# Special values of the trigonometric functions -- $$\theta=\frac{\pi}{4}$$

• Recall the Pythagoras Theorem: for a right triangle with legs of length $$a$$ and $$b$$, and hypotenuse of length $$c$$, we have $$a^2+b^2=c^2$$.
• One way you can get this is $$1^2+1^2=(\sqrt2)^2$$. This right triangle will have both non-right angles $$\frac{\pi}{4}$$. This gives us the following special values: $\cos\tfrac{\pi}{4}=\tfrac{1}{\sqrt2},\,\sin\tfrac{\pi}{4}=\tfrac{1}{\sqrt2},\,\tan\tfrac{\pi}{4}=1.$

# Special values of the trigonometric functions -- $$\theta=\frac{\pi}{6},\frac{\pi}{3}$$

• Another way you can make the Pythagoras Theorem hold is $$1^2+(\sqrt3)^2=2^2$$. This right triangle will have non-right angles $$\frac{\pi}{6}$$ and $$\frac{\pi}{3}$$. This gives us the following special values: $\cos\tfrac{\pi}{6}=\tfrac{\sqrt3}{2},\,\sin\tfrac{\pi}{6}=\tfrac12,\,\tan\tfrac{\pi}{6}=\tfrac{1}{\sqrt3}$ $\cos\tfrac{\pi}{3}=\tfrac{1}{2},\,\sin\tfrac{\pi}{6}=\tfrac{\sqrt3}{2},\,\tan\tfrac{\pi}{6}=\sqrt3$
• Unfortunately, there's no other way but to remember these. But as I said when we were talking about the graphs, you only have to remember that there's the special angle values $$\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{3}$$ and the special cos/sin values $$\frac12,\frac{\sqrt2}{2},\frac{\sqrt3}{2}$$, and that $$\cos(x)$$ is decreasing in $$[0,\frac{\pi}{2}]$$, and $$\sin(x)$$ is increasing in $$[0,\frac{\pi}{2}]$$.

# Trigonometric identities -- from symmetries

• The trigonometric identities based on symmetries can be inferred from considering the parametrization $$f(\theta)=(\cos(\theta),\sin(\theta))$$, or the graphs of the trigonometric functions.
• We have already seen that $$\cos$$ and $$\sin$$ are $$2\pi$$-periodic: $\cos(\theta+2\pi)=\cos(\theta),\,\sin(\theta+2\pi)=\sin(\theta).$
• Changing from $$f(\theta)$$ to $$f(\theta+\pi)$$ amounts to rotation by $$\pi$$ radians, that is mirroring with respect to the origin, which is turn amounts to changing the signs of both coordinate functions. $\cos(\theta+\pi)=-\cos(\theta),\,\sin(\theta+\pi)=-\sin(\theta)$
• The following formulas can be inferred from considering the graphs. $\cos(\theta+\frac{\pi}{2})=-\sin(\theta),\,\sin(\theta+\frac{\pi}{2})=\cos(\theta)$
• Finally, note that changing the sign of $$\theta$$ amounts to mirroring with respect to the $$x$$-axis. Therefore, the effect on the coordinate functions of $$f(\theta)$$ is doing nothing to the $$x$$-coordinate, and changing the sign of the $$y$$-coordinate: $\cos(-\theta)=\cos(\theta),\,\sin(-\theta)=-\sin(\theta).$
• Because of the latter, we say that $$\cos(\theta)$$ is an even function, and that $$\sin(\theta)$$ is an odd function.
• Exercises: D.26.28

# Trigonometric identities -- from the unit circle equation

• Recall that the function $$f(\theta)=(\cos(\theta),\sin(\theta))$$ parametrizes the unit circle.
• Since the equation of the unit circle is $$x^2+y^2=1$$, this means that for all $$\theta$$, we have $$$\tag{*} (\cos\theta)^2+(\sin\theta)^2=1.$$$
• Dividing both sides by $$(\cos\theta)^2$$, we get $1+(\tan\theta)^2=(\sec\theta)^2$
• Similarly, dividing both sides of (*) by $$(\sin\theta)^2$$, we get $(\cot\theta)^2+1=(\csc\theta)^2.$

# Trigonometric identities -- from the addition formulas

• The trigonometric addition formulas are the following. \begin{align*} \cos(x+y)=&\cos(x)\cos(y)-\sin(x)\sin(y) \\ \sin(x+y)=&\sin(x)\cos(y)+\cos(x)\sin(y). \end{align*}
• If you are interested in the proof of these formulas, you can find a geometric proof in Exercises D.85-87. When we talk about exponential functions, I will give a much simpler proof using complex numbers (you don't have to know complex numbers in this class, I'll just show it as a way to remember the addition formulas).
• Using that $$\cos$$ is even and that $$\sin$$ is odd, the addition formulas give the subtraction formulas: \begin{align*} \cos(x-y)=&\cos(x)\cos(y)+\sin(x)\sin(y) \\ \sin(x-y)=&\sin(x)\cos(y)-\cos(x)\sin(y). \end{align*}
• Dividing the formulas we get: $\tan(x\pm y)=\frac{\sin(x\pm y)}{\cos(x\pm y)}=\frac{\sin(x)\cos(y)\pm\cos(x)\sin(y)}{\cos(x)\cos(y)\mp\sin(x)\sin(y)}=\frac{\tan(x)\pm\tan(y)}{1\mp\tan(x)\tan(y)}.$
• Putting $$x=y$$ in the addition formulas, we get the double-angle formulas: \begin{align*} \cos(2x)=&(\cos x)^2-(\sin x)^2 \\ \sin(2x)=&2\sin(x)\cos(x) \end{align*}
• Using the identity $$(\cos x)^2+(\sin x)^2=1$$ in the double-angle formula for $$\cos$$, we get the following alternative forms. \begin{align*} \cos(2x)=&(\cos x)^2-(1-(\cos x)^2)=2(\cos x)^2-1 \\ \cos(2x)=&(1-(\sin x)^2)-(\sin x)^2=1-2(\sin x)^2 \end{align*}

# Trigonometric identities -- from the addition formulas 2

• Once again, the addition and subtraction formulas are \begin{align*} \cos(x\pm y)=&\cos(x)\cos(y)\mp\sin(x)\sin(y) \\ \sin(x\pm y)=&\sin(x)\cos(y)\pm\cos(x)\sin(y), \end{align*} and the alternative double-angle formulas for $$\cos$$ are \begin{align*} \cos(2x)=&\cos(x)^2-(1-\cos(x)^2)=2\cos(x)^2-1 \\ \cos(2x)=&(1-(\sin x)^2)-(\sin x)^2=1-2(\sin x)^2. \end{align*}
• Rearranging the latter, we get the following half-angle formulas. \begin{align*} (\cos x)^2=&\frac{\cos(2x)+1}{2} \\ (\sin x)^2=&\frac{1-\cos(2x)}{2} \end{align*}
• Finally, rearranging the addition formulas, we get the following product formulas. \begin{align*} \cos(x)\cos(y)=&\frac{\cos(x+y)+\cos(x-y)}{2} \\ \sin(x)\cos(y)=&\frac{\sin(x+y)+\sin(x-y)}{2} \\ \sin(x)\sin(y)=&\frac{\cos(x-y)-\cos(x+y)}{2}. \end{align*}
• The half-angle and product formulas will be used frequently in integral calculus, since they are linearization formulas.
• Exercises: D.45, 47, 67

# Exponentiation by rational numbers

• An exponential function is a function of the form $$f(x)=b^x$$ for some fixed positive number $$b>0$$. Let's recall what the expression $$b^x$$ means.
• If $$n$$ is a positive integer, then $$b^n=b\cdot\ldots\cdot b$$, where the product has $$n$$ terms.
• We have $$b^0=1$$, and $$b^{-n}=\frac{1}{b^n}$$.
• If $$x=\frac{p}{q}$$ is a rational number, then we have $$b^{p/q}=\sqrt[q]{b^p}=(\sqrt[q]b)^p$$.

# Irrational numbers

• An irrational number $$c$$ is such that there are no integers such that $$c=\frac{p}{q}$$.
• For example, $$c=\sqrt2$$ is an irrational number.
• This is proven by contradiction. I will show the proof, because it's interesting to see how you do such a thing, but knowing it will not be required in this class.
• This means that we show that by supposing for contradiction that $$\sqrt2=\frac{p}{q}$$ for some integers $$p,q$$, we get something impossible.
• Squaring the equation $$\sqrt2=\frac{p}{q}$$, we get $$2=\frac{p^2}{q^2}$$, which yields $$2q^2=p^2$$.
• Let's see what the highest powers of $$2$$ are, which divide the two sides. For example this is 1 for 6, since $$2\mid6$$ but $$4\nmid6$$ (that is because, $$\frac{6}{2}=3$$ is an integer, but $$\frac{6}{4}=3.5$$ is not), and 3 for 40, since $$8\mid40$$, but $$16\nmid40$$.
• In general, for a squared integer, this highest power is always an even number.
• But that implies that this highest power is even for $$p^2$$, and odd for $$2q^2$$.
• Since $$p^2=2q^2$$, that is impossible.
• This shows that it is impossible that $$\sqrt2=\frac{p}{q}$$ for some integers $$p,q$$, that is, $$\sqrt2$$ is an irrational number.
• Another example of an irrational number is $$\pi$$ (the proof is a lot harder).

# Exponentiation by irrational numbers

• Since we have seen that there are numbers which are not rational, we still need to say what the values of an exponential function $$f(x)=b^x$$ are for irrational $$x$$.
• We will do this by making $$f(x)=b^x$$ continuous. We will learn about continuity in the next week. It means that the graph of $$f(x)$$ is one curve without jumps (you could draw it without lifting your pen).
• What we can use is that irrational numbers can be approximated to arbirary precision by rational numbers. That is, we can write approximations like $$\sqrt2\approx1.41421356237$$ up to an arbitrary number of decimals.
• This means that we have $$1.41<\sqrt2<1.42$$, $$1.414<\sqrt2<1.415$$, $$1.4142<\sqrt2<1.4143$$, etc.
• These equalities cannot be strict, since that would imply that $$\sqrt2$$ is rational. For example, $$1.41=\frac{141}{100}$$.
• We give $$3^\sqrt2$$ a value so that $$3^{1.41}\le3^{\sqrt2}\le3^{1.42}$$, $$3^{1.414}\le3^{\sqrt2}\le3^{1.415}$$, $$3^{1.4142}\le3^{\sqrt2}\le3^{1.4143}$$, etc.
• It can be proven that there is exactly one number like that, which makes the function $$f(x)=3^x$$ well defined.
• We do the same for $$f(x)=b^x$$ for other values of $$b$$.