Kafli 7

$\frac{\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{2}-\sqrt{6}}{4}$

$\frac{1-\sqrt{3}}{1+\sqrt{3}}$

$\cos \left(\frac{5\pi }{14}\right)$

$\cos \left(2\alpha \right)=\frac{7}{32}$

${\cos }^4\theta -{\sin }^4\theta =\left({\cos }^2\theta +{\sin }^2\theta \right)\left({\cos }^2\theta -{\sin }^2\theta \right)=\cos \left(2\theta \right)$

$\cos \left(2\theta \right)\cos \theta =\left({\cos }^2\theta -{\sin }^2\theta \right)\cos \theta ={\cos }^3\theta -\cos \theta {\sin }^2\theta$

$$\begin{array}{ll}10{\cos }^{4}x=10{\cos }^{4}x=10{({\cos }^{2}x)}^2\hfill & \hfill \\ =10{\left[\frac{1+\cos (2x)}{2}\right]}^2\hfill & {\text{Substitute reduction formula for cos}}^{2}x.\hfill \\ =\frac{10}{4}[1+2\cos (2x)+{\cos }^2(2x)]\hfill & \hfill \\ =\frac{10}{4}+\frac{10}{2}\cos (2x)+\frac{10}{4}\left(\frac{1+\cos 2(2x)}{2}\right)\hfill & {\text{Substitute reduction formula for cos}}^{2}x.\hfill \\ =\frac{10}{4}+\frac{10}{2}\cos (2x)+\frac{10}{8}+\frac{10}{8}\cos (4x)\hfill & \hfill \\ =\frac{30}{8}+5\cos (2x)+\frac{10}{8}\cos (4x)\hfill & \hfill \\ =\frac{15}{4}+5\cos (2x)+\frac{5}{4}\cos (4x)\hfill & \hfill \end{array}$$

$-\frac{2}{\sqrt{5}}$

$\frac{1}{2}\left(\cos 6\theta +\cos 2\theta \right)$

$\frac{1}{2}\left(\sin 2x+\sin 2y\right)$

$\frac{-2-\sqrt{3}}{4}$

$2\sin \left(2\theta \right)\cos \left(\theta \right)$

$$\begin{array}{l}\tan \theta \cot \theta -{\cos }^2\theta =\left(\frac{\sin \theta }{\cos \theta }\right)\left(\frac{\cos \theta }{\sin \theta }\right)-{\cos }^2\theta \hfill \\ =1-{\cos }^2\theta \hfill \\ ={\sin }^2\theta \hfill \end{array}$$

$x=\frac{7\pi }{6},\frac{11\pi }{6}$

$\frac{\pi }{3}\pm \pi k$

$\theta \approx 1.7722\pm 2\pi k$ og $\theta \approx 4.5110\pm 2\pi k$

$\cos \theta =-1,\theta =\pi$

$\frac{\pi }{2},\frac{2\pi }{3},\frac{4\pi }{3},\frac{3\pi }{2}$

Útslagið er $3,$ og lotan er $\frac{2}{3}.$

$y=8\sin \left(\frac{\pi }{12}t\right)+32$
Hitastigið nær frostmarki á hádegi og miðnætti.

upphafsgildi displacement =6, deyfingarfasti = -6, frequency = $\frac{2}{\pi }$

$y=10e^{-0.5t}\cos \left(\pi t\right)$

$y=5\cos \left(6\pi t\right)$

Öll föllin þrjú, $F$ , $G$ og $H,$ eru jafnstæð.

Ástæðan er sú að $F\left(-x\right)=\sin \left(-x\right)\sin \left(-x\right)=\left(-\sin x\right)\left(-\sin x\right)={\sin }^{2}x=F\left(x\right),G\left(-x\right)=\cos \left(-x\right)\cos \left(-x\right)=\cos x\cos x={\cos }^{2}x=G\left(x\right)$ og $H\left(-x\right)=\tan \left(-x\right)\tan \left(-x\right)=\left(-\tan x\right)\left(-\tan x\right)={\tan }^{2}x=H\left(x\right).$

Þegar $\cos t=0,$ þá er $\sec t=\frac{1}{0},$ sem er óskilgreint.

$\sin x$

$\sec x$

$\csc t$

$\mathrm{-1}$

${\sec }^{2}x$

${\sin }^{2}x+1$

$\frac{1}{\sin x}$

$\frac{1}{\cot x}$

$\tan x$

$-4\sec x\tan x$

$\pm \sqrt{\frac{1}{{\cot }^{2}x}+1}$

$\frac{\pm \sqrt{1-{\sin }^{2}x}}{\sin x}$

$\cos x-{\cos }^{3}x=\cos x\left(1-{\cos }^{2}x\right)$
$=\cos x{\sin }^{2}x$

Svör eru mismunandi. Dæmi um sönnun:
$\frac{1+{\sin }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}+\frac{{\sin }^{2}x}{{\cos }^{2}x}={\sec }^{2}x+{\tan }^{2}x={\tan }^{2}x+1+{\tan }^{2}x=1+2{\tan }^{2}x$

Svör eru mismunandi. Dæmi um sönnun:
${\cos }^{2}x-{\tan }^{2}x=1-{\sin }^{2}x-\left({\sec }^{2}x-1\right)=1-{\sin }^{2}x-{\sec }^{2}x+1=2-{\sin }^{2}x-{\sec }^{2}x$

True $3{\sin }^2\theta +4{\cos }^2\theta =3{\sin }^2\theta +3{\cos }^2\theta +{\cos }^2\theta =3\left({\sin }^2\theta +{\cos }^2\theta \right)+{\cos }^2\theta =3+{\cos }^2\theta$

Reglur um fylliföll eiga við um fyllihorn. Ef annað hvassa hornsins í rétthyrndum þríhyrningi er $x,$ þá er hitt hornið $\frac{\pi }{2}-x.$ Þá er $\sin x=\cos \left(\frac{\pi }{2}-x\right).$ Sama gildir um hinar reglurnar um fylliföll. Lykilatriðið er að hornin eru fyllihorn.

$\sin \left(-x\right)=-\sin x,$ svo $\sin x$ er oddstætt. $\cos \left(-x\right)=\cos \left(0-x\right)=\cos x,$ svo $\cos x$ er jafnstætt.

$\frac{\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

$-2-\sqrt{3}$

$-\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x$

$-\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x$

$\csc \theta$

$\cot x$

$\tan \left(\frac{x}{10}\right)$

$\sin (a-b)=\left(\frac{4}{5}\right)\left(\frac{1}{3}\right)-\left(\frac{3}{5}\right)\left(\frac{2\sqrt{2}}{3}\right)=\frac{4-6\sqrt{2}}{15}$
$\cos (a+b)=\left(\frac{3}{5}\right)\left(\frac{1}{3}\right)-\left(\frac{4}{5}\right)\left(\frac{2\sqrt{2}}{3}\right)=\frac{3-8\sqrt{2}}{15}$

$\frac{\sqrt{2}-\sqrt{6}}{4}$

$\sin x$

$\cot \left(\frac{\pi }{6}-x\right)$

$\cot \left(\frac{\pi }{4}+x\right)$

$\frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}}$

Þau eru ólík; prófaðu $g\left(x\right)=\sin \left(9x\right)-\cos \left(3x\right)\sin \left(6x\right).$

Þau eru ólík; prófaðu $g\left(\theta \right)=\frac{2\tan \theta }{1-{\tan }^2\theta }.$

Þau eru ólík; prófaðu $g\left(x\right)=\frac{\tan x-\tan \left(2x\right)}{1+\tan x\tan \left(2x\right)}.$

$-\frac{\sqrt{3}-1}{2\sqrt{2}},\text{ or }-0.2588$

$\frac{1+\sqrt{3}}{2\sqrt{2}},$ eða 0.9659

$\begin{array}{c}\tan \left(x+\frac{\pi }{4}\right)=\\ \frac{\tan x+\tan \left(\frac{\pi }{4}\right)}{1-\tan x\tan \left(\frac{\pi }{4}\right)}=\\ \frac{\tan x+1}{1-\tan x(1)}=\frac{\tan x+1}{1-\tan x}\end{array}$

$\begin{array}{c}\frac{\cos \left(a+b\right)}{\cos a\cos b}=\\ \frac{\cos a\cos b}{\cos a\cos b}-\frac{\sin a\sin b}{\cos a\cos b}=1-\tan a\tan b\end{array}$

$\begin{array}{c}\frac{\cos \left(x+h\right)-\cos x}{h}=\\ \frac{\cos x\cosh -\sin x\sinh -\cos x}{h}=\\ \frac{\cos x(\cosh -1)-\sin x\sinh }{h}=\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}\end{array}$

Satt. Athugaðu að $\sin \left(\alpha +\beta \right)=\sin \left(\pi -\gamma \right)$ og liðaðu hægri hliðina.

$\frac{1-\cos x}{\sin x},\frac{\sin x}{1+\cos x},$ með því að margfalda teljara og nefnara með $\sqrt{1-\cos x}$ og $\sqrt{1+\cos x},$ í þessari röð.

a) $\frac{3\sqrt{7}}{32}$ b) $\frac{31}{32}$ c) $\frac{3\sqrt{7}}{31}$

a) $\frac{\sqrt{3}}{2}$ b) $-\frac{1}{2}$ c) $-\sqrt{3}$

$\cos \theta =-\frac{2\sqrt{5}}{5},\sin \theta =\frac{\sqrt{5}}{5},\tan \theta =-\frac{1}{2},\csc \theta =\sqrt{5},\sec \theta =-\frac{\sqrt{5}}{2},\cot \theta =-2$

$\sin \left(\frac{\pi }{2}\right)$

$\frac{\sqrt{2-\sqrt{2}}}{2}$

$\frac{\sqrt{2-\sqrt{3}}}{2}$

$2+\sqrt{3}$

$-1-\sqrt{2}$

a) $\frac{3\sqrt{13}}{13}$ b) $-\frac{2\sqrt{13}}{13}$ c) $-\frac{3}{2}$

a) $\frac{\sqrt{10}}{4}$ b) $\frac{\sqrt{6}}{4}$ c) $\frac{\sqrt{15}}{3}$

$\frac{120}{169},–\frac{119}{169},–\frac{120}{119}$

$\frac{2\sqrt{13}}{13},\frac{3\sqrt{13}}{13},\frac{2}{3}$

$\cos ({74}^{\circ })$

$\cos (18x)$

$3\sin (10x)$

$-2\sin \left(-x\right)\cos \left(-x\right)=-2(-\sin \left(x\right)\cos \left(x\right))=\sin \left(2x\right)$

$\begin{array}{l}\frac{\sin \left(2\theta \right)}{1+\cos \left(2\theta \right)}{\tan }^2\theta =\frac{2\sin \left(\theta \right)\cos \left(\theta \right)}{1+{\cos }^2\theta -{\sin }^2\theta }{\tan }^2\theta =\\ \frac{2\sin \left(\theta \right)\cos \left(\theta \right)}{2{\cos }^2\theta }{\tan }^2\theta =\frac{\sin \left(\theta \right)}{\cos \theta }{\tan }^2\theta =\\ \tan \theta {\tan }^2\theta ={\tan }^3\theta \end{array}$

$\frac{1+\cos (12x)}{2}$

$\frac{3+\cos (12x)-4\cos (6x)}{8}$

$\frac{2+\cos (2x)-2\cos (4x)-\cos (6x)}{32}$

$\frac{3+\cos (4x)-4\cos (2x)}{3+\cos (4x)+4\cos (2x)}$

$\frac{1-\cos (4x)}{8}$

$\frac{3+\cos (4x)-4\cos (2x)}{4(\cos (2x)+1)}$

$\frac{\left(1+\cos \left(4x\right)\right)\sin x}{2}$

$4\sin x\cos x\left({\cos }^{2}x-{\sin }^{2}x\right)$

$\frac{2\tan x}{1+{\tan }^{2}x}=\frac{\frac{2\sin x}{\cos x}}{1+\frac{{\sin }^{2}x}{{\cos }^{2}x}}=\frac{\frac{2\sin x}{\cos x}}{\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}}=$
$\frac{2\sin x}{\cos x}.\frac{{\cos }^{2}x}{1}=2\sin x\cos x=\sin (2x)$

$\frac{2\sin x\cos x}{2{\cos }^{2}x-1}=\frac{\sin (2x)}{\cos (2x)}=\tan (2x)$

$\begin{array}{l}\sin (x+2x)=\sin x\cos (2x)+\sin (2x)\cos x\hfill \\ =\sin x({\cos }^{2}x-{\sin }^{2}x)+2\sin x\cos x\cos x\hfill \\ =\sin x{\cos }^{2}x-{\sin }^{3}x+2\sin x{\cos }^{2}x\hfill \\ =3\sin x{\cos }^{2}x-{\sin }^{3}x\hfill \end{array}$

$\begin{array}{l}\frac{1+\cos (2t)}{\sin (2t)-\cos t}=\frac{1+2{\cos }^{2}t-1}{2\sin t\cos t-\cos t}\hfill \\ =\frac{2{\cos }^{2}t}{\cos t(2\sin t-1)}\hfill \\ =\frac{2\cos t}{2\sin t-1}\hfill \end{array}$

$\begin{array}{l}\left({\cos }^2(4x)-{\sin }^2(4x)-\sin (8x))({\cos }^2(4x)-{\sin }^2(4x)+\sin (8x)\right)=\hfill \\ =\left(\cos (8x)-\sin (8x))(\cos (8x)+\sin (8x)\right)\hfill \\ ={\cos }^2(8x)-{\sin }^2(8x)\hfill \\ =\cos (16x)\hfill \end{array}$

Settu $\alpha$ inn í kósínus og $\beta$ inn í sínus og reiknaðu.

Svör eru mismunandi. Sumar jöfnur innihalda summu tveggja hornafallastæða og verða auðveldari úrlausnar þegar summunni er breytt í margfeldi. Til dæmis: $\frac{\sin (3x)+\sin x}{\cos x}=1.$ Þegar teljaranum er breytt í margfeldi verður jafnan: $\frac{2\sin (2x)\cos x}{\cos x}=1$

$8\left(\cos \left(5x\right)-\cos \left(27x\right)\right)$

$\sin \left(2x\right)+\sin \left(8x\right)$

$\frac{1}{2}\left(\cos \left(6x\right)-\cos \left(4x\right)\right)$

$2\cos \left(5t\right)\cos t$

$2\cos \left(7x\right)$

$2\cos \left(6x\right)\cos \left(3x\right)$

$\frac{1}{4}\left(1+\sqrt{3}\right)$

$\frac{1}{4}\left(\sqrt{3}-2\right)$

$\frac{1}{4}\left(\sqrt{3}-1\right)$

$\cos \left(\mathrm{80°}\right)-\cos \left(\mathrm{120°}\right)$

$\frac{1}{2}(\sin (\mathrm{221°})+\sin (\mathrm{205°}))$

$\sqrt{2}\cos \left(\mathrm{31°}\right)$

$2\cos (66.5°)\sin (34.5°)$

$2\sin \left(\mathrm{-1.5°}\right)\cos \left(\mathrm{0.5°}\right)$

$$\begin{array}{l}2\sin (7x)-2\sin x=2\sin (4x+3x)-2\sin (4x-3x)=\hfill \\ 2(\sin (4x)\cos (3x)+\sin (3x)\cos (4x))-2(\sin (4x)\cos (3x)-\sin (3x)\cos (4x))=\hfill \\ 2\sin (4x)\cos (3x)+2\sin (3x)\cos (4x))-2\sin (4x)\cos (3x)+2\sin (3x)\cos (4x))=\hfill \\ 4\sin (3x)\cos (4x)\hfill \\ \hfill \end{array}$$

$\sin x+\sin \left(3x\right)=2\sin \left(\frac{4x}{2}\right)\cos \left(\frac{-2x}{2}\right)=$
$2\sin (2x)\cos x=2(2\sin x\cos x)\cos x=$
$4\sin x{\cos }^{2}x$

$2\tan x\cos \left(3x\right)=\frac{2\sin x\cos (3x)}{\cos x}=\frac{2(.5(\sin (4x)-\sin (2x)))}{\cos x}$
$=\frac{1}{\cos x}\left(\sin (4x)-\sin (2x)\right)=\sec x\left(\sin \left(4x\right)-\sin \left(2x\right)\right)$

$2\cos ({35}^{\circ })\cos ({23}^{\circ }),\text{ 1}\text{.5081}$

$-2\sin ({33}^{\circ })\sin ({11}^{\circ }),-0.2078$

$\frac{1}{2}\left(\cos ({99}^{\circ })-\cos ({71}^{\circ })\right),-0.2410$

Þetta er ekki jafngildi, en $2{\cos }^{3}x$ er það.

$\tan \left(3t\right)$

$2\cos \left(2x\right)$

$-\sin (14x)$

Byrjaðu á $\cos x+\cos y.$ Gerðu innsetningu og látum $x=\alpha +\beta$ og 𝑦 = 𝛼 − 𝛽 . $y=\alpha -\beta ,$ Þá verður $\cos x+\cos y$ að
$\cos (\alpha +\beta )+\cos (\alpha -\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta +\cos \alpha \cos \beta +\sin \alpha \sin \beta =2\cos \alpha \cos \beta$

Þar sem $x=\alpha +\beta$ og $y=\alpha -\beta ,$ getum við leyst fyrir $\alpha$ og $\beta$ með x og y og sett niðurstöðurnar inn í 2 ⁢ c o s ⁡ 𝛼 ⁢ c o s ⁡ 𝛽 . $2\cos \alpha \cos \beta$ Þá fæst $2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right).$

$\frac{\cos \left(3x\right)+\cos x}{\cos \left(3x\right)-\cos x}=\frac{2\cos \left(2x\right)\cos x}{-2\sin \left(2x\right)\sin x}=-\cot \left(2x\right)\cot x$

$\begin{array}{l}\frac{\cos \left(2y\right)-\cos \left(4y\right)}{\sin \left(2y\right)+\sin \left(4y\right)}=\frac{-2\sin \left(3y\right)\sin \left(-y\right)}{2\sin \left(3y\right)\cos y}=\\ \frac{2\sin \left(3y\right)\sin \left(y\right)}{2\sin \left(3y\right)\cos y}=\tan y\end{array}$

$\begin{array}{l}\cos x-\cos \left(3x\right)=-2\sin (2x)\sin (-x)=\\ 2(2\sin x\cos x)\sin x=4{\sin }^{2}x\cos x\end{array}$

$\tan \left(\frac{\pi }{4}-t\right)=\frac{\tan \left(\frac{\pi }{4}\right)-\tan t}{1+\tan \left(\frac{\pi }{4}\right)\tan (t)}=\frac{1-\tan t}{1+\tan t}$

Hornafallajöfnur hafa ekki alltaf lausnir. Einfalt dæmi er $\cos (x)=\mathrm{-5.}$

$\frac{\pi }{3},\frac{2\pi }{3}$

$\frac{3\pi }{4},\frac{5\pi }{4}$

$\frac{\pi }{4},\frac{5\pi }{4}$

$\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

$\frac{\pi }{4},\frac{7\pi }{4}$

$\frac{7\pi }{6},\frac{11\pi }{6}$

$\frac{\pi }{18}$, $\frac{5\pi }{18}$, $\frac{13\pi }{18}$, $\frac{17\pi }{18}$, $\frac{25\pi }{18}$, $\frac{29\pi }{18}$

$\frac{3\pi }{12},\frac{5\pi }{12},\frac{11\pi }{12},\frac{13\pi }{12},\frac{19\pi }{12},\frac{21\pi }{12}$

$\frac{1}{6},\frac{5}{6},\frac{13}{6},\frac{17}{6},\frac{25}{6},\frac{29}{6},\frac{37}{6}$

$0,\frac{\pi }{3},\pi ,\frac{5\pi }{3}$

$\frac{\pi }{3},\pi ,\frac{5\pi }{3}$

$\frac{\pi }{3},\frac{3\pi }{2},\frac{5\pi }{3}$

$0,\pi$

$\pi -{\sin }^{-1}\left(-\frac{1}{4}\right),\frac{7\pi }{6},\frac{11\pi }{6},2\pi +{\sin }^{-1}\left(-\frac{1}{4}\right)$

$\frac{1}{3}\left({\sin }^{-1}\left(\frac{9}{10}\right)\right),$ $\frac{\pi }{3}-\frac{1}{3}\left({\sin }^{-1}\left(\frac{9}{10}\right)\right),$ $\frac{2\pi }{3}+\frac{1}{3}\left({\sin }^{-1}\left(\frac{9}{10}\right)\right),$ $\pi -\frac{1}{3}\left({\sin }^{-1}\left(\frac{9}{10}\right)\right),$ $\frac{4\pi }{3}+\frac{1}{3}\left({\sin }^{-1}\left(\frac{9}{10}\right)\right),$ $\frac{5\pi }{3}-\frac{1}{3}\left({\sin }^{-1}\left(\frac{9}{10}\right)\right)$

$0$

$\theta ={\sin }^{-1}\left(\frac{2}{3}\right),$ $\mathrm{\pi }-{\sin }^{-1}\left(\frac{2}{3}\right),$ $\mathrm{\pi }+{\sin }^{-1}\left(\frac{2}{3}\right),$ $2\mathrm{\pi }-{\sin }^{-1}\left(\frac{2}{3}\right)$

$\frac{3\pi }{2},\frac{\pi }{6},\frac{5\pi }{6}$