# 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: $\cos'(x)+i\sin'(x)=-\sin(x)+i\cos(x).$
• Voilà! 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