Hoofdstuk 11 Goniometrie *: Goniometrie in de rechthoekige driehoek *
Periodicity of trigonometric functions
The following laws for sine and cosine follow directly from the definition and the symmetry of the circle.
|
Period #2\pi# |
#\cos(t+2\pi) = \cos(t)# | #\sin(t+2\pi) = \sin(t)# |
| Reflection along the #x#-axis | #\cos(-t) = \cos(t)# |
#\sin(-t) = -\sin(t)# |
| Reflection along the #y#-axis | #\cos(\pi-t) = -\cos(t)# |
#\sin(\pi-t) = \sin(t)# |
| Reflection along the line #x=y# |
#\cos(\frac{\pi}{2}-t) = \sin(t)# |
#\sin(\frac{\pi}{2}-t) = \cos(t)# |
The period #2\pi# is a direct result of the fact that the point that is reached by moving counterclockwise a distance of #2\pi# from #\rv{\cos(t),\sin(t)}# on the unit circle, is the point #\rv{\cos(t),\sin(t)}# itself. Because, this means #\rv{\cos(t+2\pi),\sin(t+2\pi)}=\rv{\cos(t),\sin(t)}#.
Reflection along the #x#-axis: The unit circle transforms in itself if we reflect along the #x#-axis. In this reflection, the point #\rv{\cos(t),\sin(t)}# goes to #\rv{\cos(t),-\sin(t)}#, on the other hand this is the point reached by moving a distance #-t# from #\rv{0,0}# on the unit circle, hence equal to #\rv{\cos(-t),\sin(-t)}#. Comparing these two coordinates gives the two equations #\cos(t) = \cos(-t)# and #-\sin(t) = \sin(-t)#.
Reflection along the #y#-axis: The unit circle transforms in itself if we reflect along the #y#-axis. In this reflection, the point #\rv{\cos(t),\sin(t)}# goes to #\rv{-\cos(t),\sin(t)}#, on the other hand this is the point reached by moving a distance #\pi-t# from #\rv{0,0}# on the unit circle, hence equal to #\rv{\cos(\pi-t),\sin(\pi-t)}#. Comparing thse two coordinates gives the two equations #-\cos(t) = \cos(\pi-t)# and #\sin(t) = \sin(\pi-t)#.
Reflection along the line #x=y#: The unit circle transforms in itself if we reflect along the line #x=y#. In this reflection, the point #\rv{\cos(t),\sin(t)}# goes to #\rv{\sin(t),\cos(t)}#, on the other hand this is the point reached by moving a distance #\dfrac{\pi}{2}-t# from #\rv{0,0}# on the unit circle, hence equal to #\rv{\cos(\dfrac{\pi}{2}-t),\sin(\dfrac{\pi}{2}-t)}#. Comparing thse two coordinates gives the two equations #\sin(t) =\cos(\dfrac{\pi}{2}-t)# and #\cos(t) = \sin(\dfrac{\pi}{2}-t)#.
The result is that we can calculate the value of any cosine and sine if we know their values for #t# in the interval #\left[0,\dfrac{\pi}{4}\right]#. Since, if #t\not\in \left[0,2\pi\right)#, then, because of the period #2\pi#, we can add a multitude of #2\pi# to reduce the argument to #t# in #\ivco{0}{2 \pi}#. If #t\in \ivco{\pi}{2\pi}#, then, because of reflection in the #y#-axis and reflection in the #x#-axis, we can replace #t# by (first #\pi-t# and next) #t-\pi# to reduce the argument #t# to #\ivco{0}{\pi}#. If #t\in \ivco{\dfrac{\pi}{2}}{\pi}#, then, because reflection along the line #x=y# and reflection along the #x#-axis, we can replace #t# by (first #\dfrac{\pi}{2}-t# and next) #t-\dfrac{\pi}{2}# to reduce the argument #t# to #\ivco{0}{\dfrac{\pi}{2}}#. Finally, if #t\in \ivco{\dfrac{\pi}{4}}{\dfrac{\pi}{2}}#, then, because of reflection along the line #x=y#, we can replace #t# by #\dfrac{\pi}{2}-t# to reduce the argument #t# to #\ivco{0}{\dfrac{\pi}{4}}#.
If #t=\dfrac{35\pi}{6}# and we follow the steps above, we find\[\begin{array}{rclcl}\cos\left(\dfrac{35\pi}{6}\right) &=& \cos\left(\dfrac{11\pi}{6}\right)&\phantom{xx}& \color{blue}{\text{period }2\pi}\\ &=& -\cos\left(\dfrac{-5\pi}{6}\right) =- \cos\left(\dfrac{5\pi}{6}\right)&\phantom{xx}& \color{blue}{\text{reflection }y\text{-axis and }x\text{-axis}}\\ &=& -\sin\left(\dfrac{-2\pi}{6}\right) = \sin\left(\dfrac{\pi}{3}\right)&\phantom{xx}& \color{blue}{\text{reflection }x=y\text{ and }x\text{-axis}}\\ &=& \cos\left(\dfrac{\pi}{6}\right)&\phantom{xx}& \color{blue}{\text{reflection }x=y\tiny.}\end{array}\]
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