# Derivatives of trigonometric functions

# Correction about \(G(t)=\frac{1-2t}{3+t}\).

- I should have really calculated those infinite limits
- Let's look at the situation as \(t\to(-3)^-\).
- The nominator \(1-2t\) has a nonzero, positive value at \(t=-3\).
- As \(t\) approaches \((-3)\) from the left, the denominator \(3+t\) gets negative values indefinitely close to zero.
- Therefore, the fractional \(G(t)=\frac{1-2t}{3+t}\) gets indefinitely small as \(t\to(-3)^-\), which means \(\lim_{t\to(-3)^-}G(t)=-\infty\).
- The shape of the graph confirms this.

# Derivatives of \(\sin(x)\) and \(\cos(x)\)

- If you're interested, you can find a geometric argument for these derivatives in the textbook.
- We'll apply Euler formula magic, which will be much quicker!
- Of course, you just have to remember the formulas themselves, but you have to see how much shorter this proof is.
- Recall that we have the imaginary number \(i\), which has the wonderful property \(i^2=-1\).
*Euler's formula* was the following. For a real number \(x\), we have that \[
e^{ix}=\cos(x)+i\sin(x).
\]
- Let's derivate this exponential function.
- In general, for a constant \(a\), we have \(\frac{\mathrm d}{\mathrm dx}e^{ax}=ae^{ax}\).
- In this particular case, we get \[
(e^{ix})'=ie^{ix}.
\]
- Applying Euler's formula on the left side, we get \[
(e^{ix})'=(\cos(x)+i\sin(x))'=\cos'(x)+i\sin'(x).
\]
- On the right side, we get \[
ie^{ix}=i(\cos(x)+i\sin(x))=i\cos(x)+i^2\sin(x)=i\cos(x)-\sin(x).
\]

# Derivatives of \(\sin(x)\) and \(\cos(x)\)

- We equate what we've gotten on the two sides to get: \[
\cos'(x)+i\sin'(x)=-\sin(x)+i\cos(x).
\]
- VoilĂ ! Equating the real parts gives \[
\cos'(x)=-\sin(x),
\] and equating the imaginary parts gives \[
\sin'(x)=\cos(x).
\]

- Note that these derivatives give limits we couldn't verify before.
- Recall that we've made the guess \(\lim_{x\to0}\frac{\sin(x)}{x}=1\) based on numerical evidence.
- Note that we have \[
\lim_{x\to 0}\frac{\sin(x)}{x}=\sin'(x=0).
\]
- And now we know that. \[
\sin'(x=0)=\cos(x=0)=1.
\]
- Exercise. Find \(\lim_{x\to0}\frac{\cos(x)-1}{x}\).

# Derivatives of the other trigonometric functions.

- Recall the quotient rule: \[
\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}.
\]
- Using this, we get the following. \[
\tan'(x)=\left(\frac{\sin(x)}{\cos(x)}\right)'=\frac{\sin'(x)\cos(x)-\sin(x)\cos'(x)}{\cos(x)^2}
=\frac{\cos(x)^2+\sin(x)^2}{\cos(x)^2}=\frac{1}{\cos(x)^2}.
\]
- Exercises. 3.3. 2, 4, 12, 16, 34, 35, 40, 42, 44, 50