A) \(\displaystyle{ sin 30° + cos 120°=}\)
B) \(\displaystyle{ \frac{sin 120° + cos300°}{tg(-225°)}=}\)
C) \(\displaystyle{ \frac{sin 125° + 2cos-60°}{tg(-225°)}=}\)
D) \(\displaystyle{ sin\frac{13}{6}pi + cos\frac{7}{3}pi =}\)
E) \(\displaystyle{ sin\frac{ \frac{37}{6}pi - 2 cos \frac{pi}{6} }{-tg \frac{5}{4}pi }}\)
Wie ktoś jak to rozwiązać?
Kąt obrotu
- anibod
- Użytkownik
- Posty: 188
- Rejestracja: 12 wrz 2008, o 10:27
- Płeć: Kobieta
- Lokalizacja: Sulejówek
- Pomógł: 58 razy
Kąt obrotu
korzystasz ze wzorów redukcyjnych
a)
\(\displaystyle{ \sin{30^{0}}+\cos{120^{0}}= \sin(90^{0}-60^{0})+\cos(90^{0}+30^{0})=\sin(90^{0}+(-60^{0}))-\sin{30^{0}}=\cos(-60^{0})-\frac{1}{2}=\cos{60^{0}}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0}\)
[ Dodano: 28 Września 2008, 17:53 ]
b) \(\displaystyle{ \frac{\sin{120^{0}}+ \cos{300^{0}}}{tg(-225^{0})}=\frac{ \sin(90^{0}+30^{0})+\cos(270^{0}+30^{0})}{-\tg{225^{0}}}=\frac{\cos30^{0}+sin30^{0}}{-tg(180^{0}+45^{0})}=\frac{\frac{\sqrt{3}}{2}+\frac{1}{2}}{-tg45^{0}}=-\frac{\sqrt{3}+1}{2}}\)
[ Dodano: 28 Września 2008, 18:16 ]
d) \(\displaystyle{ \pi=180^{0}}\)
\(\displaystyle{ \sin{\frac{13}{6} \pi+ \cos{{\frac{7}{3} \pi}}=\sin{390^{0}}+\cos{420^{0}}=\sin(360^{0}+30^{0})+\cos(360^{0}+60^{0})=\sin30^{0}+\cos60^{0}=\frac{1}{2}+\frac{1}{2}=1}\)
[ Dodano: 28 Września 2008, 18:35 ]
e) \(\displaystyle{ \frac{\sin{\frac{37}{6} \pi}-2\cos{\frac{\pi}{6}}}{-tg{\frac{5}{4}}\pi} =\frac{\sin(3 360^{0}+30^{0})-2 \frac{\sqrt{3}}{2}}{-tg(180^{0}+45^{0})}=\frac{\sin{30^{0}}-\sqrt{3}}{-tg{45^{0}}}=-\frac{1}{2}+\sqrt{3}}\)
a)
\(\displaystyle{ \sin{30^{0}}+\cos{120^{0}}= \sin(90^{0}-60^{0})+\cos(90^{0}+30^{0})=\sin(90^{0}+(-60^{0}))-\sin{30^{0}}=\cos(-60^{0})-\frac{1}{2}=\cos{60^{0}}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0}\)
[ Dodano: 28 Września 2008, 17:53 ]
b) \(\displaystyle{ \frac{\sin{120^{0}}+ \cos{300^{0}}}{tg(-225^{0})}=\frac{ \sin(90^{0}+30^{0})+\cos(270^{0}+30^{0})}{-\tg{225^{0}}}=\frac{\cos30^{0}+sin30^{0}}{-tg(180^{0}+45^{0})}=\frac{\frac{\sqrt{3}}{2}+\frac{1}{2}}{-tg45^{0}}=-\frac{\sqrt{3}+1}{2}}\)
[ Dodano: 28 Września 2008, 18:16 ]
d) \(\displaystyle{ \pi=180^{0}}\)
\(\displaystyle{ \sin{\frac{13}{6} \pi+ \cos{{\frac{7}{3} \pi}}=\sin{390^{0}}+\cos{420^{0}}=\sin(360^{0}+30^{0})+\cos(360^{0}+60^{0})=\sin30^{0}+\cos60^{0}=\frac{1}{2}+\frac{1}{2}=1}\)
[ Dodano: 28 Września 2008, 18:35 ]
e) \(\displaystyle{ \frac{\sin{\frac{37}{6} \pi}-2\cos{\frac{\pi}{6}}}{-tg{\frac{5}{4}}\pi} =\frac{\sin(3 360^{0}+30^{0})-2 \frac{\sqrt{3}}{2}}{-tg(180^{0}+45^{0})}=\frac{\sin{30^{0}}-\sqrt{3}}{-tg{45^{0}}}=-\frac{1}{2}+\sqrt{3}}\)