Trigonometry 2

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 \[\begin{equation}\tag{*} (\cos\theta)^2+(\sin\theta)^2=1. \end{equation}\]
    • 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

Exponential functions

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 \(b\).