Sinus og cosinus til summar og differansar av vinklar

$\begin{array}{rcl}\sin \left(30°-v\right) & =& \sin 30°·\cos v-\cos 30°·\sin v\\ & =& \frac{1}{2}·\cos v-\frac{1}{2}\sqrt{3}\sin v\\ & =& \frac{1}{2}\left(\cos v-\sqrt{3}\sin v\right)\end{array}$

$\begin{array}{rcl}\cos \left(60°-v\right) & =& \cos 60°·\cos v+\sin 60°·\sin v\\ & =& \frac{1}{2}\cos v+\frac{1}{2}\sqrt{3}\sin v\\ & =& \frac{1}{2}\left(\cos v+\sqrt{3}\sin v\right)\end{array}$

$\begin{array}{rcl}\sin \left(v+45°\right) & =& \sin v·\cos 45°+\cos v·\sin 45°\\ & =& \frac{1}{2}\sqrt{2}\sin v+\frac{1}{2}\sqrt{2}\cos v\\ & =& \frac{1}{2}\sqrt{2}\left(\sin v+\cos v\right)\end{array}$

$\begin{array}{rcl}\cos \left(v+45°\right) & =& \cos v·\cos 45°-\sin v·\sin 45°\\ & =& \frac{1}{2}\sqrt{2}\cos v-\frac{1}{2}\sqrt{2}\sin v\\ & =& \frac{1}{2}\sqrt{2}\left(\cos v-\sin v\right)\end{array}$

$\begin{array}{rcl}2\cos \left(v-\frac{\pi }{3}\right) & =& 2\left(\cos v·\cos \frac{\pi }{3}+\sin v·\sin \frac{\pi }{3}\right) \\ & =& 2\left(\frac{1}{2}\cos v+\frac{1}{2}\sqrt{3}\sin v\right)\\ & =& \cos v+\sqrt{3}\sin v\end{array}$

$\begin{array}{rcl}3\sin \left(x-\pi \right) & =& 3\left(\sin x·\cos \pi -\cos x·\sin \pi \right)\\ & =& 3\left(-1·\sin x-\cos x·0\right)\\ & =& -3\sin x\end{array}$

$\begin{array}{rcl}\sqrt{3}\cos \left(\frac{\pi }{6}-x\right) & =& \sqrt{3}\left(\cos \frac{\pi }{6}·\cos x+\sin \frac{\pi }{6}·\sin x\right)\\ & =& \sqrt{3}\left(\frac{1}{2}\sqrt{3}\cos x+\frac{1}{2}\sin x\right)\\ & =& \frac{3}{2}\cos x+\frac{\sqrt{3}}{2}\sin x\\ & =& \frac{1}{2}\left(3\cos x+\sqrt{3}\sin x\right)\end{array}$

$\begin{array}{rcl}\sqrt{2}\sin \left(x-\frac{\pi }{4}\right) & =& \sqrt{2}\left(\sin x·\cos \frac{\pi }{4}-\cos x·\sin \frac{\pi }{4}\right)\\ & =& \sqrt{2}\left(\frac{1}{2}\sqrt{2}·\sin x-\frac{1}{2}\sqrt{2}·\cos x\right)\\ & =& \sin x-\cos x\end{array}$

Prøv med ulike kombinasjonar av summar og differansar av dei tre vinklane.

Vi kan finne eksakte verdiar til vinklane

$45°-30°=15°$
$45°+30°=75°$

$\begin{array}{rcl}\sin 15° & =& \sin \left(45°-30°\right)\\ & =& \sin 45°·\cos 30°-\cos 45°·\sin 30°\\ & =& \frac{1}{2}\sqrt{2}·\frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}·\frac{1}{2}\\ & =& \frac{\sqrt{6}-\sqrt{2}}{4}\end{array}$

$\begin{array}{rcl}\cos 15° & =& \cos \left(45°-30°\right)\\ & =& \cos 45°·\cos 30°+\sin 45°·\sin 30°\\ & =& \frac{1}{2}\sqrt{2}·\frac{1}{2}\sqrt{3}+\frac{1}{2}\sqrt{2}·\frac{1}{2}\\ & =& \frac{\sqrt{6}+\sqrt{2}}{4}\end{array}$

$\sin 75°$ og $\cos 75°$ kan vi finne på tilsvarande måte ved å bruke at $45°+30°=75°$, men det er enklare å bruke at 75° er komplementvinkelen til 15° fordi $15°+75°=90°$. Vi får

$\sin 75°=\cos 15°=\frac{\sqrt{6}+\sqrt{2}}{4}$

$\cos 75°=\sin 15°=\frac{\sqrt{6}-\sqrt{2}}{4}$

Vi kan finne eksakte verdiar for sinus og cosinus til supplementvinklane til 15° og 75°, nemleg

$\begin{array}{rcl}180°-15° & =& 165°\\ 180°-75° & =& 105°\end{array}$

$\begin{array}{rcl}\sin 3v & =& \sin \left(2v+v\right)\\ & =& \sin 2v·\cos v+\cos 2v·\sin v\\ & =& 2\sin v·\cos v·\cos v+\left({\cos }^{2}v-{\sin }^{2}v\right)·\sin v\\ & =& 2\sin v·{\cos }^{2}v+\sin v·{\cos }^{2}v-{\sin }^{3}v\\ & =& 3\sin v·{\cos }^{2}v-{\sin }^{3}v\end{array}$

$\begin{array}{rcl}\cos 3v & =& \cos \left(2v+v\right)\\ & =& \cos 2v·\cos v-\sin 2v·\sin v\\ & =& \left({\cos }^{2}v-{\sin }^{2}v\right)·\cos v-2\sin v·\cos v·\sin v\\ & =& {\cos }^{3}v-{\sin }^{2}v·\cos v-2{\sin }^{2}v·\cos v\\ & =& {\cos }^{3}v-3{\sin }^{2}v·\cos v\end{array}$

Vi bruker resultatet frå oppgåvene a) og b).

$\begin{array}{rcl}\tan 3v & =& \frac{\sin 3v}{\cos 3v}\\ & =& \frac{3\sin v·{\cos }^{2}v-{\sin }^{3}v}{{\cos }^{3}v-3{\sin }^{2}v·\cos v}\\ & =& \frac{{\displaystyle \frac{3\sin v·{\cos }^{2}v-{\sin }^{3}v}{{\cos }^{3}v}}}{{\displaystyle \frac{{\cos }^{3}v-3{\sin }^{2}v·\cos v}{{\cos }^{3}v}}}\\ & =& \frac{3\tan v-{\tan }^{3}v}{1-3{\tan }^{2}v}\end{array}$

$\begin{array}{rcl}\tan \left(u+v\right) & =& \frac{\sin \left(u+v\right)}{\cos \left(u+v\right)}\\ & =& \frac{\sin u·\cos v+\cos u·\sin v}{\cos u·\cos v-\sin u·\sin v}\\ & =& \frac{{\displaystyle \frac{\sin u·\cos v+\cos u·\sin v}{\cos u·\cos v}}}{{\displaystyle \frac{\cos u·\cos v-\sin u·\sin v}{\cos u·\cos v}}}\\ & =& \frac{\tan u+\tan v}{1-\tan u·\tan v}\end{array}$

$\begin{array}{rcl}\tan \left(u-v\right) & =& \frac{\sin \left(u-v\right)}{\cos \left(u-v\right)}\\ & =& \frac{\sin u·\cos v-\cos u·\sin v}{\cos u·\cos v+\sin u·\sin v}\\ & =& \frac{{\displaystyle \frac{\sin u·\cos v-\cos u·\sin v}{\cos u·\cos v}}}{{\displaystyle \frac{\cos u·\cos v+\sin u·\sin v}{\cos u·\cos v}}}\\ & =& \frac{\tan u-\tan v}{1+\tan u·\tan v}\end{array}$

Vi løyser einingsformelen med omsyn på ${\sin }^{2}v$.

$\begin{array}{rcl}{\cos }^{2}v+{\sin }^{2}v & =& 1\\ {\sin }^{2}v & =& 1-{\cos }^{2}v\end{array}$

Då får vi

$\begin{array}{rcl}\cos 2v & =& {\cos }^{2}v-{\sin }^{2}v\\ & =& {\cos }^{2}v-\left(1-{\cos }^{2}v\right)\\ & =& 2{\cos }^{2}v-1\end{array}$

Vi løyser einingsformelen med omsyn på ${\cos }^{2}v$.

$\begin{array}{rcl}{\cos }^{2}v+{\sin }^{2}v & =& 1\\ {\cos }^{2}v & =& 1-{\sin }^{2}v\end{array}$

Då får vi

$\begin{array}{rcl}\cos 2v & =& {\cos }^{2}v-{\sin }^{2}v\\ & =& 1-{\sin }^{2}v-{\sin }^{2}v\\ & =& 1-2{\sin }^{2}v\end{array}$

$\begin{array}{rcl}\sin \left(\frac{\pi }{2}-v\right) & =& \sin \frac{\pi }{2}·\cos v-\cos \frac{\pi }{2}·\sin v\\ & =& 1·\cos v-0·\sin v\\ & =& \cos v\end{array}$

$\begin{array}{rcl}\cos \left(\frac{\pi }{2}-v\right) & =& \cos \frac{\pi }{2}·\cos v+\sin \frac{\pi }{2}·\sin v\\ & =& 0·\cos v+1·\sin v\\ & =& \sin v\end{array}$

Setninga om supplementvinklar er $\sin \left(\pi -v\right)=\sin v$. Vi får

$\begin{array}{rcl}\sin \left(\pi -v\right) & =& \sin \pi ·\cos v-\cos \pi ·\sin v\\ & =& 0·\cos v-\left(-1\right)·\sin v\\ & =& \sin v\end{array}$

Set inn ein 0.

$\begin{array}{rcl}\cos \left(-v\right) & =& \cos \left(0-v\right)\\ & =& \cos 0·\cos v+\sin 0·\sin v\\ & =& 1·\cos v+0·\sin v\\ & =& \cos v\end{array}$

Vi kan bevise desse formlane frå teorisida "Einingsformelen, supplement- og komplementvinklar":

$-\cos v=\cos \left(\pi -v\right)$ ,   $-\cos v=\cos \left(\pi +v\right)$ ,    $-\sin v=\sin \left(\pi +v\right)$ ,   $-\sin v=\sin \left(-v\right)$