Ответы и объяснения

2013-09-06T21:59:46+04:00
По действиям:
1). tg 42^{\circ} = \frac{sin 42^{\circ}}{cos 42^{\circ}}= \frac{sin(45^{\circ}-3^{\circ})}{cos(45^{\circ}-3^{\circ})} = \frac{sin 45^{\circ}*cos 3^{\circ}-cos 45^{\circ}*sin 3^{\circ}}{cos 45^{\circ}*cos 3^{\circ}+sin 45^{\circ}*sin 3^{\circ}}=
= \frac{\frac{\sqrt{2}}{2}*cos 3^{\circ}-\frac{\sqrt{2}}{2}*sin 3^{\circ}}{\frac{\sqrt{2}}{2}*cos 3^{\circ}+\frac{\sqrt{2}}{2}*sin 3^{\circ}}= \frac{cos 3^{\circ}-sin 3^{\circ}}{cos 3^{\circ}+sin 3^{\circ}}

2). cos 6^{\circ}* \frac{cos 3^{\circ}-sin 3^{\circ}}{cos 3^{\circ}+sin 3^{\circ}}= (cos^{2} 3^{\circ}- sin ^{2} 3^{\circ})* \frac{cos 3^{\circ}-sin 3^{\circ}}{cos 3^{\circ}+sin 3^{\circ}}=
= (cos 3^{\circ}- sin 3^{\circ})*(cos 3^{\circ}+ sin 3^{\circ})* \frac{cos 3^{\circ}-sin 3^{\circ}}{cos 3^{\circ}+sin 3^{\circ}}= (cos 3^{\circ}- sin 3^{\circ})^{2}=
= cos^{2} 3^{\circ} -2*cos 3^{\circ}*sin 3^{\circ} + sin^{2} 3^{\circ}= 1-sin 6^{\circ}

3). sin 6^{\circ}+1-sin 6^{\circ} = 1

Ответ: 1.
можно было сделать гораздо-гораздо иррациональней и легче!)