Trigonometri

[Janhaa]

Finn en eksakt verdi for:

[tex]\cos(22,5^o)\cdot \cos(45^o)\cdot \cos(67,5^o)[/tex]

uten bruk av kalkulator - sjølsagt...

[espen180]

Finn en eksakt verdi for:

[tex]\cos(22,5^o)\cdot \cos(45^o)\cdot \cos(67,5^o)[/tex]

uten bruk av kalkulator - sjølsagt...

[tex]\cos(22.5^\circ)=\sqrt{\frac{1+\cos(45^\circ)}{2}}=\sqrt{\frac{2+\sqrt{2}}{4}}=\frac{\sqrt{2+\sqrt{2}}}{2} \\ \cos(67.5^\circ)=\sqrt{\frac{1+\cos(135^\circ}{2}} \\ \cos(135^\circ)=-\cos(45^\circ)=-\frac{\sqrt{2}}{2} \\ \cos(67.5^\circ)=\sqrt{\frac{1-\cos(45^\circ)}{2}}=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}=\sqrt{\frac{2-\sqrt{2}}{4}}=\frac{\sqrt{2-\sqrt{2}}}{2} \\ \cos(22.5^\circ)\cdot\cos(45^\circ)=\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{4+2\sqrt{2}}}{4} \\ \cos(22.5^\circ)\cdot\cos(45^\circ)\cdot\cos(67.5^\circ)=\frac{\sqrt{4+2\sqrt{2}}}{4}\cdot\frac{\sqrt{2-\sqrt{2}}}{2}=\frac{\sqrt{\left(4+2\sqrt{2}\right)\left(2-sqrt{2}\right)}}{8} \\ \left(4+2\sqrt{2}\right)\left(2-sqrt{2}\right)=\frac{\sqrt{8-4\sqrt{2}+4\sqrt{2}-4}}{8}=\frac{\sqrt{4}}{8}=\frac{2}{8}=\underline{\underline{\frac{1}{4}}}[/tex]

[drgz]

[tex]\cos(22.5^o)\cos(67.5^o)\cos(45^o)=cos(45^o)\cdot\frac{1}{2}(\cos(45^o)+\cos(90^o))=\frac{1}{2}\cos(45^o)\cos(45^o)=\frac{1}{2}(\frac{1}{2}(\cos(0^o)+\cos(90^o))=\frac{1}{4}(1+0)=\frac{1}{4}[/tex]

der en kun benytter
[tex]\cos(a)\cos(b)=\frac{1}{2}(\cos(a-b)+\cos(a+b))[/tex]

[Janhaa]

Fint arbeid på begge to. Her er en 3. løsning:

[tex]A=\cos(22,5^o)\cdot {\sqrt2\over 2}\cdot \cos(67,5^o)[/tex]

[tex]A=\cos(22,5^o)\cdot {\sqrt2\over 2}\cdot \sin(90^o-67,5^o)[/tex]
der
[tex]\sin(90^o-67,5^o)=\sin(22,5^o)=\cos(67,5^o)[/tex]
slik at
[tex]A=\cos(22,5^o)\cdot {\sqrt2\over 2}\cdot \sin(22,5^o)[/tex]

[tex]A={\sqrt2\over 2}\cdot {1\over 2}\sin(45^o)[/tex]

[tex]A=({\sqrt2\over 2})^2\cdot {1\over 2}={1\over 4}[/tex]