Ответы и объяснения

2014-03-06T14:18:50+04:00
3.
\cos\alpha=\frac{4}{5};\ \ \frac{3\pi}{2}<\alpha<2\pi;\\&#10;\sin\alpha-ctg\alpha-?\\&#10;\alpha\in(\frac{3\pi}{2};2\pi);==>-1<\sin\alpha<0;\ \ ctg\alpha<0;\\&#10;\sin\alpha=-\sqrt{1-\cos^2\alpha};\\&#10;ctg\alpha=\frac{\cos\alpha}{\sin\alpha}=\frac{\cos\alpha}{-\sqrt{1-\cos^2\alpha}}=-\frac{\cos\alpha}{\sqrt{1-\cos^2\alpha}};\\&#10;\sin\alpha-ctg\alpha=-\sqrt{1-\cos^2\alpha}+\frac{\cos\alpha}{\sqrt{1-\cos^2\alpha}}=\\&#10;
=\frac{\cos^2\alpha+\cos\alpha-1}{\sqrt{1-\cos^2\alpha}}=\frac{(\frac{4}{5})^2+\frac{4}{5}-1}{\sqrt{1-(\frac{4}{5})^2}}=\frac{\frac{16}{25}-\frac{25}{25}+\frac{20}{25}}{\sqrt{1-\frac{16}{25}}}=\\&#10;=\frac{\frac{16+20-25}{25}}{\sqrt{\frac{25-16}{25}}}=\frac{\frac{11}{25}}{\sqrt{\frac{9}{25}}}=\frac{\frac{11}{25}}{\frac{3}{5}}=\frac{11\cdot5}{25\cdot3}=\frac{11}{15};\\
Вариант в)
 \frac{11}{15}



4.
\frac{\sin(\frac{3\pi}{2}-\alpha)\cos(\pi+\alpha)ctg(\frac{3\pi}{2}+\alpha)}{tg(2\pi-\alpha)}=\\&#10;=\frac{\left(\sin\frac{3\pi}{2}\cos\alpha-\cos\frac{3\pi}{2}\sin\alpha\right)\cdot\left(\cos\pi\cos\alpha-\sin\pi\sin\alpha\right)\cdot\frac{\cos(\frac{3\pi}{2}+\alpha)}{\sin(\frac{3\pi}{2}+\alpha)}}{\frac{\sin(2\pi-\alpha)}{\cos(2\pi-\alpha)}}=\\&#10;|&#10;sin\frac{3\pi}{2}=-1;\ \ \cos\pi=-1;\ \ \cos\frac{3\pi}{2}=\sin\pi=0|\\&#10;
=\frac{\cos^2\alpha\cos(\frac{3\pi}{2}+\alpha)\cos(2\pi-\alpha)}{\sin(\frac{3\pi}{2}+\alpha)\sin(2\pi-\alpha)}=\\&#10;=\frac{\cos^2\alpha(\cos\frac{3\pi}{2}\cos\alpha-\sin\frac{3\pi}{2}\sin\alpha)(\cos2\pi\cos\alpha+\sin2\pi\sin\alpha)}{(\sin\frac{3\pi}{2}\cos\alpha+\cos\frac{3\pi}{2}\sin\alpha)(\sin2\pi\cos\alpha-\cos2\pi\sin\alpha)}=\\&#10;=\frac{\cos^2\alpha\cdot\sin\alpha\cdot\cos\alpha}{-\cos\alpha\cdot(-\sin\alpha)}=\cos^2\alpha;\\&#10;
Вариант а) сos²α



5.
\frac{ctg\alpha\cdot\sin\alpha}{1-(\sin\alpha+\cos\alpha)^2}=\frac{\frac{\cos\alpha}{\sin\alpha}\cdot\sin\alpha}{1-\cos^2\alpha-2\cos\alpha\sin\alpha-\sin^2\alpha}=\frac{\cos\alpha}{1-1-2\cos\alpha\sin\alpha}=\\&#10;=\frac{\cos\alpha}{-2\sin\alpha\cos\alpha}=-\frac{1}{2\sin\alpha};
Вариант: г)
- \frac{1}{2\sin\alpha}



6.
\frac{\sin(158^0-\alpha)}{\cos(68^0-\alpha)}=\frac{\sin(90^0+(68^0-\alpha))}{\cos(68^0-\alpha)}=\\&#10;=\frac{\sin90^0\cos(68^0-\alpha)+\cos90^0\sin(68^0-\alpha)}{\cos(68^0-\alpha)}=\\&#10;|\sin90^0=1;\ \ \cos90^0=0|\\&#10;=\frac{1\cdot\cos(68^0-\alpha)+0\cdot\sin(68^0-\alpha)}{\cos(68^0-\alpha)}=\frac{\cos(68^0-\alpha)}{\cos(68^0-\alpha)}=1
Вариант: в) 1