Trigonometric Pythagoras

Geometric illustration of the "trigonometric Pythagoras"

As a " trigonometric Pythagoras " the identity

${\displaystyle \sin ^{2}\alpha +\cos ^{2}\alpha =1}$

designated. Here stands for and for . The validity of this identity can be shown on the unit circle with the help of the Pythagorean theorem , which is also eponymous for this frequently used theorem of trigonometry . ${\displaystyle \sin ^{2}\alpha }$ ${\displaystyle (\sin \alpha )^{2}}$ ${\displaystyle \cos ^{2}\alpha }$ ${\displaystyle (\cos \alpha )^{2}}$

Geometric derivation

The Pythagorean theorem serves as the basis . It says that in a right triangle with the hypotenuse and the cathetus and ${\displaystyle c}$ ${\displaystyle a}$ ${\displaystyle b}$

${\displaystyle a^{2}+b^{2}=c^{2}}$

applies. If the angle in the right-angled triangle is chosen so that its opposite side and its adjacent side is, then the following applies in general ${\displaystyle \alpha }$ ${\displaystyle a}$ ${\displaystyle b}$

${\displaystyle a=\sin \alpha \cdot c}$,

${\displaystyle b=\cos \alpha \cdot c}$.

Substituting both equations into the Pythagorean theorem then gives

${\displaystyle (\sin ^{2}\alpha +\cos ^{2}\alpha )\cdot c^{2}=c^{2}}$,

${\displaystyle \sin ^{2}\alpha +\cos ^{2}\alpha =1}$.

Geometric illustration

The adjacent sketch shows the unit circle , i.e. a circle with radius 1, and a right-angled triangle with hypotenuse length 1 in the unit circle. The Pythagorean theorem applies here to any value of the angle in the unit circle and immediately shows the validity of the "trigonometric Pythagoras". ${\displaystyle \alpha }$

Analytical derivation

The evidential value of the perception is problematic for obtuse and hyper-obtuse angles , since for such (at least) one angle function has negative values; what are "negative sides" of a right triangle? An analytical proof shows that the trigonometric Pythagoras holds for any real arguments of the used trigonometric functions. ${\displaystyle \alpha }$ ${\displaystyle \alpha }$

With the imaginary unit and the third binomial formula we can factorize:

${\displaystyle \cos ^{2}(\alpha )+\sin ^{2}(\alpha )=\cos ^{2}(\alpha )-i^{2}\cdot \sin ^{2}(\alpha )=(\cos(\alpha )+i\cdot \sin(\alpha ))\cdot (\cos(\alpha )-i\cdot \sin(\alpha ));}$

since the cosine is an even function and the sine is an odd function , Euler's formula continues:

${\displaystyle (\cos(\alpha )+i\cdot \sin(\alpha ))\cdot (\cos(\alpha )-i\cdot \sin(\alpha ))=(\cos(\alpha )+i\cdot \sin(\alpha ))\cdot (\cos(-\alpha )+i\cdot \sin(-\alpha ))=e^{i\alpha }\cdot e^{-i\alpha }=e^{0}=1,\quad }$qed

Individual evidence

1. ^ Lothar Papula: Mathematics for Engineers and Natural Scientists Volume 1 . Springer,, ISBN 978-3-8348-1749-5 , p. 251.
2. ↑ Hans Kreul, Harald Ziebarth: Mathematics made easy . Harri Deutsch Verlag, August 2009, ISBN 978-3-8171-1836-6 , p. 94.