# Trigonometriske identiteter

Følgende identiteter brukes mye innenfor trigonometri, og du vil se at du får bruk for disse når du løser slike oppgaver.

Formel

### Trigonometriskeidentiteter

1.
${\mathrm{cos}}^{2}\alpha +{\mathrm{sin}}^{2}\alpha =1$
2.
$\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\alpha +\frac{\pi }{2}\right)=\mathrm{cos}\alpha$
3.
$\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\alpha +\frac{\pi }{2}\right)=-\mathrm{sin}\alpha$
4.
$\mathrm{sin}2\alpha =2\mathrm{sin}\alpha \mathrm{cos}\alpha$
5.
$\begin{array}{ccc}\hfill \mathrm{cos}2\alpha & ={\mathrm{cos}}^{2}\alpha -{\mathrm{sin}}^{2}\alpha \hfill & \hfill \\ \hfill & =2{\mathrm{cos}}^{2}\alpha -1\hfill \\ \hfill & =1-2{\mathrm{sin}}^{2}\alpha \hfill \end{array}$
$\mathrm{cos}2\alpha ={\mathrm{cos}}^{2}\alpha -{\mathrm{sin}}^{2}\alpha =2{\mathrm{cos}}^{2}\alpha -1=1-2{\mathrm{sin}}^{2}\alpha$
6.
$\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\beta +\mathrm{cos}\alpha \mathrm{sin}\beta$
7.
$\mathrm{sin}\left(\alpha -\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\beta -\mathrm{cos}\alpha \mathrm{sin}\beta$
8.
$\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\alpha \mathrm{cos}\beta -\mathrm{sin}\alpha \mathrm{sin}\beta$
9.
$\mathrm{cos}\left(\alpha -\beta \right)=\mathrm{cos}\alpha \mathrm{cos}\beta +\mathrm{sin}\alpha \mathrm{sin}\beta$
10.
$\mathrm{tan}\alpha =\frac{\mathrm{sin}\alpha }{\mathrm{cos}\alpha }$

Eksempel 1

Vis at $\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{\pi }{4}+v\right)=\frac{\sqrt{2}}{2}\left(\mathrm{cos}v-\mathrm{sin}v\right)$

For å regne ut dette bruker du formelen

 $\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\alpha \mathrm{cos}\beta -\mathrm{sin}\alpha \mathrm{sin}\beta .$

Da blir utregningen: $\begin{array}{llll}\hfill \mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{\pi }{4}+v\right)& =\mathrm{cos}\frac{\pi }{4}\mathrm{cos}v-\mathrm{sin}\frac{\pi }{4}\mathrm{sin}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\sqrt{2}}{2}\mathrm{cos}v-\frac{\sqrt{2}}{2}\mathrm{sin}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\sqrt{2}}{2}\left(\mathrm{cos}v-\mathrm{sin}v\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Eksempel 2

Finn eksaktverdien til $\mathrm{sin}\frac{\pi }{12}$

For å løse denne oppgaven skriver du at $\mathrm{sin}\left(\alpha \right)=\mathrm{sin}\left(\pi -\alpha \right)$ og bruker sammenhengen

 $\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\alpha \mathrm{cos}\beta +\mathrm{cos}\alpha \mathrm{sin}\beta .$

Da blir utregningen:

$\begin{array}{llll}\hfill \mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{\pi }{12}\right)& =\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\pi -\frac{\pi }{12}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{11\pi }{12}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{\pi }{6}+\frac{3\pi }{4}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{\pi }{6}\right)\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3\pi }{4}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{\pi }{6}\right)\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3\pi }{4}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{2}\cdot \phantom{\rule{-0.17em}{0ex}}\left(-\frac{\sqrt{2}}{2}\right)+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\sqrt{6}-\sqrt{2}}{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{\pi }{12}\right)& =\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\pi -\frac{\pi }{12}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{11\pi }{12}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{\pi }{6}+\frac{3\pi }{4}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{\pi }{6}\right)\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3\pi }{4}\right)+\mathrm{cos}\phantom{\rule{-0.17em}{0ex}}\left(\frac{\pi }{6}\right)\mathrm{sin}\phantom{\rule{-0.17em}{0ex}}\left(\frac{3\pi }{4}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{1}{2}\cdot \phantom{\rule{-0.17em}{0ex}}\left(-\frac{\sqrt{2}}{2}\right)+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{\sqrt{6}-\sqrt{2}}{4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Eksempel 3

Gitt $\mathrm{sin}v=\frac{\sqrt{3}}{2}$, finn $\mathrm{cos}v$

For å regne ut dette bruker du formelen

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

Da blir utregningen: $\begin{array}{llll}\hfill {\mathrm{cos}}^{2}v+{\mathrm{sin}}^{2}v& =1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\mathrm{cos}}^{2}v& =1-{\mathrm{sin}}^{2}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Dermed er $\begin{array}{llll}\hfill \mathrm{cos}v& =±\sqrt{1-{\mathrm{sin}}^{2}v}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =±\sqrt{1-\phantom{\rule{-0.17em}{0ex}}{\left(\frac{\sqrt{3}}{2}\right)}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =±\sqrt{1-\frac{3}{4}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =±\sqrt{\frac{1}{4}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =±\frac{1}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Eksempel 4

Gitt ${\mathrm{cos}}^{2}v+{\mathrm{sin}}^{2}v={\mathrm{tan}}^{2}v$, finn $\mathrm{sin}v$

For å løse denne må du bruke flere av sammenhengene over for så å løse ut $\mathrm{sin}v$:

$\begin{array}{llll}\hfill 2{\mathrm{cos}}^{2}v+{\mathrm{sin}}^{2}v& ={\mathrm{tan}}^{2}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 1-{\mathrm{sin}}^{2}v+{\mathrm{sin}}^{2}v& =\frac{{\mathrm{sin}}^{2}v}{{\mathrm{cos}}^{2}v}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 1& =\frac{{\mathrm{sin}}^{2}v}{{\mathrm{cos}}^{2}v}\phantom{\rule{1em}{0ex}}|\cdot {\mathrm{cos}}^{2}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\mathrm{cos}}^{2}v& ={\mathrm{sin}}^{2}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 1-{\mathrm{sin}}^{2}v& ={\mathrm{sin}}^{2}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 1& =2{\mathrm{sin}}^{2}v\phantom{\rule{1em}{0ex}}|:2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{1}{2}& ={\mathrm{sin}}^{2}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llllll}\hfill {\mathrm{cos}}^{2}v+{\mathrm{sin}}^{2}v& ={\mathrm{tan}}^{2}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 1-{\mathrm{sin}}^{2}v+{\mathrm{sin}}^{2}v& =\frac{{\mathrm{sin}}^{2}v}{{\mathrm{cos}}^{2}v}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 1& =\frac{{\mathrm{sin}}^{2}v}{{\mathrm{cos}}^{2}v}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{1em}{0ex}}|\cdot {\mathrm{cos}}^{2}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {\mathrm{cos}}^{2}v& ={\mathrm{sin}}^{2}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 1-{\mathrm{sin}}^{2}v& ={\mathrm{sin}}^{2}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 1& =2{\mathrm{sin}}^{2}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{1em}{0ex}}|:2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{1}{2}& ={\mathrm{sin}}^{2}v\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Dermed er
 $\mathrm{sin}v=±\sqrt{\frac{1}{2}}=±\frac{1}{\sqrt{2}}=±\frac{\sqrt{2}}{2}.$

Eksempel 5

Vis at $\mathrm{cos}\left(2\alpha \right)={\mathrm{cos}}^{2}\alpha -{\mathrm{sin}}^{2}\alpha$

Når du skal vise slike sammenhenger så vil du lage en ekvivalens fra venstresiden av likheten til høyresiden av likheten ved hjelp av logiske steg: $\begin{array}{llll}\hfill \mathrm{cos}\left(2\alpha \right)& =\mathrm{cos}\left(\alpha +\alpha \right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{cos}\alpha \mathrm{cos}\alpha -\mathrm{sin}\alpha \mathrm{sin}\alpha \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\mathrm{cos}}^{2}\alpha -{\mathrm{sin}}^{2}\alpha \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Q.E.D

Eksempel 6

Vis at $\mathrm{sin}\left(2\alpha \right)=2\mathrm{sin}\alpha \mathrm{cos}\alpha$

Når du skal vise slike sammenhenger så vil du lage en ekvivalens fra venstresiden av likheten til høyresiden av likheten ved hjelp av logiske steg: $\begin{array}{llll}\hfill \mathrm{sin}\left(2\alpha \right)& =\mathrm{sin}\left(\alpha +\alpha \right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\mathrm{sin}\alpha \mathrm{cos}\alpha +\mathrm{cos}\alpha \mathrm{sin}\alpha \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\mathrm{sin}\alpha \mathrm{sin}\alpha \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Q.E.D

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