MATHHX A
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8.4 Beviser - Trigonometriske funktioner
Vi starter med at se på nogle nyttige formler vi får brug for i det efterfølgende.
Overgangsformler
-
Bevis
Vi starter med vinklerne fra de to første formler:
På ovenstående tegning ses vinklen \(x\) tegnet med rød og vinklen \(-x\) tegnet med grøn. Det er tydeligt at de har samme cosinusværdi, og at deres sinusværdier er lige store men med omvendt fortegn.
Lad os se på vinklerne i de to sidste formler:
På ovenstående tegning ses vinklen \(x\) tegnet med rød og vinklen \(\pi -x\) tegnet med grøn. Vinklen \(\pi -x\) er selvfølgelig tegnet ved at starte med en vinkel på \(\pi \) og så gå \(x\) baglæns. Det er tydeligt at de to
vinkler har samme sinusværdi, og at deres cosinusværdier er lige store, men med omvendt fortegn.
Perioden for tangens
Vi vil nu bevise, at tangens er periodisk med periode \(\pi \).
-
Bevis
At tagens er periodiske med \(\pi \) betyder at den begynder at gentage sig selv, efter vi er gået \(\pi \) ud af \(x\)-aksen. Det må være det samme som at funktionsværdien i et vilkårligt
\(x\) er den samme som funktionsværdien i \(x+\pi \):
Vi skal altså vise at \(\tan (x)=\tan (x+\pi )\).
\(\seteqnumber{0}{8.}{0}\)
\begin{align*}
\tan (x+\pi ) & = \frac {\sin (x+\pi )}{\cos (x+\pi )} && \text {pr. definition af tangens}\\ & = \frac {\sin (\pi -(-x))}{\cos (\pi -(-x))} && \text {da $x=-(-x)$}\\
& = \frac {\sin (-x)}{-\cos (-x)} && \text {overgangsformel 3 og 4}\\ & = \frac {-\sin (x)}{-\cos (x)} && \text {overgangsformel 1 og 2}\\ & = \frac {\sin (x)}{\cos
(x)} && \text {forkortet med $-1$}\\ & = \tan (x)\\
\end{align*}
Vi har nu vist at \(\tan (x)=\tan (x+\pi )\) og tangens er altså periodisk med periode \(\pi \).
Harmoniske svingninger
Sammenhæng mellem periode og vinkelfrekvens
Man kan regne perioden ud fra vinkelfrekvensen med formlen
\[T=\frac {2\pi }{b}\]
Vi vil nu udlede denne formel. Vi har den harmoniske svingning
\[f(x)=a\cdot \sin (bx+c)\]
Perioden er den (mindste) konstant \(T\) som opfylder:
\[f(x)=f(x+T)\]
Vi indsætter forskriften for \(f\):
\[a\cdot \sin (bx+c)=a\cdot \sin (b(x+T)+c)\]
Vi regner parentesen inde i sinus på højresiden:
\[a\cdot \sin (bx+c)=a\cdot \sin (bx+bT+c)\]
Vi kan se at der står det samme på begge sider bortset fra størrelsen \(bT\), som er lagt til inde i sinus funktionen på højresiden. Da sinus er periodisk med \(2\pi \), skal denne størrelse give \(2\pi \) før at vi får det samme på
begge sider af lighedstegnet. Dvs.
\[bT=2\pi \]
Vi deler med \(b\):
\[T=\frac {2\pi }{b}\]
Faseforskydning
Faseforskydningen forskyder grafen langs \(x\)-aksen. Mere præcist: Grafen for funktionen \(f(x)=a\cdot \sin (bx+c)\) fremkommer ud fra grafen for \(g(x)=a\cdot \sin (bx)\) ved at forskyde \(g\) med \(\frac {c}{b}\)
til venstre, hvis \(c>0\). Vi vil nu vise dette. Vi starter med en illustration af påstanden:
Tegning viser at \(f\) er opstået ud fra \(g\) ved en vandret forskydning til venstre med \(\frac {c}{b}\), hvis \(f\) opfylder \(f(x-\frac {c}{b})=f(x)\). Så vi skal altså eftervise at \(f(x-\frac {c}{b})=f(x)\).
\(\seteqnumber{0}{8.}{0}\)
\begin{equation*}
\begin{split} f(x-\frac {c}{b}) & = a\cdot \sin (b(x-\frac {c}{b})+c)\\ &= a\cdot \sin (bx-b\frac {c}{b}+c)\\ &= a\cdot \sin (bx-c+c)\\ &= a\cdot \sin (bx)\\ &= g(x) \end
{split}
\end{equation*}
Så jo, den var god nok. Grafen for \(f\) er opstået ud fra grafen for \(g\) ved at forskyde \(g\) med \(\frac {c}{b}\) til venstre.
Øvelse 8.4.1 (Svær)
I ovenstående har vi taget udgangspunkt i en positiv faseforskydning, når vi tegnede grafen for \(f\).
Ekstra
Der er et problem i vores bevis for perioden for tangens. Vi har vist, at
\[\tan (x)=\tan (x+\pi )\]
Men det betyder faktisk kun at tangens er periodisk med højst \(\pi \) (hvorfor?). Vi kan dog nemt tjekke, at perioden ikke kan være mindre end \(\pi \). Vi skal altså se, om der findes et tal \(T<\pi \) så
\[\tan (x)=\tan (x+T)\]
Det skal gælde for alle \(x\), så lad os bare tage udgangspunkt i \(x=0\). Vi skal altså se om der findes et tal \(T<\pi \) så
\[\tan (0)=\tan (T)\]
Vi regner \(\tan (0)\):
\[\tan (0)= \frac {\sin (0)}{\cos (0)}=\frac {0}{1}=0\]
Så \(\tan (0)=0\). Hvor langt mon vi skal gå ud før tangens bliver nul igen? Ja, hvis \(\tan (T)\) skal være nul, så skal \(\sin (T)\) give nul, og næste gang sinus giver nul er ved \(x=\pi \). Så perioden \(T\) kan ikke være
mindre end \(\pi \). Vi viste først, at perioden var højst \(\pi \), og nu har vi argumenteret for, at den er mindst \(\pi \). Det betyder at perioden er \(\pi \).
