Drill 6.3.1.
Use the Quotient Rule to show that \(\dx{(\cot(x))} = -\csc^2(x)\dx{x}. \)
Section 6.3 The Differentials of the Other Trigonometric Functions
Once the differential of \(\sin(x)\) is known, the differentials of the other trigonometric functions are easily computed. Since our current goal is simply to develop the tools we will need later we will not spend any more time on this than necessary.
Observe that by definition \(\tan(x)=\frac{\sin(x)}{\cos(x)}\text{,}\) so the Quotient Rule applies:
\begin{align*}
\dx\left(\tan(x)\right) \amp = \dx{\left(\frac{\sin(x)}{\cos(x)}\right)}\\
\amp = \frac{\cos(x)\dx(\sin(x))-\sin(x)\dx(\cos(x))}{\cos^2(x)}\\
\amp = \frac{\cos^2(x)\dx{x}
+\sin^2(x)\dx{x}}{\cos^2(x)}\\
\amp = \frac{1}{\cos^2(x)}\dx{x}
\end{align*}
and since \(\sec(x) = 1/\cos(x)\) we have
\begin{equation*}
\dx{\left(\tan(x)\right)} = \sec^2(x)\dx{x}.
\end{equation*}
Observe that by definition \(\sec(x) = \frac{1}{\cos x} = \inverse{(\cos(x))}\text{.}\) Differentiating, we have
\begin{equation*}
\dx(\sec(x)) = \dx\left(\inverse{(\underbrace{\cos(
x)}_{=z})}\right),
\end{equation*}
and making the substitution \(z=\cos(x)\) as indicated we have \(\dx(\sec(x)) = \dx(\inverse{z})\text{.}\) By the Power Rule this is
\begin{equation*}
\dx(\sec(x)) = -z^{-2}\dx{z} =-\left(\cos(x)\right)^{-2}\dx(\cos(x)) =
-(\cos(x))^{-2}(-\sin(x))\dx{x}.
\end{equation*}
Simplifying gives
\begin{equation*}
\dx(\sec(x)) = \frac{\sin(x)}{\cos^2(x)}\dx{x}
= \frac{1}{\cos(x)}\cdot\frac{\sin(x)}{\cos(x)}\dx x
\end{equation*}
so that, finally we have
\begin{equation*}
\dx{(\sec(x))} = \sec(x)\tan(x)\dx{x}.
\end{equation*}
Drill 6.3.2.
By similar means show that \(\dx{(\csc(x))} = -\csc(x)\cot(x)\dx{x}\text{.}\)
Remember all of those identities you had to memorize in Trigonometry? You can reduce that memorization burden a bit by using Calculus. If two variable quantities are equal then their differentials must be equal too. So if you differentiate both sides an identity you get another identity!
Example 6.3.3.
Consider the double angle formula for the sine function:
\begin{equation*}
\sin(2x)=2\sin(x)\cos(x)\text{.}
\end{equation*}
Differentiating gives
\begin{equation*}
\cos(2x)\dx x = \cos^2(x)\dx x -\sin^2(x)\dx x
\end{equation*}
or
\begin{equation}
\cos(2x)
=\cos^2(x)-\sin^2(x).\tag{6.11}
\end{equation}
which is the double angle formula for the cosine function.
Example 6.3.4.
We know that \(\sin^2(x)+\cos^2(x)=1\) no matter what value \(x\) has, so we’d expect that differential of \(\sin^2(x)+\cos^2(x)\) to be zero. Let’s check.
By the Sum Rule
\begin{equation*}
\dx(\sin^2(x)+\cos^2(x)) = \dx(\sin^2(x)) + \dx(\cos^2(x))
\end{equation*}
and by the Power Rule we have
\begin{align*}
\dx(\sin^2(x)+\cos^2(x)) \amp = 2\sin(x)\dx(\sin(x)) + 2\cos(x)\dx(\cos(x))\\
\amp = 2\sin(x)\cos(x)\dx{x} + 2\cos(x)(-\sin(x))\dx x\\
\amp = 2(\sin(x)\cos(x) - \sin(x)\cos(x))\dx x\\
\amp = 0.
\end{align*}
Drill 6.3.5.
Since the sine and cosine functions are both differentiable and \(\dx{(\sin^2(x)+\cos^2(x))}=0\) we can conclude that \(\sin^2(x)+\cos^2(x)\) is (probably) equal to some constant. If we didn’t already know, how could we conclude that the constant is \(1?\)
We now know the differentials of all of the trigonometric functions. These are shown in the table below. Memorize them.
| Function | Differential |
| \(\sin(x)\) | \(\cos(x)\dx{x}\) |
| \(\cos(x) \) | \(-\sin(x)\dx{x} \) |
| \(\tan(x) \) | \(\sec^2(x)\dx{x} \) |
| \(\cot(x) \) | \(-\csc^2(x)\dx{x} \) |
| \(\sec(x) \) | \(\sec(x)\tan(x)\dx{x} \) |
| \(\csc(x) \) | \(-\csc(x)\cot(x)\dx{x} \) |
Drill 6.3.7.
Differentiate both sides of each trigonometric identity to get another identity. Verify each identity (including the one you compute) by graphing the expression on both sides of the equals sign.
- \(\displaystyle \cos(2\theta) =\cos^2(\theta)-\sin^2(\theta)\)
- \(\displaystyle \tan(2\theta)=\frac{2\tan(\theta)}{1-\tan^2(\theta)}\)
- \(\displaystyle \sec(2\theta)=\frac{1}{\cos^4(\theta)-\sin^4(\theta)}\)
- \(\displaystyle \cos^2\left(\frac{\theta}{2}\right)= \frac12(1+\cos(\theta))\)
- \(\displaystyle \sin^2\left(\frac{\theta}{2}\right)=\frac12(1-\cos(\theta))\)
- \(\displaystyle \tan\left(\frac{\theta}{2}\right)=\frac{\sin(\theta)}{1+\cos(\theta)}\)
Drill 6.3.8.
If possible find an equation of the line tangent to the graphs of \(y=\tan(x)\) and \(y=\sec(x)\) at each of the points below. If no such line exists explain why not.
- \(\displaystyle x=0\)
- \(\displaystyle x=\pm\frac{\pi}{6}\)
- \(\displaystyle x=\pm\frac{\pi}{4}\)
- \(\displaystyle x=\pm\frac{\pi}{3}\)
- \(\displaystyle x=\pm\frac{\pi}{2}\)
- \(\displaystyle x=\pm\frac{2\pi}{3}\)
Problem 6.3.9.
Show that differentiating each of the identities below leads to the other. Assume \(A\) is a constant.
(a)
\(\sin(\theta+ A)=\sin(\theta)\cos(A)+ \cos(\theta)\sin(A)\)
(b)
\(\cos(\theta+ A)=\cos(\theta)\cos(A)- \sin(\theta)\sin(A)\)
Problem 6.3.10.
Compute \(\dx{y},\) for each of the following functions:
(a)
\(y=\tan(10x)\)
(b)
\(y=\cot(\pi x)\)
(c)
\(y=\sec(2\pi x)\)
(d)
\(y=\csc(-\pi x)\)
(e)
\(y=\sqrt{x^2+1}\sec(x^2+x)\)
(f)
\(y=\csc \left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)\)
(g)
\(\sin(y)=\tan(x)\)
(h)
\(\sec^2(y)=\frac{1}{x^{3/2}} +\cos(y)\)
(i)
\(y=\sqrt{\csc^2(x)-1}\)
(j)
\(y=\tan \left( \sqrt{3x-7} \right)\cot \left( \sqrt{3x-7}
\right)\)
Problem 6.3.11.
Assume that \(x=x(t)\) and \(y=y(t)\text{.}\) Find an equation relating \(\dx{x}\) and \(\dx{y}\text{.}\) Use this to compute \(\dfdx{y}{x}\text{,}\) \(\dfdx{y}{t}\text{,}\) \(\dfdx{x}{y}\text{,}\) and \(\dfdx{x}{t}\text{.}\)
(a)
\(\tan(x^2+y)=y\sec(x)\)
(b)
\(y^2\sqrt{1+\csc^2(x)}=y^2+x\)
(c)
\(\cot(y)=x\)
(d)
\(\sec^2(y+x)=y\csc(x^2)\)
Problem 6.3.12.
Find an equation of the tangent line to each curve at the indicated point.
(a)
\(\tan(y)=x^2-x+1\) at \((1,\pi/4)\)
(b)
\(\cot^2(y)=x^2+x+3\) at \((0,\pi/6)\)
(c)
\(\csc(y)=x^2+x+2\) at \((0, \pi/6)\)
Problem 6.3.13.
(a)
Show that the line tangent to the curve \(y=\tan(x)\) at \((x_0,y_0)\) is parallel to the line tangent to the curve at \((-x_0,-y_0).\)
(b)
Show that the line tangent to the curve \(y=\cot(x)\) at \((x_0,y_0)\) is parallel to the line tangent to the curve at \((-x_0,-y_0).\)
Drill 6.3.14.
(a)
Show that there is no line tangent to the graph of \(y=\tan(x)\) which is parallel to any tangent line of the graph of \(y=\cot(x).\)
(b)
Show that this is not true of the graphs of \(y=x^3\) and \(y=-x^3\text{.}\)
Problem 6.3.15.
A camera located at \(C\) at ground level is tracking a rocket \(R\) which is traveling vertically and took off from a spot \(500\) meters from the camera.
(a)
How fast is the angle of elevation of the camera changing (in radians per second) when the rocket is \(1000\) meters high and traveling at \(250 \frac{\text{meters}}{\text{second}}\text{?}\)
(b)
Now suppose the rocket is climbing at an angle \(\pi/6\) radians off of vertical as shown in the diagram below.
How fast is the angle of elevation of the camera changing (in radians per second) when the rocket is \(1000\) meters high and traveling at \(250 \frac{\text{meters}}{\text{second}}\text{?}\)
Problem 6.3.16.
(a)
Use the Difference formulas for the sine and cosine from Trigonometry:
\begin{align*}
\sin(\alpha- \beta)\amp =\sin(\alpha)\cos(\beta)- \cos(\alpha)\sin(\beta)\\
\cos(\alpha- \beta)\amp =\cos(\alpha)\cos(\beta)+ \sin(\alpha)\sin(\beta),
\end{align*}
to show that
\begin{equation*}
\tan(\alpha-\beta)=\frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}.
\end{equation*}
(b)
Consider two points, \(P\) and \(Q\text{,}\) moving upward on the line \(x=1\text{,}\) with \(P\) above \(Q\) as seen in the sketch below:
- Suppose \(P\) is moving up at a rate of \(5\) units per second and \(Q\) is moving up at a rate of \(3\) units per second. How fast is \(\theta\) increasing when \(p=20\) and \(q=10?\)
- Suppose \(Q\) is moving up at a rate of \(3\) units per second and we wanted the value of \(\theta\) to remain constant. How fast must \(P\) move?
