2. Powers, logarithms and trigonometry

Part 1.: problems, solutions.
Part 2.: problems, solutions.
Part 3.: problems, solutions.

Exponentiation

Everybody knows what is a^n, where n is a natural number. It is simply multiplying a by itself n times (the only doubt may be caused by n=0, we set a^0=1). If z is a negative integer, we let a^z=1/a^{|z|}. Finally let a^{1/n}=\sqrt[n]{a}, is a number such that when we take it to power of n we get a. Notice that, if n is an even number, we have to assume that a\geq 0. To sum up, easily for a\geq 0 we can define a^q, where q is a rational q=t/s, t,s\in \mathbb{Z}, as \sqrt[s]{a^t}, if q\geq 0 and \frac{1}{\sqrt[s]{a^t}} in the other case.

These definitions result in many natural properties of exponentiation, which we know already when using only natural powers, namely for a,b>0 and x,y\in \mathbb{Q} we have:

  1. 1^x=1,
  2. a^{x+y}=a^x\cdot a ^y,
  3. a^{x-y}=a^x/ a ^y,
  4. a^0=1, a^1=a,
  5. \left(a^x\right)^y=a^{xy},
  6. a^{-x}=1/a^x,
  7. (ab)^x=a^xb^x,
  8. jeśli a>1 i x<y, to a^x<a^y,
  9. jeśli 0<a<1 i x<y, to a^x>a^y.

Moreover, the last two properties allow us to define a^x not only for x\in\mathbb{Q}, but for any real number x. Roughly speaking, x can be approximated from below and from above by rational numbers q_0 and q_1. Then a^{q_0} and a^{q_1} approximate (from above and from below or from below and from above) a^x. So we define a^x is such a way that this holds. After including necessary details, one can in particular prove that this definition also satisfies the properties mentioned above. And also the following property — if a\neq 1, then for every y\in \mathbb{R} there exists exactly one x\in \mathbb{R}, such that y=a^x.

Logarithms

The property, that for every y\in \mathbb{R} exists exactly one x\in \mathbb{R}, such that y=a^x allows us to define the logarithm. Simply this number x (it is the power such that a raised to this power gives y) is called a logarithm of y with base a and is denoted by \log_a x. In other words, a^{\log_a x}=x=\log_{a} a^x. So the logarithm is an inverse function to exponentiation.

Notice also that \log_a x makes sense only if a,x>0 and a\neq 1.

Since 10, 2, and (and least in the future) number e (whatever it is, we will explain it later) take a special place in our heart, we denote \log_{10} x=\log x, \log_2 x= \lg x and \log_{e}x = \ln x.

All the mentioned above properties of exponentiation can be easily translated into properties of logarithms:
for all a,b>0, a,b\neq 1 and x,y\in \mathbb{R} we have:

  1. x,y>0, then \log_a xy=\log_a x+\log_a y,
  2. x,y>0, then \log_a \frac{x}{y}=\log_a x-\log_a y,
  3. \log_a 1=0, \log_a a=1,
  4. x>0, then \log_a x^y=y\log_a x,
  5. x>0, then \log_a \frac{1}{x}=-\log_a x,
  6. x>0, then \log_a x=\frac{\log_b x}{\log_b a}, i.e. \log_a x \cdot \log_b a=\log_b x,
  7. if a>1 and 0<x<y, then \log_a x<\log_a y,
  8. if 0<a<1 and 0<x<y, then \log_a x>\log_a y.

And also, for any y and a>0, a\neq 1 there exists exactly one x, such that y=\log_a x.

Due to these properties logarithms were the main tool in scientifical computation before computers. Now, they are also a very useful tool. E.g. in chemistry pH is defined as minus logarithm of the concentration of hydrogen ions. So increasing this concentration tenfold changes pH by one, which makes it easier to describe and imagine (we look at the order of a number instead of the number itself). You can also find such a scale in the definition of brightness of stars, Richter scale, and also some graphs related to covid…

Trigonometry

Except for degrees there is one more unit of angles, more widely used in sciences, i.e. radians. Radians describe the length of a section of a unit circle related to the given angle. So 360^\circ is 2\pi, 90^\circ is \pi/2, and e.g. \pi/3 is 60^\circ. In case of radians we do not use any suffix like ^\circ.

Notice that in a triangle with fixed angles regardless of its size the ratio of the sides remains constant. In particular in a right triangle with sides a,b (by the right angle) and c and the angle opposite to a equal \alpha the ratio a/c, b/c, a/b, and b/a depend only on this angle. They are thus functions of \alpha. And are called \sin \alpha, \cos \alpha, \text{tg}\alpha and \text{ctg }\alpha respectively. Immediately we notice that \text{tg} \alpha = \sin \alpha/\cos \alpha and \text{ctg} \alpha = \cos \alpha/ \sin\alpha=1/\text{tg}\alpha.

Imagining a point (x,y) on the unit circle we see that (x,y)=(\cos \alpha, \sin \alpha), where \alpha is the angle between the radius of the circle in this point and axis x. This implies many properties and formulas concerning these functions.

For example, each of these functions is periodic, i.e. \sin \alpha = \sin (\alpha+2k\pi) and \cos \alpha = \cos (\alpha+2k\pi) for any k\in \mathbb{Z}.

From Pythagoras Theorem we get that \sin^2 \alpha+\cos^2 \alpha = 1 (which is called the trigonometric unit, \sin^2\alpha denotes (\sin \alpha)^2).

Looking at the unit triangle we can also easily notice that \sin(\alpha)=-\sin(-\alpha) and \cos \alpha=\cos (-\alpha).

When rotated by \pi/2 the point (x,y) becomes (-y,x), so \cos\alpha=\sin(\alpha+\pi/2) and \sin \alpha=-\cos(\alpha+\pi/2).

When rotated by \pi the point (x,y) becomes (-x,-y), so \cos\alpha=-\cos(\alpha+\pi) and \sin \alpha=-\sin(\alpha+\pi).

Notice that the section of the unit circle between the point described by angle \alpha-\beta and the point described by angle \alpha and the section of the unit circle between the points described by angles 0 and \beta are both of angle \beta, so they are of the same length. So the cords are also of the same length, and writing out their length from Pythagoras Theorem, we get:

    \[(\cos (\alpha-\beta)-1)^2+(\sin(\alpha-\beta)-0)^2=(\cos\alpha-\cos \beta)^2+(\sin \alpha-\sin \beta)^2.\]

Which gets us to:

    \[\cos (\alpha-\beta)=\cos \alpha\cos \beta + \sin \alpha \sin \beta,\]

so also

    \[\cos (\alpha+\beta)=\cos \alpha\cos \beta - \sin \alpha \sin \beta,\]

which implies

    \[\sin (\alpha+\beta)=\sin \alpha \cos \beta +\cos \alpha \sin \beta\]

and

    \[\sin (\alpha+\beta)=\sin \alpha \cos \beta -\cos \alpha \sin \beta.\]

Thus, \sin 2\alpha=2\sin\alpha\cos \alpha and \cos 2\alpha =\cos^2\alpha-\sin^2 \alpha.