Question

Помогите с алгеброй.​

Answer 1

$\displaystyle\bf\\1)\\\\Cos85^\circ\cdot Cos25^\circ=\frac{1}{2} \Big[Cos(85^\circ-25^\circ) +Cos(85^\circ+25^\circ) \Big]=\\\\\\=\frac{1}{2} \Big(Cos60^\circ+Cos110^\circ\Big)=\frac{1}{2} \Big(\frac{1}{2} +Cos110^\circ\Big)=\frac{1}{4} +\frac{Cos(90^\circ +20^\circ)}{2} =\\\\\\=\frac{2-Sin20 ^\circ }{4}$

$\displaystyle\bf\\2)\\\\Sin\frac{\pi }{12} \cdot Cos\frac{5\pi }{12} =\frac{1}{2} \Big[Sin\Big(\frac{\pi }{12}-\frac{5\pi }{12} \Big) +Sin\Big(\frac{\pi }{12}+\frac{5\pi }{12} \Big) \Big]=\\\\\\=\frac{1}{2} \Big(-Sin\frac{\pi }{3}+Sin\frac{\pi }{2} \Big)=\frac{1}{2} \Big(-\frac{\sqrt{3} }{2}+1\Big)=\frac{1}{2} \cdot\frac{2-\sqrt{3} }{2} =\frac{2-\sqrt{3} }{4}$

$\displaystyle\bf\\3)\\\\\frac{Sin16^\circ+Sin74^\circ }{Cos16^\circ+Cos74^\circ} =\frac{2Sin\dfrac{16^\circ+74^\circ }{2} Cos\dfrac{16^\circ-74^\circ }{2} }{2Cos\dfrac{16^\circ+74^\circ }{2} Cos\dfrac{16^\circ-74^\circ }{2}} =\\\\\\=\frac{Sin45^\circ\cdot Cos29^\circ}{Cos45^\circ\cdot Cos29^\circ} =\frac{\frac{\sqrt{2} }{2} }{\frac{\sqrt{2} }{2} } =1$