Решить ур-ие:sin6 + cos6*tg42(6,42-в градусах)
Решить ур-ие:
sin6 + cos6*tg42
(6,42-в градусах)
Ответ(ы) на вопрос:
Гость
По действиям:
1). [latex]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}}=[/latex]
[latex]= \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}}[/latex]
2). [latex]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}}=[/latex]
[latex]= (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}=[/latex]
[latex]= cos^{2} 3^{\circ} -2*cos 3^{\circ}*sin 3^{\circ} + sin^{2} 3^{\circ}= 1-sin 6^{\circ}[/latex]
3). [latex]sin 6^{\circ}+1-sin 6^{\circ} = 1[/latex]
Ответ: 1.
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