15.3 Differential Laws
Let's take a look at differential laws for cosine and sine.
Laws.
The differential law for cosine states
\[
\begin{align}
\diff (\cos \ang) &= - \sin \ang \mult \diff \ang
\end{align}
\]
and the differential law for sine states
\[
\begin{align}
\diff (\sin \ang) &= \cos \ang \mult \diff \ang
\end{align}
\]
where \( \ang : \Number \) is an angle.
Why?
Consider the following path that travels counterclockwise around the unit circle.
\[
\begin{align}
p &: \Path_{(x, y)}
\\
&\left\{\begin{aligned} x &= \cos t \\ y &= \sin t \end{aligned}\right.
\end{align}
\]
Since we write our angles in radians, our path's tracing point moves with a constant speed of one. Let's draw the position vector and the velocity vector in the plane. The velocity vector must be tangent to the circle, have magnitude one, and be pointing in the counterclockwise direction.
A little geometry allows us to relate the parts of the velocity vector to the parts of the position vector.
\[
\begin{align}
\position p &: \Vector_{(x, y)}{}
&
&
&
\velocity p &: \Vector_{(x, y)}{}
\\
\position p &= x \mult \vec{\standard}_{x} + y \mult \vec{\standard}_{y}
&
&\andSpaced
&
\velocity p &= - y \mult \vec{\standard}_{x} + x \mult \vec{\standard}_{y}
\end{align}
\]
The velocity vector is the \( t \)-derivative of the position vector, and so
\[
\begin{align}
\overset{t}{\diff} x &= - y
&
&\andSpaced
&
\overset{t}{\diff} y &= x.
\end{align}
\]
The differential laws follow.
Let's use our differential laws for sine and cosine to calculate something!
Example.
Let's find the derivative for the function
\[
\begin{align}
z &\depends \ang
\\
z &= \frac{\sin \ang}{\cos \ang}.
\end{align}
\]
We compute.
\[
\begin{align}
\overset{\ang}{\diff} \biggl(\frac{\sin \ang}{\cos \ang}\biggr) &= \frac{1}{\cos \ang} \mult \overset{\ang}{\diff} (\sin \ang) - \frac{\sin \ang}{(\cos \ang)^2} \mult \overset{\ang}{\diff} (\cos \ang)
\\
&= \frac{1}{\cos \ang} \mult \cos \ang \mult \overset{\ang}{\diff} \ang - \frac{\sin \ang}{(\cos \ang)^2} \mult (- \sin \ang) \mult \overset{\ang}{\diff} \ang
\\
&= 1 + \frac{(\sin \ang)^2}{(\cos \ang)^2}
\\
&= \frac{1}{(\cos \ang)^2}
\end{align}
\]
We've just calculated the derivative of another famous function.
\[
\begin{align}
\tan \ang &= \frac{\sin \ang}{\cos \ang}
\end{align}
\]
Although they have similar names, the trigonometric tangent function, \( \tan \ang \), should not be confused with the tangent metric.