Aus Online Mathematik Brückenkurs 1

Using the formula for double angles on sin \displaystyle 160^{\circ } gives

\displaystyle \sin 160^{\circ }=2\cos 80^{\circ }\sin 80^{\circ }

On the right-hand side, we see that the factor \displaystyle \cos 80^{\circ } has appeared, and if we use the formula for double angles on the second factor ( \displaystyle \sin 80^{\circ } ),

\displaystyle 2\cos 80^{\circ }\sin 80^{\circ }=2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\sin 40^{\circ }

we obtain a further factor \displaystyle \cos 40^{\circ }. A final application of the formula for double angles on \displaystyle \sin 40^{\circ } gives us all three cosine factors:

\displaystyle 2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\centerdot \sin 40^{\circ }=2\cos 80^{\circ }\centerdot 2\cos 40^{\circ }\centerdot 2\cos 20^{\circ }\sin 20^{\circ }

We have thus succeeded in showing that

\displaystyle \sin 160^{\circ }=8\cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }\sin 20^{\circ }

which can also be written as

\displaystyle \cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}

If we draw the unit circle, we see that \displaystyle 160^{\circ } makes an angle of \displaystyle 20^{\circ } with the negative \displaystyle x -axis, and therefore the angles \displaystyle 20^{\circ } and \displaystyle 160^{\circ } have the same \displaystyle y -coordinate in the unit circle, i.e.

\displaystyle \sin 20^{\circ }=\sin 160^{\circ }.

This shows that

\displaystyle \cos 80^{\circ }\centerdot \cos 40^{\circ }\centerdot \cos 20^{\circ }=\frac{\sin 160^{\circ }}{8\sin 20^{\circ }}=\frac{1}{8}