4.3 Trigonometrische Eigenschaften

Aus Online Mathematik Brückenkurs 1

Version vom 08:12, 22. Okt. 2008

Introduction

There is a variety of trigonometric formulas to use if one wishes to transform between the sine, cosine or tangent for an angle or multiples of an angle. They are usually known as the trigonometric identities, since they only lead to different ways to describe a single expression using a variety of trigonometric functions. Here we will give some of these trigonometric relationships. There are many more than we can deal with in this course. Most can be derived from the so-called Pythagorean identity and the addition and subtraction formulas or identities (see below), which are important to know by heart.

The Pythagorean identity

Symmetries

With the help of the unit circle and reflection, and exploiting the symmetries of the trigonometric functions one obtains a large amount of relationships between the cosine and sine functions.

 \begin{align*}
   \cos (-v) &= \cos v\vphantom{\Bigl(}\\
   \sin (-v) &= - \sin v\vphantom{\Bigl(}\\
   \cos (\pi-v) &= - \cos v\vphantom{\Bigl(}\\
   \sin (\pi-v) &= \sin v\vphantom{\Bigl(}\\
 \end{align*}
 \qquad\quad
 \begin{align*}
   \cos \Bigl(\displaystyle \frac{\pi}{2} -v \Bigr) &= \sin v\\
   \sin \Bigl(\displaystyle \frac{\pi}{2} -v \Bigr) &= \cos v\\
   \cos \Bigl(v + \displaystyle \frac{\pi}{2} \Bigr) &= - \sin v\\
   \sin \Bigl( v + \displaystyle \frac{\pi}{2} \Bigr) &= \cos v\\
 \end{align*}

Instead of trying to learn all of these relationships by heart, it might be better to learn how to derive them from the unit circle.

Reflection in the x-axis

   \cos(-v) &= \cos v\,\mbox{,}\\
   \sin (-v) &= - \sin v\,\mbox{.}\\
 \end{align*}

Reflection in the y-axis

   \cos(\pi-v) &= -\cos v\,\mbox{,}\\
   \sin (\pi-v) &= \sin v\,\mbox{.}\\
 \end{align*}

Reflection in the line y = x

   \cos \Bigl(\frac{\pi}{2} - v \Bigr) &= \sin v\,\mbox{.}\\
   \sin \Bigl(\frac{\pi}{2} - v \Bigr) &= \cos v\,\mbox{.}\\
 \end{align*}

Rotation by an angle of \displaystyle \mathbf{\pi/2}

   \cos \Bigl(v+\frac{\pi}{2}\Bigr) &= -\sin v\,\mbox{,}\\
   \sin \Bigl(v+\frac{\pi}{2} \Bigr) &= \cos v\,\mbox{.}
 \end{align*}

Alternatively, one can get these relationships by reflecting and / or displacing graphs. For instance, if we want to have a formula in which \displaystyle \cos v is expressed in terms of a sine one can displace the graph for cosine to fit the sine curve. This can be done in several ways, but the most natural is to write \displaystyle \cos v = \sin (v + \pi / 2). To avoid mistakes, one can check that this is true for several different values of \displaystyle v.

Check: \displaystyle \ \cos 0 = \sin (0 + \pi / 2)=1.

The addition and subtraction formulas and double-angle and half-angle formulas

One often needs to deal with expressions in which two or more angles are involved, such as \displaystyle \sin(u+v). One will then need the so-called "addition formulas" . For sine and cosine the formulas are

   \sin(u + v) &= \sin u\,\cos v + \cos u\,\sin v\,\mbox{,}\\
   \sin(u – v) &= \sin u\,\cos v – \cos u\,\sin v\,\mbox{,}\\
   \cos(u + v) &= \cos u\,\cos v – \sin u\,\sin v\,\mbox{,}\\
   \cos(u – v) &= \cos u\,\cos v + \sin u\,\sin v\,\mbox{.}\\
 \end{align*}

If one wants to know the sine or cosine of a double angle, that is \displaystyle \sin 2v or \displaystyle \cos 2v, one can write these expressions as \displaystyle \sin(v + v) or \displaystyle \cos(v + v) and use the addition formulas above and get the double-angle formulas

   \sin 2v &= 2 \sin v \cos v\,\mbox{,}\\
   \cos 2v &= \cos^2\!v – \sin^2\!v \,\mbox{.}\\
 \end{align*}

From these relationships, one can then get the formulas for half angles. By replacing \displaystyle 2v by \displaystyle v, and consequently \displaystyle v by \displaystyle v/2, in the formula for \displaystyle \cos 2v one gets that

 \cos v = \cos^2\!\frac{v}{2} – \sin^2\!\frac{v}{2}\,\mbox{.}

If we want a formula for \displaystyle \sin(v/2) we use the Pythagorean identity to get rid of \displaystyle \cos^2(v/2)

 \cos v = 1 – \sin^2\!\frac{v}{2} – \sin^2\!\frac{v}{2}
        = 1 – 2\sin^2\!\frac{v}{2}

i.e.

 \sin^2\!\frac{v}{2} = \frac{1 – \cos v}{2}\,\mbox{.}

Similarly, we can use the Pythagorean identity to get rid of \displaystyle \sin^2(v/2). Then we will have instead

 \cos^2\!\frac{v}{2} = \frac{1 + \cos v}{2}\,\mbox{.}

Exercises