4.3 Trigonometric relations

From Förberedande kurs i matematik 1

Revision as of 19:04, 10 November 2008

Introduction

There are a variety of trigonometric formulas which relate the sine, cosine or tangent for an angle or multiples of an angle. They are usually known as the trigonometric identities. Here we will give some of these trigonometric relationships, and show how to derive them. There are many more than we can deal with in this course, but most can be derived from the so-called Pythagorean identity and the addition and subtraction formulas (see below), which are important to know by heart.

The Pythagorean identity

Symmetries

With the help of the unit circle by exploiting the symmetries of the trigonometric functions one obtains a large number 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 is 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, we can derive these relationships by reflecting and/or shifting the graph of sine and cosine. For instance, if we want to express \displaystyle \cos v as the sine of an angle, we can shift the graph of \displaystyle \sin v along in the \displaystyle v direction, so that it coincides with the graph of \displaystyle \cos v. This can be done in several ways, but the most natural is to write \displaystyle \cos v = \sin (v + \pi / 2). To avoid mistakes, we can check that this is true for several different values of \displaystyle v.

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

The addition, subtraction and double-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