Kern Wiskunde B deel 2 - Hoofdstuk 10 - Goniometrische formules oefentoetsen & antwoorden

Deze bewering is onjuist. De somformules zijn:$\sin \left( t+u \right)=\sin \left( t \right)\cos \left( u \right)+\cos \left( t \right)\sin \left( u \right)$ $\cos \left( t+u \right)=\cos \left( t \right)\cos \left( u \right)-\sin \left( t \right)\sin \left( u \right)$Deze bewering is onjuist. De verschilformules zijn:$\sin \left( t-u \right)=\sin \left( t \right)\cos \left( u \right)-\cos \left( t \right)\sin \left( u \right)$ $\cos \left( t-u \right)=\cos \left( t \right)\cos \left( u \right)+\sin \left( t \right)\sin \left( u \right)$ Voor de frequentie $f$ en de trillingstijd $T$ geldt: $f=\frac{1}{T}$.$f=\frac{1}{0,2}=5$.De frequentie is 5 Hz.We beginnen met de formule $u=a\cdot \sin \left( 2\pi ft \right)$Een amplitude van 7 geeft $u=7\sin \left( 2\pi ft \right)$Een frequentie van 5 Hz geeft $u=7\sin \left( 2\pi \cdot 5t \right)$Op $t=\tfrac{1}{15}$ wordt de evenwichtsstand stijgend gepasseerd geeft $u=7\sin \left( 2\pi \cdot 5\left( t-\tfrac{1}{15} \right) \right)$$u=7\sin \left( 2\pi \cdot 5\left( t-\tfrac{1}{15} \right) \right)=7\sin \left( 10\pi \left( t-\tfrac{1}{15} \right) \right)=7\sin \left( 10\pi t-\tfrac{2}{3} \right)$ $\sin (3x)=\tfrac{1}{2}\sqrt{3}$$3x=\tfrac{1}{3}\pi +k\cdot 2\pi \,\,\,\vee \,\,\,3x=\pi -\tfrac{1}{3}\pi +k\cdot 2\pi $$x=\tfrac{1}{9}\pi +k\cdot \tfrac{2}{3}\pi \,\,\,\vee \,\,\,x=\tfrac{2}{9}\pi +k\cdot \tfrac{2}{3}\pi $ ${{\cos }^{3}}(x)={{\cos }^{2}}(x)$${{\cos }^{3}}(x)-{{\cos }^{2}}(x)=0$${{\cos }^{2}}(x)\left( \cos (x)-1 \right)=0$${{\cos }^{2}}(x)=0\,\,\,\vee \,\,\,\cos (x)-1=0$$\cos (x)=0\,\,\,\vee \,\,\,\cos (x)=1$$x=\tfrac{1}{2}\pi +k\cdot \pi \,\,\,\vee \,\,\,x=0+k\cdot 2\pi $$\sin (\pi +x)=\cos \left( \tfrac{1}{2}\pi -x \right)$We gebruiken $\sin (x)=\cos \left( x-\tfrac{1}{2}\pi  \right)$ , dus $\sin (\pi +x)=\cos \left( x+\tfrac{1}{2}\pi  \right)$ dit geeft:$\cos \left( x+\tfrac{1}{2}\pi  \right)=\cos \left( \tfrac{1}{2}\pi -x \right)$$x+\tfrac{1}{2}\pi =\tfrac{1}{2}\pi -x+k\cdot 2\pi \,\,\,\vee \,\,\,x+\tfrac{1}{2}\pi =-\left( \tfrac{1}{2}\pi -x \right)+k\cdot 2\pi $$2x=0+k\cdot 2\pi \,\,\,\vee \,\,\,0=-\pi +k\cdot 2\pi $$x=0+k\cdot 2\pi $ ($\,0=-\pi +k\cdot 2\pi $ is een tegenstrijdigheid en heeft geen oplossingen)$2{{\sin }^{2}}(x)=2{{\cos }^{2}}\left( x \right)+4$$2\left( 1-{{\cos }^{2}}(x) \right)=2{{\cos }^{2}}\left( x \right)+4$$2-2{{\cos }^{2}}(x)=2{{\cos }^{2}}\left( x \right)+4$$-4{{\cos }^{2}}(x)=2$${{\cos }^{2}}(x)=-\tfrac{1}{2}$Deze vergelijking heeft geen oplossingen, want ${{\cos }^{2}}(x)\ge 0$ $3\tan (\pi x)=\sqrt{3}$$\tan (\pi x)=\tfrac{1}{3}\sqrt{3}$$\pi x=\tfrac{1}{6}\pi +k\cdot \pi $ $x=\tfrac{1}{6}+k$$1-\sin \left( \tfrac{3}{4}x \right)=\cos \left( \tfrac{3}{4}x \right)+1$$-\sin \left( \tfrac{3}{4}x \right)=\cos \left( \tfrac{3}{4}x \right)$$-\frac{\sin \left( \tfrac{3}{4}x \right)}{\cos \left( \tfrac{3}{4}x \right)}=1$$\tan \left( \tfrac{3}{4}x \right)=-1$$\tfrac{3}{4}x=-\tfrac{1}{4}\pi +k\cdot \pi $$x=-\tfrac{1}{3}\pi +\tfrac{4}{3}k\cdot \pi $${{\tan }^{2}}(\tfrac{1}{2}\pi +x)=\sqrt{3}\cdot \tan (\tfrac{1}{2}\pi +x)$${{\tan }^{2}}(\tfrac{1}{2}\pi +x)-\sqrt{3}\cdot \tan (\tfrac{1}{2}\pi +x)=0$$\tan (\tfrac{1}{2}\pi +x)\left( \tan (\tfrac{1}{2}\pi +x)-\sqrt{3} \right)=0$$\tan (\tfrac{1}{2}\pi +x)=0\,\,\,\vee \,\,\,\tan (\tfrac{1}{2}\pi +x)=\sqrt{3}$$\tfrac{1}{2}\pi +x=0+k\cdot \pi \,\,\,\vee \,\,\,\tfrac{1}{2}\pi +x=\tfrac{1}{3}\pi +k\cdot \pi $$x=-\tfrac{1}{2}\pi +k\cdot \pi \,\,\,\vee \,\,\,x=-\tfrac{1}{6}\pi +k\cdot \pi $ We gebruiken een somformule voor $\sin \left( x+\frac{\pi }{4} \right)$. Dit geeft:$\sin \left( x+\frac{\pi }{4} \right)=\sin \left( x \right)\cos \left( \frac{\pi }{4} \right)+\cos \left( x \right)\sin \left( \frac{\pi }{4} \right)$$\cos \left( \frac{\pi }{4} \right)=\tfrac{1}{2}\sqrt{2}$ en $\sin \left( \frac{\pi }{4} \right)=\tfrac{1}{2}\sqrt{2}$ geeft:$\sin \left( x \right)\cos \left( \frac{\pi }{4} \right)+\cos \left( x \right)\sin \left( \frac{\pi }{4} \right)=\sin \left( x \right)\cdot \tfrac{1}{2}\sqrt{2}+\cos \left( x \right)\cdot \tfrac{1}{2}\sqrt{2}$Uit $\sin \left( x \right)\cdot \tfrac{1}{2}\sqrt{2}+\cos \left( x \right)\cdot \tfrac{1}{2}\sqrt{2}=\tfrac{1}{2}\sqrt{2}\left( \sin \left( x \right)+\cos \left( x \right) \right)$ volgt:$\sin \left( x+\frac{\pi }{4} \right)=\tfrac{1}{2}\sqrt{2}\left( \sin \left( x \right)+\cos \left( x \right) \right)$We starten met de rechterkant:$\frac{\sin (2x)}{1+\cos (2x)}$We gebruiken de verdubbelingsformules $\sin (2x)=2\sin (x)\cos (x)$ en $\cos (2x)=2{{\cos }^{2}}(x)-1$ (het kan ook met $\cos (2x)={{\cos }^{2}}(x)-{{\sin }^{2}}(x)$, maar dat is iets meer werk).Dit geeft $\frac{\sin (2x)}{1+\cos (2x)}=\frac{2\sin (x)\cos (x)}{1+2{{\cos }^{2}}(x)-1}=\frac{2\sin (x)\cos (x)}{2{{\cos }^{2}}(x)}$Vereenvoudigen geeft:$\frac{2\sin (x)\cos (x)}{2{{\cos }^{2}}(x)}=\frac{\sin (x)}{\cos (x)}$Hieruit volgt:$\frac{\sin (x)}{\cos (x)}=\tan (x)$ In punt $P$ geldt $f\left( p \right)=g\left( p \right)$$4{{\cos }^{2}}\left( p-\tfrac{1}{2}\pi  \right)=\frac{1}{{{\sin }^{2}}\left( p-\tfrac{1}{2}\pi  \right)}-1$$4{{\cos }^{2}}\left( p-\tfrac{1}{2}\pi  \right)\cdot {{\sin }^{2}}\left( p-\tfrac{1}{2}\pi  \right)=1-{{\sin }^{2}}\left( p-\tfrac{1}{2}\pi  \right)$$4{{\cos }^{2}}\left( p-\tfrac{1}{2}\pi  \right)\cdot {{\sin }^{2}}\left( p-\tfrac{1}{2}\pi  \right)={{\cos }^{2}}\left( p-\tfrac{1}{2}\pi  \right)$${{\cos }^{2}}\left( p-\tfrac{1}{2}\pi  \right)=0\,\,\,\vee \,\,\,4\cdot {{\sin }^{2}}\left( p-\tfrac{1}{2}\pi  \right)=1$$\cos \left( p-\tfrac{1}{2}\pi  \right)=0\,\,\,\vee \,\,\,{{\sin }^{2}}\left( p-\tfrac{1}{2}\pi  \right)=\tfrac{1}{4}$$\cos \left( p-\tfrac{1}{2}\pi  \right)=0$ geeft$p-\tfrac{1}{2}\pi =\tfrac{1}{2}\pi +k\pi $$p=\pi +k\pi $${{\sin }^{2}}\left( p-\tfrac{1}{2}\pi  \right)=\tfrac{1}{4}$ geeft$\sin \left( p-\tfrac{1}{2}\pi  \right)=-\tfrac{1}{2}\,\,\,\vee \,\,\,\sin \left( p-\tfrac{1}{2}\pi  \right)=\tfrac{1}{2}$$p-\tfrac{1}{2}\pi =-\tfrac{1}{6}\pi +2k\pi \,\,\,\vee \,\,\,p-\tfrac{1}{2}\pi =\tfrac{7}{6}\pi +2k\pi \,\,\,\vee \,\,\,p-\tfrac{1}{2}\pi =\tfrac{1}{6}\pi +2k\pi \,\,\,\vee \,\,\,p-\tfrac{1}{2}\pi =\tfrac{5}{6}\pi +2k\pi $$p=\tfrac{1}{3}\pi +2k\pi \,\,\,\vee \,\,\,p=\tfrac{5}{3}\pi +2k\pi \,\,\,\vee \,\,\,p=\tfrac{2}{3}\pi +2k\pi \,\,\,\vee \,\,\,p=\tfrac{4}{3}\pi +2k\pi $Er moet gelden $0 \lt p\le \tfrac{1}{2}\pi $, dus $p=\tfrac{1}{3}\pi $Er geldt: $\left| AB \right|=L\left( a \right)=f\left( a \right)-g\left( a \right)$Als $\left| AB \right|$ is maximaal, dan ${L}'\left( a \right)=0$$L\left( a \right)=f\left( a \right)-g\left( a \right)$$=4{{\cos }^{2}}\left( a-\tfrac{1}{2}\pi  \right)-\left( \frac{1}{{{\sin }^{2}}\left( a-\tfrac{1}{2}\pi  \right)}-1 \right)=4{{\cos }^{2}}\left( a-\tfrac{1}{2}\pi  \right)-\frac{1}{{{\sin }^{2}}\left( a-\tfrac{1}{2}\pi  \right)}+1$$=4{{\cos }^{2}}\left( a-\tfrac{1}{2}\pi  \right)-{{\sin }^{-2}}\left( a-\tfrac{1}{2}\pi  \right)+1$${L}'\left( a \right)=8\cos \left( a-\tfrac{1}{2}\pi  \right)\cdot \sin \left( a-\tfrac{1}{2}\pi  \right)+2{{\sin }^{-3}}\left( a-\tfrac{1}{2}\pi  \right)\cdot -\cos \left( a-\tfrac{1}{2}\pi  \right)$$=8\cos \left( a-\tfrac{1}{2}\pi  \right)\cdot \sin \left( a-\tfrac{1}{2}\pi  \right)-\frac{2\cos \left( a-\tfrac{1}{2}\pi  \right)}{{{\sin }^{3}}\left( a-\tfrac{1}{2}\pi  \right)}$${L}'\left( a \right)=0$ geeft:$8\cos \left( a-\tfrac{1}{2}\pi  \right)\cdot \sin \left( a-\tfrac{1}{2}\pi  \right)-\frac{2\cos \left( a-\tfrac{1}{2}\pi  \right)}{{{\sin }^{3}}\left( a-\tfrac{1}{2}\pi  \right)}=0$$8\cos \left( a-\tfrac{1}{2}\pi  \right)\cdot \sin \left( a-\tfrac{1}{2}\pi  \right)=\frac{2\cos \left( a-\tfrac{1}{2}\pi  \right)}{{{\sin }^{3}}\left( a-\tfrac{1}{2}\pi  \right)}$$8\cos \left( a-\tfrac{1}{2}\pi  \right)\cdot {{\sin }^{4}}\left( a-\tfrac{1}{2}\pi  \right)=2\cos \left( a-\tfrac{1}{2}\pi  \right)$$\cos \left( a-\tfrac{1}{2}\pi  \right)=0\,\,\,\vee \,\,\,4{{\sin }^{4}}\left( a-\tfrac{1}{2}\pi  \right)=1$$\cos \left( a-\tfrac{1}{2}\pi  \right)=0$ geeft:$a-\tfrac{1}{2}\pi =\tfrac{1}{2}\pi +k\pi $$a=\pi +k\pi $Geen oplossing op $0 \lt a \lt Asymptoot als $\cos (\tfrac{1}{2}x)=0$$\cos (\tfrac{1}{2}x)=0$$\tfrac{1}{2}x=\tfrac{1}{2}\pi +k\cdot \pi $$x=\pi +k\cdot 2\pi $Voor $0\le x\le 2\pi $ is een exacte vergelijking van de asymptoot $x=\pi $.Stel $l:\,\,\,y=ax+b$$a={f}'\left( \tfrac{1}{2}\pi  \right)$${f}'\left( x \right)=\left( 1+{{\tan }^{2}}\left( \tfrac{1}{2}x \right) \right)\cdot \tfrac{1}{2}$ of ${f}'\left( x \right)=\frac{1}{2{{\cos }^{2}}\left( \tfrac{1}{2}x \right)}$$a={f}'\left( \tfrac{1}{2}\pi  \right)=\left( 1+{{\tan }^{2}}\left( \tfrac{1}{4}\pi  \right) \right)\cdot \tfrac{1}{2}=\left( 1+{{1}^{2}} \right)\cdot \tfrac{1}{2}=1$ Dus $l:\,\,\,y=x+b$Om te berekenen hebben we een punt nodig dat op de raaklijn ligt.$f\left( \tfrac{1}{2}\pi  \right)=\tan \left( \tfrac{1}{4}\pi  \right)-1=0$, dus $\left( \tfrac{1}{2}\pi ,0 \right)$ ligt op de raaklijn.$\left( \tfrac{1}{2}\pi ,0 \right)$ substitueren in $l:\,\,\,y=x+b$ geeft: $0=\tfrac{1}{2}\pi +b\,\,\,\Rightarrow \,\,\,b=-\tfrac{1}{2}\pi $Dus $l:\,\,\,y=x-\tfrac{1}{2}\pi $Het $x$-coördinaat van $P$ is $x=\pi $.Het $y$-coördinaat van $P$ is $y=\pi -\tfrac{1}{2}\pi =\tfrac{1}{2}\pi $, dus $P\left( \pi ,\tfrac{1}{2}\pi  \right)$.Als $P$ op $g$ ligt, moet gelden $g\left( \pi  \right)=\tfrac{1}{2}\pi $$g\left( \pi  \right)=\tfrac{1}{2}\pi \sin \left( \pi -\tfrac{1}{2}\pi  \right)=\tfrac{1}{2}\pi \sin \left( \tfrac{1}{2}\pi  \right)=\tfrac{1}{2}\pi \cdot 1=\tfrac{1}{2}\pi $Punt $P$ ligt dus op de grafiek van $g$. $\angle \left( \overrightarrow{OA},\overrightarrow{OB} \right)={{45}^{\circ }}\,\,\,\Rightarrow \,\,\,\cos \angle \left( \overrightarrow{OA},\overrightarrow{OB} \right)=\cos \left( {{45}^{\circ }} \right)=\tfrac{1}{2}\sqrt{2}$$\overrightarrow{OA}=\left( \begin{matrix} 6 \\  18 \\ \end{matrix} \right)$, $\overrightarrow{OB}=\left( \begin{matrix}  p \\ q \\  \end{matrix} \right)$ en $\left| \overrightarrow{OB} \right|=4\sqrt{5}$.$\cos \angle \left( \overrightarrow{OA},\overrightarrow{OB} \right)=\frac{\left( \begin{matrix}  6 \\  18 \\ \end{matrix} \right)\cdot \left( \begin{matrix} p \\  q \\ \end{matrix} \right)}{\left| \left( \begin{matrix}   6 \\   18 \\ \end{matrix} \right) \right|\cdot \left| \left( \begin{matrix}  p \\   q \\ \end{matrix} \right) \right|}=\tfrac{1}{2}\sqrt{2}$$\frac{6p+18q}{\sqrt{360}\cdot 4\sqrt{5}}=\tfrac{1}{2}\sqrt{2}$$\frac{6p+18q}{4\sqrt{1800}}=\tfrac{1}{2}\sqrt{2}$$6p+18q=2\sqrt{3600}$$6p+18q=120$$\left| \overrightarrow{OB} \right|=\left|  \begin{pmatrix}   p \\   q \\ \end{pmatrix}  \right|=4\sqrt{5}$ geeft ${{p}^{2}}+{{q}^{2}}={{\left( 4\sqrt{5} \right)}^{2}}=80$$\left\{ \begin{pmatrix}   6p+18q=120 \\  {{p}^{2}}+{{q}^{2}}=80 \\ \end{pmatrix} \right.\Rightarrow \left\{ \begin{pmatrix}    p=-3q+20 \\   {{p}^{2}}+{{q}^{2}}=80 \\ \end{pmatrix} \right.\Rightarrow {{\left( -3q+20 \right)}^{2}}+{{q}^{2}}=80$ ${{\left( -3q+20 \right)}^{2}}+{{q}^{2}}=80$$9{{q}^{2}}-120q+400+{{q}^{2}}=80$$10{{q}^{2}}-120q+320=0$${{q}^{2}}-12q+32=0$$\left( q-4 \right)\left( q-8 \right)=0$$q=4\,\,\,\vee \,\,\,q=8$$q=4$ geeft $6p+18\cdot 4=120\Rightarrow 6p+72=120\Rightarrow 6p=48\Rightarrow p=8$$q=8$ geeft $6p+18\cdot 8=120\Rightarrow 6p+144=120\Rightarrow 6p=-24\Rightarrow p=-4$De mogelijke coördinaten zijn $\left( 8,4 \right)$ en $\left( -4,8 \right)$. $p=3$ geeft $l:\left( \begin{matrix}  x  \\  y  \\ \end{matrix} \right)=\left(  \begin{matrix}  2  \\  1  \\ \end{matrix} \right)+\lambda \left( \begin{matrix}  3  \\    1  \\ \end{matrix} \right)$.We bereken de hoek met behulp van normaalvectoren.Een normaalvector van $l$ is $\overrightarrow{{{n}_{l}}}=\left( \begin{matrix}  -1 \\   3 \\ \end{matrix} \right)$.Een normaalvector van $m$ is $\overrightarrow{{{n}_{m}}}=\left( \begin{matrix}  2 \\   1 \\ \end{matrix} \right)$.We berekenen de hoek met de formule $\cos \left( \angle \left( l,m \right) \right)=\frac{\left| \overrightarrow{{{n}_{l}}}\cdot \overrightarrow{{{n}_{m}}} \right|}{\left| \overrightarrow{{{n}_{l}}} \right|\cdot \left| \overrightarrow{{{n}_{m}}} \right|}$ $\cos \left( \angle \left( l,m \right) \right)=\frac{\left| \left( \begin{matrix}  -1 \\   3 \\ \end{matrix} \right)\cdot \left( \begin{matrix}  2 \\   1 \\ \end{matrix} \right) \right|}{\left| \left( \begin{matrix}  -1 \\   3 \\ \end{matrix} \right) \right|\cdot \left| \left( \begin{matrix}  2 \\   1 \\ \end{matrix} \right) \right|}=\frac{\left| -1\cdot 2+3\cdot 1 \right|}{\sqrt{10}\cdot \sqrt{5}}=\frac{1}{\sqrt{50}}=0,141\ldots $ $\angle \left( l,m \right)\approx {{82}^{\circ }}$De lijnen staan loodrecht op elkaar als geldt $\overrightarrow{{{n}_{l}}}\cdot \overrightarrow{{{n}_{m}}}=0$$\overrightarrow{{{n}_{l}}}=\left( \begin{matrix}   -1 \\   p \\ \end{matrix} \right)$ en $\overrightarrow{{{n}_{m}}}=\left( \begin{matrix}  2 \\   1 \\ \end{matrix} \right)$$\overrightarrow{{{n}_{l}}}\cdot \overrightarrow{{{n}_{m}}}=\left( \begin{matrix}  -1 \\   p \\ \end{matrix} \right)\cdot \left( \begin{matrix}  2 \\   1 \\ \end{matrix} \right)=-2+p=0\,\,\,\Rightarrow \,\,\,p=2$