^\theta$.
<p>(15) Find the maximum or minimum of $y=\theta \cos \theta$.
<p><h3 class="answers">Answers</h3>
<p>(1) (i) $\dfrac{dy}{d\theta} = A \cos \left( \theta - \dfrac{\pi}{2} \right)$;
<p>(ii) $\dfrac{dy}{d\theta} = 2\sin\theta \cos\theta = \sin2\theta$ and $\dfrac{dy}{d\theta} = 2\cos2\theta$;
<p>(iii) $\dfrac{dy}{d\theta} = 3\sin^2 \theta \cos\theta$ and $\dfrac{dy}{d\theta} = 3\cos3\theta$.
<p>(2) $\theta = 45°$ or $\dfrac{\pi}{4}$ radians.
<p>(3) $\dfrac{dy}{dt} = -n \sin 2\pi nt$.
<p>(4) $a^x \log_\epsilon a \cos a^x$.
<p>(5) $\dfrac{\cos x}{\sin x} = \text{cotan}\; x$
<p>(6) $18.2 \cos \left(x + 26° \right)$.
<p>(7) The slope is $\dfrac{dy}{d\theta} = 100\cos\left(\theta - 15° \right)$, which is a maximum
when $(\theta -15°) = 0$, or $\theta = 15°$; the value of the slope
being then ${}= 100$. When $\theta = 75°$ the slope is
$100\cos(75° - 15°) = 100\cos 60° = 100 × \frac{1}{2} = 50$.
<p>(8) $\begin{aligned}[t]
\cos\theta \sin2\theta + 2\cos2\theta \sin\theta
&= 2\sin\theta\left(\cos^2 \theta + \cos2\theta\right) \\
&= 2\sin\theta\left(3\cos^2 \theta - 1\right).
\end{aligned}$
<p>(9) $amn\theta^{n-1} \tan^{m-1}\left(\theta^n\right)\sec^2 \theta^n$.
<p>(10) $\epsilon^x \left(\sin^2 x + \sin2x\right)$; $\epsilon^x \left(\sin^2 x + 2\sin2x + 2\cos2x\right)$.
<p>(11) $\left(i\right) \dfrac{dy}{dx} = \dfrac{ab}{\left(x + b\right)^2}$;
(ii) $\dfrac{a}{b} \epsilon^{-\frac{x}{b}}$;
(iii) $\dfrac{1}{90}° × \dfrac{ab}{\left(b^2 + x^2\right)}$.
<p>(12) (i) $\dfrac{dy}{dx} = \sec x \tan x$;
(ii) $\dfrac{dy}{dx} = - \dfrac{1}{\sqrt{ 1 - x^2}}$;
(iii) $\dfrac{dy}{dx} = \dfrac{1}{ 1 + x^2}$;
(iv) $\dfrac{dy}{dx} = \dfrac{1}{x \sqrt{ x^2 - 1}}$;
(v) $\dfrac{dy}{dx} = \dfrac{\sqrt{ 3\sec x} \left(3\sec^2 x - 1\right)}{2}$.
<p>(13) $\dfrac{dy}{d\theta} = 4.6\left(2\theta + 3\right)^{1.3} \cos\left(2\theta + 3\right)^{2.3}$.
<p>(14) $\dfrac{dy}{d\theta} = 3\theta^2 + 3\cos \left( \theta + 3 \right) - \log_\epsilon 3 \left( \cos\theta × 3^{\sin\theta} + 3\theta \right)$.
<p>(15) $\theta = \cot\theta; \theta = ±0.86$; is max. for $+\theta$, min. for $-\theta$.
<br>
<hr>
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