Sinus, cosinus og tangens gir desimaltal mellom -1 og 1, men noen vinkler har eksakte verdier:
| $x$ (grader) $x$ (radianer) | $0^o$ $0$ | $30^o$ $\frac{\pi}{6}$ | $45^o$ $\frac{\pi}{4}$ | $60^o$ $\frac{\pi}{3}$ | $90^o$ $\frac{\pi}{2}$ | $180^o$ $\pi$ |
| sin $x$ cos $x$ tan $x$ | $\frac{\textcolor{blue}{\sqrt{0}}}{2} = 0$ $\frac{\textcolor{blue}{\sqrt{4}}}{2} = 1$ $0$ | $\frac{\textcolor{blue}{\sqrt{1}}}{2} = \frac{1}{2}$ $\frac{\textcolor{blue}{\sqrt{3}}}{2}$ $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ | $\frac{\textcolor{blue}{\sqrt{2}}}{2}$ $\frac{\textcolor{blue}{\sqrt{2}}}{2}$ $1$ | $\frac{\textcolor{blue}{\sqrt{3}}}{2}$ $\frac{\textcolor{blue}{\sqrt{1}}}{2} = \frac{1}{2}$ $\sqrt{3}$ | $\frac{\textcolor{blue}{\sqrt{4}}}{2} = 1$ $\frac{\textcolor{blue}{\sqrt{0}}}{2} = 0$ $-$ | 0 1 0 |
Legg merke til:
- $\tan x$ = $\frac{\sin x}{\cos x}$ og derfor får $\tan x$ et brudd når $\cos x$ = 0
- cos 0 = 0 og sin 0 = 1
- $\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}$ er fordelt fint utover på $\sin x$ og $\cos x$.
+ Eksempel: Finn eksakt verdi til $\sin(120^o)$ og $\cos(120^o)$?
Alternativ 1: Først finner vi $120^o$ i enhetssirkelen. Vinkelen ligger i andre kvadrant siden den er større enn $90^o$ og mindre enn $180^o$:
Sinus-verdien (på den vertikale aksen) for $120^o$ er den samme som for $60^o$ siden $180^o - 60^o = 120^o$. Derfor kan vi bruke den eksakte verdien til $60^o$:
b = \sin(120^o) = \sin(60^o) = \frac{\sqrt{3}}{2}Cosinus-verdien (på den horisontale aksen) for $120^o$ er den samme som for $60^o$, men med motsatt fortegn. Derfor kan vi bruke den eksakte verdien til $60^o$ og sette minus foran:
a = \cos(120^o) = -\cos(60^o) = -\frac{1}{2}Og, vips, er vi ferdige!
Alternativ 2: Bruk formlene for summen av to vinkler:
\begin{aligned}
\textnormal{Formel: } & \sin(\textcolor{red}{u} + \textcolor{blue}{v}) = \sin(\textcolor{red}{u}) \cos(\textcolor{blue}{v}) + \sin(\textcolor{blue}{v}) \cos(\textcolor{red}{u}) \\
\sin(120^o) & = \sin(\textcolor{red}{90^o} + \textcolor{blue}{30^o}) \\
\Rightarrow \quad \sin(120^o)
&= \sin(\textcolor{red}{90^o}) \cos(\textcolor{blue}{30^o}) + \sin(\textcolor{blue}{30^o}) \cos(\textcolor{red}{90^o}) \\
\Rightarrow \quad \sin(120^o)
&= \textcolor{red}{1} \cdot \textcolor{blue}{\frac{\sqrt{3}}{2}} + \textcolor{blue}{\frac{1}{2}} \cdot \textcolor{red}{0} \\
\Rightarrow \quad \sin(120^o)
&= \frac{\sqrt{3}}{2}
\end{aligned}\begin{aligned}
\textnormal{Formel: } & \cos(\textcolor{red}{u} + \textcolor{blue}{v}) = \cos(\textcolor{red}{u}) \cos(\textcolor{blue}{v}) - \sin(\textcolor{red}{u}) \sin(\textcolor{blue}{v}) \\
\cos(120^o) & = \cos(\textcolor{red}{90^o} + \textcolor{blue}{30^o}) \\
\Rightarrow \quad \cos(120^o)
&= \cos(\textcolor{red}{90^o}) \cos(\textcolor{blue}{30^o}) + \sin(\textcolor{red}{90^o}) \sin(\textcolor{blue}{30^o}) \\
\Rightarrow \quad \sin(120^o)
&= \textcolor{red}{0} \cdot \textcolor{blue}{\frac{\sqrt{3}}{2}} + \textcolor{red}{1} \cdot \textcolor{blue}{\frac{1}{2}} \\
\Rightarrow \quad \sin(120^o)
&= - \frac{1}{2}
\end{aligned}+ Eksempel: Finn eksakt verdi til $\sin(240^o)$ og $\cos(240^o)$?
Alternativ 1: Først finner vi $240^o$ i enhetssirkelen. Den ligger i tredje kvadrant siden den er større enn $180^o$ og mindre enn $270^o$:
Sinus-verdien (på den vertikale aksen) for $120^o$ er den samme som for $60^o$, men med motsatt fortegn siden $180^o + 60^o = 240^o$. Derfor kan vi bruke den eksakte verdien til $60^o$ og sette minus foran:
b = \sin(240^o) = - \sin(60^o) = - \frac{\sqrt{3}}{2}Cosinus-verdien (på den horisontale aksen) for $120^o$ er den samme som for $60^o$, men med motsatt fortegn. Derfor kan vi bruke den eksakte verdien til $60^o$ og sette minus foran:
a = \cos(240^o) = -\cos(60^o) = -\frac{1}{2}Og, vips, er vi ferdige!
Alternativ 2: Bruk formlene for summen av to vinkler:
\begin{aligned}
\textnormal{Formel: } & \sin(\textcolor{red}{u} + \textcolor{blue}{v}) = \sin(\textcolor{red}{u}) \cos(\textcolor{blue}{v}) + \sin(\textcolor{blue}{v}) \cos(\textcolor{red}{u}) \\
\sin(240^o) & = \sin(\textcolor{red}{180^o} + \textcolor{blue}{60^o}) \\
\Rightarrow \quad \sin(240^o)
&= \sin(\textcolor{red}{180^o}) \cos(\textcolor{blue}{60^o}) + \sin(\textcolor{blue}{60^o}) \cos(\textcolor{red}{180^o}) \\
\Rightarrow \quad \sin(240^o)
&= \textcolor{red}{0} \cdot \textcolor{blue}{\frac{1}{2}} + \textcolor{blue}{\frac{\sqrt{3}}{2}} \cdot (\textcolor{red}{-1}) \\
\Rightarrow \quad \sin(240^o)
&= -\frac{\sqrt{3}}{2}
\end{aligned}\begin{aligned}
\textnormal{Formel: } & \cos(\textcolor{red}{u} + \textcolor{blue}{v}) = \cos(\textcolor{red}{u}) \cos(\textcolor{blue}{v}) - \sin(\textcolor{red}{u}) \sin(\textcolor{blue}{v}) \\
\cos(240^o) & = \cos(\textcolor{red}{180^o} + \textcolor{blue}{60^o}) \\
\Rightarrow \quad \cos(240^o)
&= \cos(\textcolor{red}{180^o}) \cos(\textcolor{blue}{60^o}) + \sin(\textcolor{red}{180^o}) \sin(\textcolor{blue}{60^o}) \\
\Rightarrow \quad \sin(240^o)
&= (\textcolor{red}{-1}) \cdot \textcolor{blue}{\frac{1}{2}} + \textcolor{red}{0} \cdot \textcolor{blue}{\frac{\sqrt{3}}{2}} \\
\Rightarrow \quad \sin(240^o)
&= - \frac{1}{2}
\end{aligned}